Stochastic Calculus Solution Manual

Stochastic Calculus for Finance, Volume I and II by Yan Zeng Last updated: August 20, 2007 This is a solution manual for the two-volume textbook Stochastic calculus for ? nance, by Steven Shreve. If you have any comments or ? nd any typos/errors, please email me at [email  protected] edu. The current version omits the following problems. Volume I: 1. 5, 3. 3, 3. 4, 5. 7; Volume II: 3. 9, 7. 1, 7. 2, 7. 5–7. 9, 10. 8, 10. 9, 10. 10. Acknowledgment I thank Hua Li (a graduate student at Brown University) for reading through this solution manual and communicating to me several mistakes/typos. 1. 1. Stochastic Calculus for Finance I: The Binomial Asset Pricing Model 1. The Binomial No-Arbitrage Pricing Model Proof. If we get the up sate, then X1 = X1 (H) = ? 0 uS0 + (1 + r)(X0 ? ?0 S0 ); if we get the down state, then X1 = X1 (T ) = ? 0 dS0 + (1 + r)(X0 ? ?0 S0 ). If X1 has a positive probability of being strictly positive, then we must either have X1 (H) > 0 or X1 (T ) > 0. (i) If X1 (H) > 0, then ? 0 uS0 + (1 + r)(X0 ? ?0 S0 ) > 0. Plug in X0 = 0, we get u? 0 > (1 + r)? 0 . By condition d < 1 + r < u, we conclude ? 0 > 0.In this case, X1 (T ) = ? 0 dS0 + (1 + r)(X0 ? ?0 S0 ) = ? 0 S0 [d ? (1 + r)] < 0. (ii) If X1 (T ) > 0, then we can similarly deduce ? 0 < 0 and hence X1 (H) < 0. So we cannot have X1 strictly positive with positive probability unless X1 is strictly negative with positive probability as well, regardless the choice of the number ? 0 . Remark: Here the condition X0 = 0 is not essential, as far as a property de? nition of arbitrage for arbitrary X0 can be given. Indeed, for the one-period binomial model, we can de? ne arbitrage as a trading strategy such that P (X1 ?X0 (1 + r)) = 1 and P (X1 > X0 (1 + r)) > 0. First, this is a generalization of the case X0 = 0; second, it is â€Å"proper† because it is comparing the result of an arbitrary investment involving money and stock markets with that of a safe investment involving only money market. This can also be seen by regarding X0 as borrowed from money market account. Then at time 1, we have to pay back X0 (1 + r) to the money market account. In summary, arbitrage is a trading strategy that beats â€Å"safe† investment. Accordingly, we revise the proof of Exercise 1. 1. as follows.If X1 has a positive probability of being strictly larger than X0 (1 + r), the either X1 (H) > X0 (1 + r) or X1 (T ) > X0 (1 + r). The ? rst case yields ? 0 S0 (u ? 1 ? r) > 0, i. e. ?0 > 0. So X1 (T ) = (1 + r)X0 + ? 0 S0 (d ? 1 ? r) < (1 + r)X0 . The second case can be similarly analyzed. Hence we cannot have X1 strictly greater than X0 (1 + r) with positive probability unless X1 is strictly smaller than X0 (1 + r) with positive probability as well. Finally, we comment that the above formulation of arbitrage is equivalent to the one in the textbook. For details, see Shreve [7], Exercise 5. . 1. 2. 1 5 Proof. X1 (u) = ? 0 ? 8 + ? 0 ? 3 ? 5 (4? 0 + 1. 20? 0 ) = 3? 0 + 1. 5? 0 , a nd X1 (d) = ? 0 ? 2 ? 4 (4? 0 + 1. 20? 0 ) = 4 ? 3? 0 ? 1. 5? 0 . That is, X1 (u) = ? X1 (d). So if there is a positive probability that X1 is positive, then there is a positive probability that X1 is negative. Remark: Note the above relation X1 (u) = ? X1 (d) is not a coincidence. In general, let V1 denote the ? ? payo? of the derivative security at time 1. Suppose X0 and ? 0 are chosen in such a way that V1 can be ? 0 ? ?0 S0 ) + ? 0 S1 = V1 . Using the notation of the problem, suppose an agent begins ? replicated: (1 + r)(X with 0 wealth and at time zero buys ? 0 shares of stock and ? 0 options. He then puts his cash position ? 0 S0 ? ?0 X0 in a money market account. At time one, the value of the agent’s portfolio of stock, option and money market assets is ? X1 = ? 0 S1 + ? 0 V1 ? (1 + r)(? 0 S0 + ? 0 X0 ). Plug in the expression of V1 and sort out terms, we have ? X1 = S0 (? 0 + ? 0 ? 0 )( S1 ? (1 + r)). S0 ? Since d < (1 + r) < u, X1 (u) and X1 (d) have opposite signs. So if the price of the option at time zero is X0 , then there will no arbitrage. 1. 3. S0 1 Proof. V0 = 1+r 1+r? d S1 (H) + u? ? r S1 (T ) = 1+r 1+r? d u + u? 1? r d = S0 . This is not surprising, since u? d u? d u? d u? d this is exactly the cost of replicating S1 . Remark: This illustrates an important point. The â€Å"fair price† of a stock cannot be determined by the risk-neutral pricing, as seen below. Suppose S1 (H) and S1 (T ) are given, we could have two current prices, S0 and S0 . Correspondingly, we can get u, d and u , d . Because they are determined by S0 and S0 , respectively, it’s not surprising that risk-neutral pricing formula always holds, in both cases. That is, 1+r? d u? d S1 (H) S0 = + u? 1? r u? d S1 (T ) 1+r S0 = 1+r? d u ? d S1 (H) + u ? 1? r u ? d S1 (T ) 1+r . Essentially, this is because risk-neutral pricing relies on fair price=replication cost. Stock as a replicating component cannot determine its own â€Å"fair† price via the risk-n eutral pricing formula. 1. 4. Proof. Xn+1 (T ) = = ? n dSn + (1 + r)(Xn ? ?n Sn ) ?n Sn (d ? 1 ? r) + (1 + r)Vn pVn+1 (H) + q Vn+1 (T ) ? ? Vn+1 (H) ? Vn+1 (T ) (d ? 1 ? r) + (1 + r) = u? d 1+r = p(Vn+1 (T ) ? Vn+1 (H)) + pVn+1 (H) + q Vn+1 (T ) ? ? ? = pVn+1 (T ) + q Vn+1 (T ) ? ? = Vn+1 (T ). 1. 6. 2 Proof. The bank’s trader should set up a replicating portfolio whose payo? s the opposite of the option’s payo?. More precisely, we solve the equation (1 + r)(X0 ? ?0 S0 ) + ? 0 S1 = ? (S1 ? K)+ . 1 Then X0 = ? 1. 20 and ? 0 = ? 2 . This means the trader should sell short 0. 5 share of stock, put the income 2 into a money market account, and then transfer 1. 20 into a separate money market account. At time one, the portfolio consisting of a short position in stock and 0. 8(1 + r) in money market account will cancel out with the option’s payo?. Therefore we end up with 1. 20(1 + r) in the separate money market account. Remark: This problem illustrates why we are in terested in hedging a long position.In case the stock price goes down at time one, the option will expire without any payo?. The initial money 1. 20 we paid at time zero will be wasted. By hedging, we convert the option back into liquid assets (cash and stock) which guarantees a sure payo? at time one. Also, cf. page 7, paragraph 2. As to why we hedge a short position (as a writer), see Wilmott [8], page 11-13. 1. 7. Proof. The idea is the same as Problem 1. 6. The bank’s trader only needs to set up the reverse of the replicating trading strategy described in Example 1. 2. 4. More precisely, he should short sell 0. 1733 share of stock, invest the income 0. 933 into money market account, and transfer 1. 376 into a separate money market account. The portfolio consisting a short position in stock and 0. 6933-1. 376 in money market account will replicate the opposite of the option’s payo?. After they cancel out, we end up with 1. 376(1 + r)3 in the separate money market ac count. 1. 8. (i) 2 s s Proof. vn (s, y) = 5 (vn+1 (2s, y + 2s) + vn+1 ( 2 , y + 2 )). (ii) Proof. 1. 696. (iii) Proof. ?n (s, y) = vn+1 (us, y + us) ? vn+1 (ds, y + ds) . (u ? d)s 1. 9. (i) Proof. Similar to Theorem 1. 2. 2, but replace r, u and d everywhere with rn , un and dn .More precisely, set pn = 1+rn ? dn and qn = 1 ? pn . Then un ? dn Vn = pn Vn+1 (H) + qn Vn+1 (T ) . 1 + rn (ii) Proof. ?n = (iii) 3 Vn+1 (H)? Vn+1 (T ) Sn+1 (H)? Sn+1 (T ) = Vn+1 (H)? Vn+1 (T ) . (un ? dn )Sn 10 10 Proof. un = Sn+1 (H) = Sn +10 = 1+ Sn and dn = Sn+1 (T ) = Sn ? 10 = 1? Sn . So the risk-neutral probabilities Sn Sn Sn Sn at time n are pn = u1? dnn = 1 and qn = 1 . Risk-neutral pricing implies the price of this call at time zero is ? ? 2 2 n ? d 9. 375. 2. Probability Theory on Coin Toss Space 2. 1. (i) Proof. P (Ac ) + P (A) = (ii) Proof. By induction, it su? ces to work on the case N = 2.When A1 and A2 are disjoint, P (A1 ? A2 ) = A1 ? A2 P (? ) = A1 P (? ) + A2 P (? ) = P (A1 ) + P (A2 ). When A1 and A2 are arbitrary, using the result when they are disjoint, we have P (A1 ? A2 ) = P ((A1 ? A2 ) ? A2 ) = P (A1 ? A2 ) + P (A2 ) ? P (A1 ) + P (A2 ). 2. 2. (i) 1 3 1 Proof. P (S3 = 32) = p3 = 8 , P (S3 = 8) = 3p2 q = 3 , P (S3 = 2) = 3pq 2 = 8 , and P (S3 = 0. 5) = q 3 = 8 . 8 Ac P (? ) + A P (? ) = P (? ) = 1. (ii) Proof. E[S1 ] = 8P (S1 = 8) + 2P (S1 = 2) = 8p + 2q = 5, E[S2 ] = 16p2 + 4  · 2pq + 1  · q 2 = 6. 25, and 3 1 E[S3 ] = 32  · 1 + 8  · 8 + 2  · 3 + 0.  · 8 = 7. 8125. So the average rates of growth of the stock price under P 8 8 5 are, respectively: r0 = 4 ? 1 = 0. 25, r1 = 6. 25 ? 1 = 0. 25 and r2 = 7. 8125 ? 1 = 0. 25. 5 6. 25 (iii) 8 1 Proof. P (S3 = 32) = ( 2 )3 = 27 , P (S3 = 8) = 3  · ( 2 )2  · 1 = 4 , P (S3 = 2) = 2  · 1 = 2 , and P (S3 = 0. 5) = 27 . 3 3 3 9 9 9 Accordingly, E[S1 ] = 6, E[S2 ] = 9 and E[S3 ] = 13. 5. So the average rates of growth of the stock price 9 6 under P are, respectively: r0 = 4 ? 1 = 0. 5, r1 = 6 ? 1 = 0. 5, and r2 = 13. 5 ? 1 = 0. 5. 9 2. 3. Proof. Apply conditional Jensen’s inequality. 2. 4. (i) Proof.En [Mn+1 ] = Mn + En [Xn+1 ] = Mn + E[Xn+1 ] = Mn . (ii) 2 n+1 Proof. En [ SSn ] = En [e? Xn+1 e? +e ] = 2 ? Xn+1 ] e? +e E[e = 1. 2. 5. (i) 2 2 Proof. 2In = 2 j=0 Mj (Mj+1 ? Mj ) = 2 j=0 Mj Mj+1 ? j=1 Mj ? j=1 Mj = 2 j=0 Mj Mj+1 + n? 1 n? 1 n? 1 n? 1 2 2 2 2 2 2 2 2 Mn ? j=0 Mj+1 ? j=0 Mj = Mn ? j=0 (Mj+1 ? Mj ) = Mn ? j=0 Xj+1 = Mn ? n. n? 1 n? 1 n? 1 n? 1 n? 1 (ii) Proof. En [f (In+1 )] = En [f (In + Mn (Mn+1 ? Mn ))] = En [f (In + Mn Xn+1 )] = 1 [f (In + Mn ) + f (In ? Mn )] = 2 v v v g(In ), where g(x) = 1 [f (x + 2x + n) + f (x ? 2x + n)], since 2In + n = |Mn |. 2 2. 6. 4 Proof. En [In+1 ?In ] = En [? n (Mn+1 ? Mn )] = ? n En [Mn+1 ? Mn ] = 0. 2. 7. Proof. We denote by Xn the result of n-th coin toss, where Head is represented by X = 1 and Tail is 1 represented by X = ? 1. We also suppose P (X = 1) = P (X = ? 1) = 2 . De? ne S1 = X1 and Sn+1 = n Sn +bn (X1 ,  ·  ·  · , Xn )Xn+1 , where bn ( ·) is a bounded function on {? 1, 1} , to be determined later on. Clearly (Sn )n? 1 is an adapted stochastic process, and we can show it is a martingale. Indeed, En [Sn+1 ? Sn ] = bn (X1 ,  ·  ·  · , Xn )En [Xn+1 ] = 0. For any arbitrary function f , En [f (Sn+1 )] = 1 [f (Sn + bn (X1 ,  ·  ·  · , Xn )) + f (Sn ? n (X1 ,  ·  ·  · , Xn ))]. Then 2 intuitively, En [f (Sn+1 ] cannot be solely dependent upon Sn when bn ’s are properly chosen. Therefore in general, (Sn )n? 1 cannot be a Markov process. Remark: If Xn is regarded as the gain/loss of n-th bet in a gambling game, then Sn would be the wealth at time n. bn is therefore the wager for the (n+1)-th bet and is devised according to past gambling results. 2. 8. (i) Proof. Note Mn = En [MN ] and Mn = En [MN ]. (ii) Proof. In the proof of Theorem 1. 2. 2, we proved by induction that Xn = Vn where Xn is de? ned by (1. 2. 14) of Chapter 1. In other words, the sequence (Vn )0? n?N can be realized as the value process of a portfolio, Xn which consists of stock and money market accounts. Since ( (1+r)n )0? n? N is a martingale under P (Theorem Vn 2. 4. 5), ( (1+r)n )0? n? N is a martingale under P . (iii) Proof. (iv) Proof. Combine (ii) and (iii), then use (i). 2. 9. (i) (H) S1 (H) 1 = 2, d0 = S1S0 = 2 , S0 (T and d1 (T ) = S21 (TT)) = 1. S 1 1 0 ? d So p0 = 1+r? d0 0 = 2 , q0 = 2 , p1 (H) u0 5 q1 (T ) = 6 . Therefore P (HH) = p0 p1 (H) = 1 , 4 5 q0 q1 (T ) = 12 . Vn (1+r)n = En VN (1+r)N , so V0 , V1 1+r ,  ·Ã‚ ·Ã‚ ·, VN ? 1 , VN (1+r)N ? 1 (1+r)N is a martingale under P . Proof. u0 = u1 (H) = =S2 (HH) S1 (H) = 1. 5, d1 (H) = S2 (HT ) S1 (H) = 1, u1 (T ) = S2 (T H) S1 (T ) =4 1+r1 (H)? d1 (H) u1 (H)? d1 (H) 1 = 1 , q1 (H) = 2 , p1 (T ) = 2 1 4, 1+r1 (T )? d1 (T ) u1 (T )? d1 (T ) 1 12 1 = 6 , and P (HT ) = p0 q1 (H) = P (T H) = q0 p1 (T ) = and P (T T ) = The proofs of Theorem 2. 4. 4, Theorem 2. 4. 5 and Theorem 2. 4. 7 still work for the random interest rate m odel, with proper modi? cations (i. e. P would be constructed according to conditional probabilities P (? n+1 = H|? 1 ,  ·  ·  · , ? n ) := pn and P (? n+1 = T |? 1 ,  ·  ·  · , ? n ) := qn . Cf. notes on page 39. ). So the time-zero value of an option that pays o?V2 at time two is given by the risk-neutral pricing formula V0 = E (1+r0V2 1 ) . )(1+r (ii) Proof. V2 (HH) = 5, V2 (HT ) = 1, V2 (T H) = 1 and V2 (T T ) = 0. So V1 (H) = 2. 4, V1 (T ) = p1 (T )V2 (T H)+q1 (T )V2 (T T ) 1+r1 (T ) p1 (H)V2 (HH)+q1 (H)V2 (HT ) 1+r1 (H) = = 1 9, and V0 = p0 V1 (H)+q0 V1 (T ) 1+r0 ? 1. 5 (iii) Proof. ?0 = (iv) Proof. ?1 (H) = 2. 10. (i) Xn+1 Proof. En [ (1+r)n+1 ] = En [ ? n Yn+1 Sn + (1+r)n+1 (1+r)(Xn n Sn ) ] (1+r)n+1 Xn (1+r)n . V2 (HH)? V2 (HT ) S2 (HH)? S2 (HT ) V1 (H)? V1 (T ) S1 (H)? S1 (T ) = 1 2. 4? 9 8? 2 = 0. 4 ? 1 54 ? 0. 3815. = 5? 1 12? 8 = 1. = ?n Sn (1+r)n+1 En [Yn+1 ] + Xn Sn (1+r)n = ?n Sn (1+r)n+1 (up + dq) + Xn n Sn (1+r)n = ?n Sn +Xn n Sn (1+r)n = (ii) Proof . From (2. 8. 2), we have ? n uSn + (1 + r)(Xn ? ?n Sn ) = Xn+1 (H) ? n dSn + (1 + r)(Xn ? ?n Sn ) = Xn+1 (T ). So ? n = Xn+1 (H)? Xn+1 (T ) uSn ? dSn and Xn = En [ Xn+1 ]. To make the portfolio replicate the payo? at time N , we 1+r VN X must have XN = VN . So Xn = En [ (1+r)N ? n ] = En [ (1+r)N ? n ]. Since (Xn )0? n? N is the value process of the N unique replicating portfolio (uniqueness is guaranteed by the uniqueness of the solution to the above linear VN equations), the no-arbitrage price of VN at time n is Vn = Xn = En [ (1+r)N ? ]. (iii) Proof. En [ Sn+1 ] (1 + r)n+1 = = < = 1 En [(1 ? An+1 )Yn+1 Sn ] (1 + r)n+1 Sn [p(1 ? An+1 (H))u + q(1 ? An+1 (T ))d] (1 + r)n+1 Sn [pu + qd] (1 + r)n+1 Sn . (1 + r)n Sn (1+r)n+1 (1? a)(pu+qd) Sn+1 If An+1 is a constant a, then En [ (1+r)n+1 ] = Sn (1+r)n (1? a)n . = Sn (1+r)n (1? a). Sn+1 So En [ (1+r)n+1 (1? a)n+1 ] = 2. 11. (i) Proof. FN + PN = SN ? K + (K ? SN )+ = (SN ? K)+ = CN . (ii) CN FN PN Proof. Cn = En [ (1+r)N ? n ] = En [ (1+ r)N ? n ] + En [ (1+r)N ? n ] = Fn + Pn . (iii) FN Proof. F0 = E[ (1+r)N ] = 1 (1+r)N E[SN ? K] = S0 ? K (1+r)N . (iv) 6 Proof.At time zero, the trader has F0 = S0 in money market account and one share of stock. At time N , the trader has a wealth of (F0 ? S0 )(1 + r)N + SN = ? K + SN = FN . (v) Proof. By (ii), C0 = F0 + P0 . Since F0 = S0 ? (vi) SN ? K Proof. By (ii), Cn = Pn if and only if Fn = 0. Note Fn = En [ (1+r)N ?n ] = Sn ? So Fn is not necessarily zero and Cn = Pn is not necessarily true for n ? 1. (1+r)N S0 (1+r)N ? n (1+r)N S0 (1+r)N = 0, C0 = P0 . = Sn ? S0 (1 + r)n . 2. 12. Proof. First, the no-arbitrage price of the chooser option at time m must be max(C, P ), where C=E (SN ? K)+ (K ? SN )+ , and P = E . (1 + r)N ? m (1 + r)N ? That is, C is the no-arbitrage price of a call option at time m and P is the no-arbitrage price of a put option at time m. Both of them have maturity date N and strike price K. Suppose the market is liquid, then the chooser option is equivalent to receiving a payo? of max(C, P ) at time m. Therefore, its current no-arbitrage price should be E[ max(C,P ) ]. (1+r)m K K By the put-call parity, C = Sm ? (1+r)N ? m + P . So max(C, P ) = P + (Sm ? (1+r)N ? m )+ . Therefore, the time-zero price of a chooser option is E K (Sm ? (1+r)N ? m )+ P +E (1 + r)m (1 + r)m =E K (Sm ? (1+r)N ? m )+ (K ? SN )+ . +E (1 + r)N (1 + r)mThe ? rst term stands for the time-zero price of a put, expiring at time N and having strike price K, and the K second term stands for the time-zero price of a call, expiring at time m and having strike price (1+r)N ? m . If we feel unconvinced by the above argument that the chooser option’s no-arbitrage price is E[ max(C,P ) ], (1+r)m due to the economical argument involved (like â€Å"the chooser option is equivalent to receiving a payo? of max(C, P ) at time m†), then we have the following mathematically rigorous argument. First, we can construct a portfolio ? 0 ,  ·  ·  · , ? m? 1 , whose payo? at time m is max(C, P ).Fix ? , if C(? ) > P (? ), we can construct a portfolio ? m ,  ·  ·  · , ? N ? 1 whose payo? at time N is (SN ? K)+ ; if C(? ) < P (? ), we can construct a portfolio ? m ,  ·  ·  · , ? N ? 1 whose payo? at time N is (K ? SN )+ . By de? ning (m ? k ? N ? 1) ? k (? ) = ? k (? ) ? k (? ) if C(? ) > P (? ) if C(? ) < P (? ), we get a portfolio (? n )0? n? N ? 1 whose payo? is the same as that of the chooser option. So the no-arbitrage price process of the chooser option must be equal to the value process of the replicating portfolio. In Xm particular, V0 = X0 = E[ (1+r)m ] = E[ max(C,P ) ]. (1+r)m 2. 13. (i) Proof.Note under both actual probability P and risk-neutral probability P , coin tosses ? n ’s are i. i. d.. So n+1 without loss of generality, we work on P . For any function g, En [g(Sn+1 , Yn+1 )] = En [g( SSn Sn , Yn + = pg(uSn , Yn + uSn ) + qg(dSn , Yn + dSn ), which is a function of (Sn , Yn ). So (Sn , Yn )0? n? N is Markov un der P . (ii) 7 Sn+1 Sn Sn )] Proof. Set vN (s, y) = f ( Ny ). Then vN (SN , YN ) = f ( +1 Vn = where En [ Vn+1 ] 1+r = n+1 En [ vn+1 (S1+r ,Yn+1 ) ] N n=0 Sn N +1 ) = VN . Suppose vn+1 is given, then = 1 1+r [pvn+1 (uSn , Yn + uSn ) + qvn+1 (dSn , Yn + dSn )] = vn (Sn , Yn ), vn (s, y) = n+1 (us, y + us) + vn+1 (ds, y + ds) . 1+r 2. 14. (i) Proof. For n ? M , (Sn , Yn ) = (Sn , 0). Since coin tosses ? n ’s are i. i. d. under P , (Sn , Yn )0? n? M is Markov under P . More precisely, for any function h, En [h(Sn+1 )] = ph(uSn ) + h(dSn ), for n = 0, 1,  ·  ·  · , M ? 1. For any function g of two variables, we have EM [g(SM +1 , YM +1 )] = EM [g(SM +1 , SM +1 )] = pg(uSM , uSM )+ n+1 n+1 qg(dSM , dSM ). And for n ? M +1, En [g(Sn+1 , Yn+1 )] = En [g( SSn Sn , Yn + SSn Sn )] = pg(uSn , Yn +uSn )+ qg(dSn , Yn + dSn ), so (Sn , Yn )0? n? N is Markov under P . (ii) y Proof. Set vN (s, y) = f ( N ? M ).Then vN (SN , YN ) = f ( N K=M +1 Sk N ? M ) = VN . Suppose vn+1 is already given. a) If n > M , then En [vn+1 (Sn+1 , Yn+1 )] = pvn+1 (uSn , Yn + uSn ) + qvn+1 (dSn , Yn + dSn ). So vn (s, y) = pvn+1 (us, y + us) + qvn+1 (ds, y + ds). b) If n = M , then EM [vM +1 (SM +1 , YM +1 )] = pvM +1 (uSM , uSM ) + vn+1 (dSM , dSM ). So vM (s) = pvM +1 (us, us) + qvM +1 (ds, ds). c) If n < M , then En [vn+1 (Sn+1 )] = pvn+1 (uSn ) + qvn+1 (dSn ). So vn (s) = pvn+1 (us) + qvn+1 (ds). 3. State Prices 3. 1. Proof. Note Z(? ) := P (? ) P (? ) = 1 Z(? ) . Apply Theorem 3. 1. 1 with P , P , Z replaced by P , P , Z, we get the nalogous of properties (i)-(iii) of Theorem 3. 1. 1. 3. 2. (i) Proof. P (? ) = (ii) Proof. E[Y ] = (iii) ? Proof. P (A) = (iv) Proof. If P (A) = A Z(? )P (? ) = 0, by P (Z > 0) = 1, we conclude P (? ) = 0 for any ? ? A. So P (A) = A P (? ) = 0. (v) Proof. P (A) = 1 P (Ac ) = 0 P (Ac ) = 0 P (A) = 1. (vi) A P (? ) = Z(? )P (? ) = E[Z] = 1. Y (? )P (? ) = Y (? )Z(? )P (? ) = E[Y Z]. Z(? )P (? ). Since P (A) = 0, P (? ) = 0 for any ? ? A. So P (A) = 0. 8 Proof. Pick ? 0 such that P (? 0 ) > 0, de? ne Z(? ) = 1 P (? 0 ) 0, 1 P (? 0 ) , if ? = ? 0 Then P (Z ? 0) = 1 and E[Z] = if ? = ? 0 .  · P (? 0 ) = 1. =? 0 Clearly P (? {? 0 }) = E[Z1? {? 0 } ] = Z(? )P (? ) = 0. But P (? {? 0 }) = 1 ? P (? 0 ) > 0 if P (? 0 ) < 1. Hence in the case 0 < P (? 0 ) < 1, P and P are not equivalent. If P (? 0 ) = 1, then E[Z] = 1 if and only if Z(? 0 ) = 1. In this case P (? 0 ) = Z(? 0 )P (? 0 ) = 1. And P and P have to be equivalent. In summary, if we can ? nd ? 0 such that 0 < P (? 0 ) < 1, then Z as constructed above would induce a probability P that is not equivalent to P . 3. 5. (i) Proof. Z(HH) = (ii) Proof. Z1 (H) = E1 [Z2 ](H) = Z2 (HH)P (? 2 = H|? 1 = H) + Z2 (HT )P (? 2 = T |? 1 = H) = 3 E1 [Z2 ](T ) = Z2 (T H)P (? 2 = H|? = T ) + Z2 (T T )P (? 2 = T |? 1 = T ) = 2 . (iii) Proof. V1 (H) = [Z2 (HH)V2 (HH)P (? 2 = H|? 1 = H) + Z2 (HT )V2 (HT )P (? 2 = T |? 1 = T )] = 2. 4, Z1 (H)(1 + r1 (H)) [Z2 (T H)V2 (T H)P (? 2 = H|? 1 = T ) + Z2 (T T )V2 (T T )P (? 2 = T |? 1 = T )] 1 = , Z1 (T )(1 + r1 (T )) 9 3 4. 9 16 , Z(HT ) = 9 , Z(T H) = 8 3 8 and Z(T T ) = 15 4 . Z1 (T ) = V1 (T ) = and V0 = Z2 (HH)V2 (HH) Z2 (HT )V2 (HT ) Z2 (T H)V2 (T H) P (HH) + P (T H) + 0 ? 1. 1 1 1 1 P (HT ) + 1 (1 + 4 )(1 + 4 ) (1 + 4 )(1 + 4 ) (1 + 4 )(1 + 1 ) 2 3. 6. Proof. U (x) = have XN = 1 x, (1+r)N ? Z so I(x) = = 1 Z] 1 x. Z (3. 3. 26) gives E[ (1+r)N 1 X0 (1 + r)n Zn En [Z  ·X0 N Z (1 + r) . 0 = Xn , where ? Hence Xn = (1+r)N ? Z X En [ (1+r)N ? n ] N ] = X0 . So ? = = En [ X0 (1+r) Z n 1 X0 . By (3. 3. 25), we 1 ] = X0 (1 + r)n En [ Z ] = the second to last â€Å"=† comes from Lemma 3. 2. 6. 3. 7. Z ? Z Proof. U (x) = xp? 1 and so I(x) = x p? 1 . By (3. 3. 26), we have E[ (1+r)N ( (1+r)N ) p? 1 ] = X0 . Solve it for ? , we get ? ?p? 1 1 1 ? ? =? ? X0 p E 1 Z p? 1 Np ? ? ? = p? 1 X0 (1 + r)N p (E[Z p? 1 ])p? 1 1 p . (1+r) p? 1 ? Z So by (3. 3. 25), XN = ( (1+r)N ) p? 1 = 1 1 Np ? p? 1 Z p? 1 N (1+r) p? 1 = X0 (1+r) p? 1 E[Z p p? 1 Z p? 1 N (1+r) p? 1 = (1+r)N X0 Z p? 1 E[Z p p? 1 1 . ] ] 3. 8. (i) 9 d d Proof. x (U (x) ? yx) = U (x) ? y. So x = I(y) is an extreme point of U (x) ? yx. Because dx2 (U (x) ? yx) = U (x) ? 0 (U is concave), x = I(y) is a maximum point. Therefore U (x) ? y(x) ? U (I(y)) ? yI(y) for every x. 2 (ii) Proof. Following the hint of the problem, we have E[U (XN )] ? E[XN ? Z ? Z ? Z ? Z ] ? E[U (I( ))] ? E[ I( )], N N N (1 + r) (1 + r) (1 + r) (1 + r)N ? ? ? ? ? i. e. E[U (XN )] ? ?X0 ? E[U (XN )] ? E[ (1+r)N XN ] = E[U (XN )] ? ?X0 . So E[U (XN )] ? E[U (XN )]. 3. 9. (i) X Proof. Xn = En [ (1+r)N ? n ]. So if XN ? 0, then Xn ? 0 for all n. N (ii) 1 Proof. a) If 0 ? x < ? and 0 < y ? ? , then U (x) ? yx = ? yx ? and U (I(y)) ? yI(y) = U (? ) ? y? = 1 ? y? ? 0. So U (x) ? yx ? U (I(y)) ? yI(y). 1 b) If 0 ? x < ? and y > ? , then U (x) ? yx = ? yx ? 0 and U (I(y)) ? yI(y) = U (0) ? y  · 0 = 0. So U (x) ? yx ? U (I(y)) ? yI(y). 1 c) If x ? ? and 0 < y ? ? , then U (x) ? yx = 1 ? yx and U (I(y)) ? yI(y) = U (? ) ? y? = 1 ? y? ? 1 ? yx. So U (x) ? yx ? U (I(y)) ? yI(y). 1 d) If x ? ? and y > ? , then U (x) ? yx = 1 ? yx < 0 and U (I(y)) ? yI(y) = U (0) ? y  · 0 = 0. So U (x) ? yx ? U (I(y)) ? yI(y). (iii) XN ? Z Proof. Using (ii) and set x = XN , y = (1+r)N , where XN is a random variable satisfying E[ (1+r)N ] = X0 , we have ?Z ? Z ? E[U (XN )] ? E[ XN ] ? E[U (XN )] ? E[ X ? ]. (1 + r)N (1 + r)N N ? ? That is, E[U (XN )] ? ?X0 ? E[U (XN )] ? ?X0 . So E[U (XN )] ? E[U (XN )]. (iv) Proof. Plug pm and ? m into (3. 6. 4), we have 2N 2N X0 = m=1 pm ? m I( m ) = m=1 1 pm ? m ? 1{ m ? ? } . So X0 ? X0 ? {m : = we are looking for positive solution ? > 0). Conversely, suppose there exists some K so that ? K < ? K+1 and K X0 1 m=1 ? m pm = ? . Then we can ? nd ? > 0, such that ? K < < ? K+1 . For such ? , we have Z ? Z 1 E[ I( )] = pm ? m 1{ m ? ? } ? = pm ? m ? = X0 . N (1 + r) (1 + r)N m=1 m=1 Hence (3. 6. 4) has a solution. 0 2N K 2N X0 1 m=1 pm ? m 1{ m ? ? } . Suppose there is a solution ? to (3. 6. 4), note ? > 0, we then can conclude 1 1 1 m ? ? } = ?. Let K = max{m : m ? ? }, then K ? ? < K+1 . So ? K < ? K+1 and K N m=1 pm ? m (Note, however, that K could be 2 . In this case, ? K+1 is interpreted as ?. Also, note = (v) ? 1 Proof. XN (? m ) = I( m ) = ? 1{ m ? ? } = ?, if m ? K . 0, if m ? K + 1 4. American Derivative Securities Before proceeding to the exercise problems, we ? rst give a brief summary of pricing American derivative securities as presented in the textbook. We shall use the notation of the book.From the buyer’s perspective: At time n, if the derivative security has not been exercised, then the buyer can choose a policy ? with ? ? Sn . The valuation formula for cash ? ow (Theorem 2. 4. 8) gives a fair price for the derivative security exercised according to ? : N Vn (? ) = k=n En 1{? =k} 1 1 Gk = En 1{? ?N } G? . (1 + r)k? n (1 + r)? ?n The buyer wants to consider all the possible ? ’s, so that he c an ? nd the least upper bound of security value, which will be the maximum price of the derivative security acceptable to him. This is the price given by 1 De? nition 4. 4. 1: Vn = max? ?Sn En [1{? ?N } (1+r)? n G? ]. From the seller’s perspective: A price process (Vn )0? n? N is acceptable to him if and only if at time n, he can construct a portfolio at cost Vn so that (i) Vn ? Gn and (ii) he needs no further investing into the portfolio as time goes by. Formally, the seller can ? nd (? n )0? n? N and (Cn )0? n? N so that Cn ? 0 and Sn Vn+1 = ? n Sn+1 + (1 + r)(Vn ? Cn ? ?n Sn ). Since ( (1+r)n )0? n? N is a martingale under the risk-neutral measure P , we conclude En Cn Vn+1 Vn =? ? 0, ? n+1 n (1 + r) (1 + r) (1 + r)n Vn i. e. ( (1+r)n )0? n? N is a supermartingale. This inspired us to check if the converse is also true.This is exactly the content of Theorem 4. 4. 4. So (Vn )0? n? N is the value process of a portfolio that needs no further investing if and only if Vn (1+r)n Vn (1+r)n is a supermartingale under P (note this is independent of the requirement 0? n? N Vn ? Gn ). In summary, a price process (Vn )0? n? N is acceptable to the seller if and only if (i) Vn ? Gn ; (ii) is a supermartingale under P . 0? n? N Theorem 4. 4. 2 shows the buyer’s upper bound is the seller’s lower bound. So it gives the price acceptable to both. Theorem 4. 4. 3 gives a speci? c algorithm for calculating the price, Theorem 4. 4. establishes the one-to-one correspondence between super-replication and supermartingale property, and ? nally, Theorem 4. 4. 5 shows how to decide on the optimal exercise policy. 4. 1. (i) Proof. V2P (HH) = 0, V2P (HT ) = V2P (T H) = 0. 8, V2P (T T ) = 3, V1P (H) = 0. 32, V1P (T ) = 2, V0P = 9. 28. (ii) Proof. V0C = 5. (iii) Proof. gS (s) = |4 ? s|. We apply Theorem 4. 4. 3 and have V2S (HH) = 12. 8, V2S (HT ) = V2S (T H) = 2. 4, V2S (T T ) = 3, V1S (H) = 6. 08, V1S (T ) = 2. 16 and V0S = 3. 296. (iv) 11 Proof. First, we note the simple inequality max(a1 , b1 ) + max(a2 , b2 ) ? max(a1 + a2 , b1 + b2 ). >† holds if and only if b1 > a1 , b2 < a2 or b1 < a1 , b2 > a2 . By induction, we can show S Vn = max gS (Sn ), S S pVn+1 + Vn+1 1+r C P P pV C + Vn+1 pVn+1 + Vn+1 + n+1 1+r 1+r C C pVn+1 + Vn+1 1+r ? max gP (Sn ) + gC (Sn ), ? max gP (Sn ), P C = Vn + Vn . P P pVn+1 + Vn+1 1+r + max gC (Sn ), S P C As to when â€Å" C C pVn+1 +qVn+1 1+r or gP (Sn ) > P P pVn+1 +qVn+1 1+r and gC (Sn ) < C C pVn+1 +qVn+1 }. 1+r 4. 2. Proof. For this problem, we need Figure 4. 2. 1, Figure 4. 4. 1 and Figure 4. 4. 2. Then ? 1 (H) = and ? 0 = V2 (HH) ? V2 (HT ) 1 V2 (T H) ? V2 (T T ) = ? , ? 1 (T ) = = ? 1, S2 (HH) ? S2 (HT ) 12 S2 (T H) ?S2 (T T ) V1 (H) ? V1 (T ) ? ?0. 433. S1 (H) ? S1 (T ) The optimal exercise time is ? = inf{n : Vn = Gn }. So ? (HH) = ? , ? (HT ) = 2, ? (T H) = ? (T T ) = 1. Therefore, the agent borrows 1. 36 at time zero and buys the put. At the same time, to hedge the long position, he needs to borr ow again and buy 0. 433 shares of stock at time zero. At time one, if the result of coin toss is tail and the stock price goes down to 2, the value of the portfolio 1 is X1 (T ) = (1 + r)(? 1. 36 ? 0. 433S0 ) + 0. 433S1 (T ) = (1 + 4 )(? 1. 36 ? 0. 433 ? 4) + 0. 433 ? 2 = ? 3. The agent should exercise the put at time one and get 3 to pay o? is debt. At time one, if the result of coin toss is head and the stock price goes up to 8, the value of the portfolio 1 is X1 (H) = (1 + r)(? 1. 36 ? 0. 433S0 ) + 0. 433S1 (H) = ? 0. 4. The agent should borrow to buy 12 shares of stock. At time two, if the result of coin toss is head and the stock price goes up to 16, the value of the 1 1 portfolio is X2 (HH) = (1 + r)(X1 (H) ? 12 S1 (H)) + 12 S2 (HH) = 0, and the agent should let the put expire. If at time two, the result of coin toss is tail and the stock price goes down to 4, the value of the portfolio is 1 1 X2 (HT ) = (1 + r)(X1 (H) ? 12 S1 (H)) + 12 S2 (HT ) = ? 1.The agent should exercise the put to get 1. This will pay o? his debt. 4. 3. Proof. We need Figure 1. 2. 2 for this problem, and calculate the intrinsic value process and price process of the put as follows. 2 For the intrinsic value process, G0 = 0, G1 (T ) = 1, G2 (T H) = 3 , G2 (T T ) = 5 , G3 (T HT ) = 1, 3 G3 (T T H) = 1. 75, G3 (T T T ) = 2. 125. All the other outcomes of G is negative. 12 2 5 For the price process, V0 = 0. 4, V1 (T ) = 1, V1 (T H) = 3 , V1 (T T ) = 3 , V3 (T HT ) = 1, V3 (T T H) = 1. 75, V3 (T T T ) = 2. 125. All the other outcomes of V is zero. Therefore the time-zero price of the derivative security is 0. and the optimal exercise time satis? es ? (? ) = ? if ? 1 = H, 1 if ? 1 = T . 4. 4. Proof. 1. 36 is the cost of super-replicating the American derivative security. It enables us to construct a portfolio su? cient to pay o? the derivative security, no matter when the derivative security is exercised. So to hedge our short position after selling the put, there is no need to charge t he insider more than 1. 36. 4. 5. Proof. The stopping times in S0 are (1) ? ? 0; (2) ? ? 1; (3) ? (HT ) = ? (HH) = 1, ? (T H), ? (T T ) ? {2, ? } (4 di? erent ones); (4) ? (HT ), ? (HH) ? {2, ? }, ? (T H) = ? (T T ) = 1 (4 di? rent ones); (5) ? (HT ), ? (HH), ? (T H), ? (T T ) ? {2, ? } (16 di? erent ones). When the option is out of money, the following stopping times do not exercise (i) ? ? 0; (ii) ? (HT ) ? {2, ? }, ? (HH) = ? , ? (T H), ? (T T ) ? {2, ? } (8 di? erent ones); (iii) ? (HT ) ? {2, ? }, ? (HH) = ? , ? (T H) = ? (T T ) = 1 (2 di? erent ones). ? 4 For (i), E[1{? ?2} ( 4 )? G? ] = G0 = 1. For (ii), E[1{? ?2} ( 5 )? G? ] ? E[1{? ? ? 2} ( 4 )? G? ? ], where ? ? (HT ) = 5 5 1 4 4 ? 2, ? ? (HH) = ? , ? ? (T H) = ? ? (T T ) = 2. So E[1{? ? ? 2} ( 5 )? G? ? ] = 4 [( 4 )2  · 1 + ( 5 )2 (1 + 4)] = 0. 96. For 5 (iii), E[1{? ?2} ( 4 )? G? has the biggest value when ? satis? es ? (HT ) = 2, ? (HH) = ? , ? (T H) = ? (T T ) = 1. 5 This value is 1. 36. 4. 6. (i) Proof. The value of the put at time N , if it is not exercised at previous times, is K ? SN . Hence VN ? 1 = VN K max{K ? SN ? 1 , EN ? 1 [ 1+r ]} = max{K ? SN ? 1 , 1+r ? SN ? 1 } = K ? SN ? 1 . The second equality comes from the fact that discounted stock price process is a martingale under risk-neutral probability. By induction, we can show Vn = K ? Sn (0 ? n ? N ). So by Theorem 4. 4. 5, the optimal exercise policy is to sell the stock at time zero and the value of this derivative security is K ?S0 . Remark: We cheated a little bit by using American algorithm and Theorem 4. 4. 5, since they are developed for the case where ? is allowed to be ?. But intuitively, results in this chapter should still hold for the case ? ? N , provided we replace â€Å"max{Gn , 0}† with â€Å"Gn †. (ii) Proof. This is because at time N , if we have to exercise the put and K ? SN < 0, we can exercise the European call to set o? the negative payo?. In e? ect, throughout the portfolio’s lifetime, the portfolio has intrinsic values greater than that of an American put stuck at K with expiration time N . So, we must have V0AP ? V0 + V0EC ? K ?S0 + V0EC . (iii) 13 Proof. Let V0EP denote the time-zero value of a European put with strike K and expiration time N . Then V0AP ? V0EP = V0EC ? E[ K SN ? K ] = V0EC ? S0 + . (1 + r)N (1 + r)N 4. 7. VN K K Proof. VN = SN ? K, VN ? 1 = max{SN ? 1 ? K, EN ? 1 [ 1+r ]} = max{SN ? 1 ? K, SN ? 1 ? 1+r } = SN ? 1 ? 1+r . K By induction, we can prove Vn = Sn ? (1+r)N ? n (0 ? n ? N ) and Vn > Gn for 0 ? n ? N ? 1. So the K time-zero value is S0 ? (1+r)N and the optimal exercise time is N . 5. Random Walk 5. 1. (i) Proof. E[ 2 ] = E[? (? 2 1 )+? 1 ] = E[? (? 2 1 ) ]E[ 1 ] = E[ 1 ]2 . (ii) Proof. If we de? ne Mn = Mn+? ? M? m (m = 1, 2,  ·  ·  · ), then (M · )m as random functions are i. i. d. with (m) distributions the same as that of M . So ? m+1 ? ?m = inf{n : Mn = 1} are i. i. d. with distributions the same as that of ? 1 . Therefore E [ m ] = E[? (? m m? 1 )+(? m? 1 m? 2 )+ ·Ã‚ ·Ã‚ ·+? 1 ] = E[ 1 ]m . (m) (m) (iii) Proof. Yes, since the argument of (ii) still works for asymmetric random walk. 5. 2. (i) Proof. f (? ) = pe? ? qe , so f (? ) > 0 if and only if ? > f (? ) > f (0) = 1 for all ? > 0. (ii) 1 1 1 n+1 Proof. En [ SSn ] = En [e? Xn+1 f (? ) ] = pe? f (? ) + qe f (? ) = 1. 1 2 (ln q ? ln p). Since 1 2 (ln q ln p) < 0, (iii) 1 Proof. By optional stopping theorem, E[Sn 1 ] = E[S0 ] = 1. Note Sn 1 = e? Mn 1 ( f (? ) )n 1 ? e?  ·1 , by bounded convergence theorem, E[1{? 1 1 for all ? > ? 0 . v (ii) 1 1 Proof. As in Exercise 5. 2, Sn = e? Mn ( f (? ) )n is a martingale, and 1 = E[S0 ] = E[Sn 1 ] = E[e? Mn 1 ( f (? ) )? 1 ? n ]. Suppose ? > ? 0 , then by bounded convergence theorem, 1 = E[ lim e? Mn 1 ( n>? 1 n 1 1 ? 1 ) ] = E[1{? 1 K} ] = P (ST > K). Moreover, by Girsanov’s Theorem, Wt = Wt + in Theorem 5. 4. 1. ) (iii) Proof. ST = xe? WT +(r? 2 ? 1 2 1 2 t ( )du 0 = Wt ? ?t is a P -Brownian motion (set ? )T = xe? WT +(r+ 2 ? 1 2 1 2 )T . So WT v > ? d+ (T, x) T = N (d+ (T, x)). P (ST > K) = P (xe? WT +(r+ 2 ? )T > K) = P 46 5. 4. First, a few typos. In the SDE for S, â€Å"? (t)dW (t)† > â€Å"? (t)S(t)dW (t)†. In the ? rst equation for c(0, S(0)), E > E. In the second equation for c(0, S(0)), the variable for BSM should be ? ? 1 T 2 1 T r(t)dt, ? (t)dt? . BSM ? T, S(0); K, T 0 T 0 (i) Proof. d ln St = X = ? is a Gaussian with X ? N ( (ii) Proof. For the standard BSM model with constant volatility ? and interest rate R, under the risk-neutral measure, we have ST = S0 eY , where Y = (R? 1 ? 2 )T +? WT ? N ((R? 1 ? )T, ? 2 T ), and E[(S0 eY ? K)+ ] = 2 2 eRT BSM (T, S0 ; K, R, ? ). Note R = 1 T (rt 0 T T dSt 1 2 1 1 2 2 St ? 2St d S t = rt dt + ? t dWt ? 2 ? t dt. So ST = S0 exp{ 0 (rt ? 2 ? t )dt + 0 T 1 2 2 ? t )dt + 0 ? t dWt . The ? rst term in the expression of X is a number and the T 2 random variable N (0, 0 ? t dt), since both r and ? ar deterministic. Th erefore, T T 2 2 (rt ? 1 ? t )dt, 0 ? t dt),. 2 0 ?t dWt }. Let second term ST = S0 eX , 1 T (E[Y ] + 1 V ar(Y )) and ? = 2 T, S0 ; K, 1 T 1 T V ar(Y ), we can get 1 V ar(Y ) . T E[(S0 eY ? K)+ ] = eE[Y ]+ 2 V ar(Y ) BSM So for the model in this problem, c(0, S0 ) = = e? ? T 0 1 E[Y ] + V ar(Y ) , 2 rt dt E[(S0 eX ? K)+ ] e BSM T, S0 ; K, 1 T T 0 T 0 1 rt dt E[X]+ 2 V ar(X) 1 T ? 1 E[X] + V ar(X) , 2 1 V ar(X) T ? = 1 BSM ? T, S0 ; K, T 0 T rt dt, 2 ? t dt? . 5. 5. (i) 1 1 Proof. Let f (x) = x , then f (x) = ? x2 and f (x) = 2 x3 . Note dZt = ? Zt ? t dWt , so d 1 Zt 1 1 1 2 2 2 ? t ? 2 t = f (Zt )dZt + f (Zt )dZt dZt = ? 2 (? Zt )? t dWt + 3 Zt ? t dt = Z dWt + Z dt. 2 Zt 2 Zt t t (ii) Proof. By Lemma 5. 2. 2. , for s, t ? 0 with s < t, Ms = E[Mt |Fs ] = E Zs Ms . So M = Z M is a P -martingale. (iii) Zt Mt Zs |Fs . That is, E[Zt Mt |Fs ] = 47 Proof. dMt = d Mt  · 1 Zt = 1 1 1 ? M t ? t M t ? 2 ? t ? t t dMt + Mt d + dMt d = dWt + dWt + dt + dt. Zt Zt Zt Zt Zt Zt Zt (iv) Proof. In part (iii), we have dMt = Let ? t = 5. 6. Proof. By Theorem 4. 6. 5, it su? ces to show Wi (t) is an Ft -martingale under P and [Wi , Wj ](t) = t? ij (i, j = 1, 2). Indeed, for i = 1, 2, Wi (t) is an Ft -martingale under P if and only if Wi (t)Zt is an Ft -martingale under P , since Wi (t)Zt E[Wi (t)|Fs ] = E |Fs . Zs By It? ’s product formula, we have o d(Wi (t)Zt ) = Wi (t)dZt + Zt dWi (t) + dZt dWi (t) = Wi (t)(? Zt )? (t)  · dWt + Zt (dWi (t) + ? i (t)dt) + (? Zt ? t  · dWt )(dWi (t) + ? i (t)dt) d t M t ? t M t ? 2 ? t ? t ? t M t ? t t dWt + dWt + dt + dt = (dWt + ? t dt) + (dWt + ? t dt). Zt Zt Zt Zt Zt Zt then dMt = ? t dWt . This proves Corollary 5. 3. 2. ?t +Mt ? t , Zt = Wi (t)(? Zt ) j=1 d ?j (t)dWj (t) + Zt (dWi (t) + ? i (t)dt) ? Zt ? i (t)dt = Wi (t)(? Zt ) j=1 ?j (t)dWj (t) + Zt dWi (t) This shows Wi (t)Zt is an Ft -martingale under P . So Wi (t) is an Ft -martingale under P . Moreover,  ·  · [Wi , Wj ](t) = Wi + 0 ?i (s)ds, Wj + 0 ?j (s)ds (t) = [Wi , Wj ](t) = t? ij . Combined, this proves the two-dimensional Girsanov’s Theorem. 5. 7. (i) Proof. Let a be any strictly positive number. We de? e X2 (t) = (a + X1 (t))D(t)? 1 . Then P X2 (T ) ? X2 (0) D(T ) = P (a + X1 (T ) ? a) = P (X1 (T ) ? 0) = 1, and P X2 (T ) > X2 (0) = P (X1 (T ) > 0) > 0, since a is arbitrary, we have proved the claim of this problem. D(T ) Remark: The intuition is that we invest the positive starting fund a into the money market account, and construct portfolio X1 from zero cost. Their sum should be able to beat the return of money market account. (ii) 48 Proof. We de? ne X1 (t) = X2 (t)D(t) ? X2 (0). Then X1 (0) = 0, P (X1 (T ) ? 0) = P X2 (T ) ? X2 (0) D(T ) = 1, P (X1 (T ) > 0) = P X2 (T ) > X2 (0) D(T ) > 0. 5. 8.The basic idea is that for any positive P -martingale M , dMt = Mt  · sentation Theorem, dMt = ? t dWt for some adapted process ? t . So martingale must be the exponential of an integral w. r. t. Brownian motion. Taking into account d iscounting factor and apply It? ’s product rule, we can show every strictly positive asset is a generalized geometric o Brownian motion. (i) Proof. Vt Dt = E[e? 0 Ru du VT |Ft ] = E[DT VT |Ft ]. So (Dt Vt )t? 0 is a P -martingale. By Martingale Represent tation Theorem, there exists an adapted process ? t , 0 ? t ? T , such that Dt Vt = 0 ? s dWs , or equivalently, ? 1 t ? 1 t ? 1 Vt = Dt 0 ? dWs . Di? erentiate both sides of the equation, we get dVt = Rt Dt 0 ? s dWs dt + Dt ? t dWt , i. e. dVt = Rt Vt dt + (ii) Proof. We prove the following more general lemma. Lemma 1. Let X be an almost surely positive random variable (i. e. X > 0 a. s. ) de? ned on the probability space (? , G, P ). Let F be a sub ? -algebra of G, then Y = E[X|F] > 0 a. s. Proof. By the property of conditional expectation Yt ? 0 a. s. Let A = {Y = 0}, we shall show P (A) = 0. In? 1 1 deed, note A ? F, 0 = E[Y IA ] = E[E[X|F]IA ] = E[XIA ] = E[X1A? {X? 1} ] + n=1 E[X1A? { n >X? n+1 } ] ? 1 1 1 1 1 P (A? {X ? 1})+ n=1 n+1 P (A? n > X ? n+1 }). So P (A? {X ? 1}) = 0 and P (A? { n > X ? n+1 }) = 0, ? 1 1 ? n ? 1. This in turn implies P (A) = P (A ? {X > 0}) = P (A ? {X ? 1}) + n=1 P (A ? { n > X ? n+1 }) = 0. ? ? t Dt dWt . T 1 Mt dMt . By Martingale Repre? dMt = Mt ( Mtt )dWt , i. e. any positive By the above lemma, it is clear that for each t ? [0, T ], Vt = E[e? t Ru du VT |Ft ] > 0 a. s.. Moreover, by a classical result of martingale theory (Revuz and Yor [4], Chapter II, Proposition (3. 4)), we have the following stronger result: for a. s. ?, Vt (? ) > 0 for any t ? [0, T ]. (iii) 1 1 Proof. By (ii), V > 0 a. s. so dVt = Vt Vt dVt = Vt Vt Rt Vt dt + ? t Dt dWt ? t = Vt Rt dt + Vt Vt Dt dWt = Rt Vt dt + T ?t Vt dWt , where ? t = 5. 9. ?t Vt Dt . This shows V follows a generalized geometric Brownian motion. Proof. c(0, T, x, K) = xN (d+ ) ? Ke? rT N (d? ) with d ± = then f (y) = ? yf (y), cK (0, T, x, K) = xf (d+ ) 1 v ? T x (ln K + (r  ± 1 ? 2 )T ). Let f (y) = 2 y v1 e? 2 2? 2 , ?d+ ? d? ? e? rT N (d? ) ? Ke? rT f (d? ) ? y ? y ? 1 1 = xf (d+ ) v ? e? rT N (d? ) + e? rT f (d? ) v , ? TK ? T 49 and cKK (0, T, x, K) x ? d? e? rT 1 ? d+ d? ? v ? e? rT f (d? ) + v (? d? )f (d? ) xf (d+ ) v f (d+ )(? d+ ) 2 ? y ? y ? y ? TK ? TK ?T x xd+ ? 1 ? 1 e? rT d? ?1 v v ? e? rT f (d? ) v ? v f (d? ) v f (d+ ) + v f (d+ ) ? T K2 ? TK K? T K? T ? T K? T x d+ e? rT f (d? ) d? v [1 ? v ] + v f (d+ ) [1 + v ] 2? T K ? T K? T ? T e? rT x f (d? )d+ ? 2 2 f (d+ )d? . K? 2 T K ? T = = = = 5. 10. (i) Proof. At time t0 , the value of the chooser option is V (t0 ) = max{C(t0 ), P (t0 )} = max{C(t0 ), C(t0 ) ? F (t0 )} = C(t0 ) + max{0, ? F (t0 )} = C(t0 ) + (e? r(T ? t0 ) K ? S(t0 ))+ . (ii) Proof. By the risk-neutral pricing formula, V (0) = E[e? rt0 V (t0 )] = E[e? rt0 C(t0 )+(e? rT K ? e? rt0 S(t0 )+ ] = C(0) + E[e? rt0 (e? r(T ? t0 ) K ? S(t0 ))+ ]. The ? st term is the value of a call expiring at time T with strike price K and the second term is the value of a put expiring at time t0 with strike price e? r(T ? t0 ) K. 5. 11. Proof. We ? rst make an analysis which leads to the hint, then we give a formal proof. (Analysis) If we want to construct a portfolio X that exactly replicates the cash ? ow, we must ? nd a solution to the backward SDE dXt = ? t dSt + Rt (Xt ? ?t St )dt ? Ct dt XT = 0. Multiply Dt on both sides of the ? rst equation and apply It? ’s product rule, we get d(Dt Xt ) = ? t d(Dt St ) ? o T T Ct Dt dt. Integrate from 0 to T , we have DT XT ? D0 X0 = 0 ? d(Dt St ) ? 0 Ct Dt dt. By the terminal T T ? 1 condition, we get X0 = D0 ( 0 Ct Dt dt ? 0 ? t d(Dt St )). X0 is the theoretical, no-arbitrage price of the cash ? ow, provided we can ? nd a trading strategy ? that solves the BSDE. Note the SDE for S ? R gives d(Dt St ) = (Dt St )? t (? t dt + dWt ), where ? t = ? t? t t . Take the proper change of measure so that Wt = t ? ds 0 s + Wt is a Brownian motion under the new measure P , we get T T T Ct Dt dt = D0 X0 + 0 T 0 ?t d(Dt St ) = D0 X0 + 0 ?t (Dt St )? t dWt . T This says the random variable 0 Ct Dt dt has a stochastic integral representation D0 X0 + 0 ? t Dt St ? dWt . T This inspires us to consider the martingale generated by 0 Ct Dt dt, so that we can apply Martingale Representation Theorem and get a formula for ? by comparison of the integrands. 50 (Formal proof) Let MT = Xt = ?1 Dt (D0 X0 T 0 Ct Dt dt, and Mt = E[MT |Ft ]. Then by Martingale Representation Theot 0 rem, we can ? nd an adapted process ? t , so that Mt = M0 + + t 0 ?t dWt . If we set ? t = T 0 ?u d(Du Su ) ? t 0 ?t Dt St ? t , we can check Cu Du du), with X0 = M0 = E[ Ct Dt dt] solves the SDE dXt = ? t dSt + Rt (Xt ? ?t St )dt ? Ct dt XT = 0. Indeed, it is easy to see that X satis? es the ? rst equation.To check the terminal condition, we note T T T XT DT = D0 X0 + 0 ? t Dt St ? t dWt ? 0 Ct Dt dt = M0 + 0 ? t dWt ? MT = 0. So XT = 0. Thus, we have found a trading strategy ? , so that the corresponding portfolio X replicates the cash ? ow and has zero T terminal value. So X0 = E[ 0 Ct Dt dt] is the no-arbitrage price of the cash ? ow at time zero. Remark: As shown in the analysis, d(Dt Xt ) = ? t d(Dt St ) ? Ct Dt dt. Integrate from t to T , we get T T 0 ? Dt Xt = t ? u d(Du Su ) ? t Cu Du du. Take conditional expectation w. r. t. Ft on both sides, we get T T ? 1 ? Dt Xt = ? E[ t Cu Du du|Ft ]. So Xt = Dt E[ t Cu Du du|Ft ].This is the no-arbitrage price of the cash ? ow at time t, and we have justi? ed formula (5. 6. 10) in the textbook. 5. 12. (i) Proof. dBi (t) = dBi (t) + ? i (t)dt = martingale. Since dBi (t)dBi (t) = P. (ii) Proof. dSi (t) = = = R(t)Si (t)dt + ? i (t)Si (t)dBi (t) + (? i (t) ? R(t))Si (t)dt ? ?i (t)Si (t)? i (t)dt d d ? ij (t) ? ij (t) d d j=1 ? i (t) ? j (t)dt = j=1 ? i (t) dWj (t) + ? ij (t)2 d e j=1 ? i (t)2 dt = dt, by L? vy’s Theorem, Bi ? ij (t) d j=1 ? i (t) dWj (t). So Bi is a is a Brownian motion under R(t)Si (t)dt + ? i (t)Si (t)dBi (t) + j=1 ?ij (t)? j (t)Si (t)dt ? Si (t) j=1 ?ij (t)? j (t)dt R(t)Si (t)dt + ? (t)Si (t)dBi (t). (iii) Proof. dBi (t)dBk (t) = (dBi (t) + ? i (t)dt)(dBj (t) + ? j (t)dt) = dBi (t)dBj (t) = ? ik (t)dt. (iv) Proof. By It? ’s product rule and martingale property, o t t t E[Bi (t)Bk (t)] = E[ 0 t Bi (s)dBk (s)] + E[ 0 t Bk (s)dBi (s)] + E[ 0 dBi (s)dBk (s)] = E[ 0 ?ik (s)ds] = 0 ?ik (s)ds. t 0 Similarly, by part (iii), we can show E[Bi (t)Bk (t)] = (v) ?ik (s)ds. 51 Proof. By It? ’s product formula, o t t E[B1 (t)B2 (t)] = E[ 0 sign(W1 (u))du] = 0 [P (W1 (u) ? 0) ? P (W1 (u) < 0)]du = 0. Meanwhile, t E[B1 (t)B2 (t)] = E[ 0 t sign(W1 (u))du [P (W1 (u) ? 0) ? P (W1 (u) < 0)]du = 0 t = 0 t [P (W1 (u) ? ) ? P (W1 (u) < u)]du 2 0 = < 0, 1 ? P (W1 (u) < u) du 2 for any t > 0. So E[B1 (t)B2 (t)] = E[B1 (t)B2 (t)] for all t > 0. 5. 13. (i) Proof. E[W1 (t)] = E[W1 (t)] = 0 and E[W2 (t)] = E[W2 (t) ? (ii) Proof. Cov[W1 (T ), W2 (T )] = E[W1 (T )W2 (T )] T T t 0 W1 (u)du] = 0, for all t ? [0, T ]. = E 0 T W1 (t)dW2 (t) + 0 W2 ( t)dW1 (t) T = E 0 W1 (t)(dW2 (t) ? W1 (t)dt) + E 0 T W2 (t)dW1 (t) = ? E 0 T W1 (t)2 dt tdt = ? 0 1 = ? T 2. 2 5. 14. Equation (5. 9. 6) can be transformed into d(e? rt Xt ) = ? t [d(e? rt St ) ? ae? rt dt] = ? t e? rt [dSt ? rSt dt ? adt]. So, to make the discounted portfolio value e? t Xt a martingale, we are motivated to change the measure t in such a way that St ? r 0 Su du? at is a martingale under the new measure. To do this, we note the SDE for S is dSt = ? t St dt+? St dWt . Hence dSt ? rSt dt? adt = [(? t ? r)St ? a]dt+? St dWt = ? St Set ? t = (? t ? r)St ? a ? St (? t ? r)St ? a dt ? St + dWt . and Wt = t ? ds 0 s + Wt , we can ? nd an equivalent probability measure P , under which S satis? es the SDE dSt = rSt dt + ? St dWt + adt and Wt is a BM. This is the rational for formula (5. 9. 7). This is a good place to pause and think about the meaning of â€Å"martingale measure. † What is to be a martingale?The new measure P should be such that the discounted value pro cess of the replicating 52 portfolio is a martingale, not the discounted price process of the underlying. First, we want Dt Xt to be a martingale under P because we suppose that X is able to replicate the derivative payo? at terminal time, XT = VT . In order to avoid arbitrage, we must have Xt = Vt for any t ? [0, T ]. The di? culty is how to calculate Xt and the magic is brought by the martingale measure in the following line of reasoning: ? 1 ? 1 Vt = Xt = Dt E[DT XT |Ft ] = Dt E[DT VT |Ft ]. You can think of martingale measure as a calculational convenience.That is all about martingale measure! Risk neutral is a just perception, referring to the actual e? ect of constructing a hedging portfolio! Second, we note when the portfolio is self-? nancing, the discounted price process of the underlying is a martingale under P , as in the classical Black-Scholes-Merton model without dividends or cost of carry. This is not a coincidence. Indeed, we have in this case the relation d(Dt Xt ) = ? t d(Dt St ). So Dt Xt being a martingale under P is more or less equivalent to Dt St being a martingale under P . However, when the underlying pays dividends, or there is cost of carry, d(Dt Xt ) = ? d(Dt St ) no longer holds, as shown in formula (5. 9. 6). The portfolio is no longer self-? nancing, but self-? nancing with consumption. What we still want to retain is the martingale property of Dt Xt , not that of Dt St . This is how we choose martingale measure in the above paragraph. Let VT be a payo? at time T , then for the martingale Mt = E[e? rT VT |Ft ], by Martingale Representation rt t Theorem, we can ? nd an adapted process ? t , so that Mt = M0 + 0 ? s dWs . If we let ? t = ? t e t , then the ? S value of the corresponding portfolio X satis? es d(e? rt Xt ) = ? t dWt . So by setting X0 = M0 = E[e? T VT ], we must have e? rt Xt = Mt , for all t ? [0, T ]. In particular, XT = VT . Thus the portfolio perfectly hedges VT . This justi? es the risk-neutral pricing of Europea n-type contingent claims in the model where cost of carry exists. Also note the risk-neutral measure is di? erent from the one in case of no cost of carry. Another perspective for perfect replication is the following. We need to solve the backward SDE dXt = ? t dSt ? a? t dt + r(Xt ? ?t St )dt XT = VT for two unknowns, X and ?. To do so, we ? nd a probability measure P , under which e? rt Xt is a martingale, t then e? rt Xt = E[e? T VT |Ft ] := Mt . Martingale Representation Theorem gives Mt = M0 + 0 ? u dWu for some adapted process ?. This would give us a theoretical representation of ? by comparison of integrands, hence a perfect replication of VT . (i) Proof. As indicated in the above analysis, if we have (5. 9. 7) under P , then d(e? rt Xt ) = ? t [d(e? rt St ) ? ae? rt dt] = ? t e? rt ? St dWt . So (e? rt Xt )t? 0 , where X is given by (5. 9. 6), is a P -martingale. (ii) 1 1 Proof. By It? ’s formula, dYt = Yt [? dWt + (r ? 2 ? 2 )dt] + 2 Yt ? 2 dt = Yt (? dWt + rdt). So d(e? rt Yt ) = o t a ? e? rt Yt dWt and e? rt Yt is a P -martingale.Moreover, if St = S0 Yt + Yt 0 Ys ds, then t dSt = S0 dYt + 0 a dsdYt + adt = Ys t S0 + 0 a ds Yt (? dWt + rdt) + adt = St (? dWt + rdt) + adt. Ys This shows S satis? es (5. 9. 7). Remark: To obtain this formula for S, we ? rst set Ut = e? rt St to remove the rSt dt term. The SDE for U is dUt = ? Ut dWt + ae? rt dt. Just like solving linear ODE, to remove U in the dWt term, we consider Vt = Ut e Wt . It? ’s product formula yields o dVt = = e Wt dUt + Ut e Wt 1 ( )dWt + ? 2 dt + dUt  · e Wt 2 1 ( )dWt + ? 2 dt 2 1 e Wt ae? rt dt ? ? 2 Vt dt. 2 53 Note V appears only in the dt term, so multiply the integration factor e 2 ? e get 1 2 1 2 d(e 2 ? t Vt ) = ae? rt Wt + 2 ? t dt. Set Yt = e? Wt +(r? 2 ? (iii) Proof. t 1 2 1 2 t on both sides of the equation, )t , we have d(St /Yt ) = adt/Yt . So St = Yt (S0 + t ads ). 0 Ys E[ST |Ft ] = S0 E[YT |Ft ] + E YT 0 t a ds + YT Ys T t T a ds|Ft Ys E YT |Ft ds Ys E[YT ? s ]ds t = S0 E[YT |Ft ] + 0 a dsE[YT |Ft ] + a Ys t t T = S0 Yt E[YT ? t ] + 0 t a dsYt E[YT ? t ] + a Ys T t = = S0 + 0 t a ds Yt er(T ? t) + a Ys ads Ys er(T ? s) ds S0 + 0 a Yt er(T ? t) ? (1 ? er(T ? t) ). r In particular, E[ST ] = S0 erT ? a (1 ? erT ). r (iv) Proof. t dE[ST |Ft ] = aer(T ? t) dt + S0 + 0 t ads Ys a (er(T ? ) dYt ? rYt er(T ? t) dt) + er(T ? t) (? r)dt r = S0 + 0 ads Ys er(T ? t) ? Yt dWt . So E[ST |Ft ] is a P -martingale. As we have argued at the beginning of the solution, risk-neutral pricing is valid even in the presence of cost of carry. So by an argument similar to that of  §5. 6. 2, the process E[ST |Ft ] is the futures price process for the commodity. (v) Proof. We solve the equation E[e? r(T ? t) (ST ? K)|Ft ] = 0 for K, and get K = E[ST |Ft ]. So F orS (t, T ) = F utS (t, T ). (vi) Proof. We follow the hint. First, we solve the SDE dXt = dSt ? adt + r(Xt ? St )dt X0 = 0. By our analysis in part (i), d(e? t Xt ) = d(e? rt St ) ? ae? rt dt. Integrate fr om 0 to t on both sides, we get Xt = St ? S0 ert + a (1 ? ert ) = St ? S0 ert ? a (ert ? 1). In particular, XT = ST ? S0 erT ? a (erT ? 1). r r r Meanwhile, F orS (t, T ) = F uts (t, T ) = E[ST |Ft ] = S0 + t ads 0 Ys Yt er(T ? t) ? a (1? er(T ? t) ). So F orS (0, T ) = r S0 erT ? a (1 ? erT ) and hence XT = ST ? F orS (0, T ). After the agent delivers the commodity, whose value r is ST , and receives the forward price F orS (0, T ), the portfolio has exactly zero value. 54 6. Connections with Partial Di? erential Equations 6. 1. (i) Proof. Zt = 1 is obvious.Note the form of Z is similar to that of a geometric Brownian motion. So by It? ’s o formula, it is easy to obtain dZu = bu Zu du + ? u Zu dWu , u ? t. (ii) Proof. If Xu = Yu Zu (u ? t), then Xt = Yt Zt = x  · 1 = x and dXu = = = = Yu dZu + Zu dYu + dYu Zu au ? ?u ? u ? u du + dWu Zu Zu [Yu bu Zu + (au ? ?u ? u ) + ? u ? u ]du + (? u Zu Yu + ? u )dWu Yu (bu Zu du + ? u Zu dWu ) + Zu (bu Xu + au )du + (? u Xu + ? u )dWu . + ? u Z u ? u du Zu Remark: To see how to ? nd the above solution, we manipulate the equation (6. 2. 4) as follows. First, to u remove the term bu Xu du, we multiply on both sides of (6. 2. 4) the integrating factor e? bv dv . Then d(Xu e? ? Let Xu = e? u t u t bv dv ) = e? u t bv dv (au du + (? u + ? u Xu )dWu ). u t bv dv Xu , au = e? ? u t bv dv au and ? u = e? ? bv dv ? ? u , then X satis? es the SDE ? ? ? dXu = au du + (? u + ? u Xu )dWu = (? u du + ? u dWu ) + ? u Xu dWu . ? ? a ? ? ? ? To deal with the term ? u Xu dWu , we consider Xu = Xu e? ? dXu = e? u t u t ?v dWv . Then ?v dWv ?v dWv ? ? [(? u du + ? u dWu ) + ? u Xu dWu ] + Xu e? a ? u t u t 1 ( u )dWu + e? 2 u t ?v dWv 2 ? u du ? +(? u + ? u Xu )( u )e? ? ?v dWv du 1 ? 2 ? ? ? = au du + ? u dWu + ? u Xu dWu ? ?u Xu dWu + Xu ? u du ? ?u (? u + ? u Xu )du ? ? ? 1 ? 2 = (? u ? ?u ? u ? Xu ? u )du + ? u dWu , a ? ? 2 where au = au e? ? ? ? 1 d Xu e 2 u t ?v dWv 2 ? v dv and ? u = ? u e? ? ? = e2 1 u t 2 ? v dv u t ?v dWv . Finally, use the integrating factor e u t 2 ? v dv u 1 2 ? dv t 2 v , we have u t 1 ? ? 1 2 (dXu + Xu  · ? u du) = e 2 2 [(? u ? ?u ? u )du + ? u dWu ]. a ? ? Write everything back into the original X, a and ? , we get d Xu e? i. e. d u t bv dv? u t 1 ? v dWv + 2 u t 2 ? v dv = e2 1 u t 2 ? v dv? u t ?v dWv ? u t bv dv [(au ? ?u ? u )du + ? u dWu ], Xu Zu = 1 [(au ? ?u ? u )du + ? u dWu ] = dYu . Zu This inspired us to try Xu = Yu Zu . 6. 2. (i) 55 Proof.The portfolio is self-? nancing, so for any t ? T1 , we have dXt = ? 1 (t)df (t, Rt , T1 ) + ? 2 (t)df (t, Rt , T2 ) + Rt (Xt ? ?1 (t)f (t, Rt , T1 ) ? ?2 (t)f (t, Rt , T2 ))dt, and d(Dt Xt ) = ? Rt Dt Xt dt + Dt dXt = Dt [? 1 (t)df (t, Rt , T1 ) + ? 2 (t)df (t, Rt , T2 ) ? Rt (? 1 (t)f (t, Rt , T1 ) + ? 2 (t)f (t, Rt , T2 ))dt] 1 = Dt [? 1 (t) ft (t, Rt , T1 )dt + fr (t, Rt , T1 )dRt + frr (t, Rt , T1 )? 2 (t, Rt )dt 2 1 +? 2 (t) ft (t, Rt , T2 )dt + fr (t, Rt , T2 )dRt + frr (t, Rt , T2 )? 2 (t, Rt )dt 2 ? Rt (? 1 (t)f (t, Rt , T1 ) + ? 2 (t)f (t, Rt , T2 ))dt] 1 = ? 1 (t)Dt [? Rt f (t, Rt , T1 ) + ft (t, Rt , T1 ) + ? t, Rt )fr (t, Rt , T1 ) + ? 2 (t, Rt )frr (t, Rt , T1 )]dt 2 1 +? 2 (t)Dt [? Rt f (t, Rt , T2 ) + ft (t, Rt , T2 ) + ? (t, Rt )fr (t, Rt , T2 ) + ? 2 (t, Rt )frr (t, Rt , T2 )]dt 2 +Dt ? (t, Rt )[Dt ? (t, Rt )[? 1 (t)fr (t, Rt , T1 ) + ? 2 (t)fr (t, Rt , T2 )]]dWt = ? 1 (t)Dt [? (t, Rt ) ? ?(t, Rt , T1 )]fr (t, Rt , T1 )dt + ? 2 (t)Dt [? (t, Rt ) ? ?(t, Rt , T2 )]fr (t, Rt , T2 )dt +Dt ? (t, Rt )[? 1 (t)fr (t, Rt , T1 ) + ? 2 (t)fr (t, Rt , T2 )]dWt . (ii) Proof. Let ? 1 (t) = St fr (t, Rt , T2 ) and ? 2 (t) = ? St fr (t, Rt , T1 ), then d(Dt Xt ) = Dt St [? (t, Rt , T2 ) ? ?(t, Rt , T1 )]fr (t, Rt , T1 )fr (t, Rt , T2 )dt = Dt |[? t, Rt , T1 ) ? ?(t, Rt , T2 )]fr (t, Rt , T1 )fr (t, Rt , T2 )|dt. Integrate from 0 to T on both sides of the above equation, we get T DT XT ? D0 X0 = 0 Dt |[? (t, Rt , T1 ) ? ?(t, Rt , T2 )]fr (t, Rt , T1 )fr (t, Rt , T2 )|dt. If ? (t, Rt , T1 ) = ? (t, Rt , T 2 ) for some t ? [0, T ], under the assumption that fr (t, r, T ) = 0 for all values of r and 0 ? t ? T , DT XT ? D0 X0 > 0. To avoid arbitrage (see, for example, Exercise 5. 7), we must have for a. s. ?, ? (t, Rt , T1 ) = ? (t, Rt , T2 ), ? t ? [0, T ]. This implies ? (t, r, T ) does not depend on T . (iii) Proof. In (6. 9. 4), let ? 1 (t) = ? (t), T1 = T and ? (t) = 0, we get d(Dt Xt ) = 1 ? (t)Dt ? Rt f (t, Rt , T ) + ft (t, Rt , T ) + ? (t, Rt )fr (t, Rt , T ) + ? 2 (t, Rt )frr (t, Rt , T ) dt 2 +Dt ? (t, Rt )? (t)fr (t, Rt , T )dWt . This is formula (6. 9. 5). 1 If fr (t, r, T ) = 0, then d(Dt Xt ) = ? (t)Dt ? Rt f (t, Rt , T ) + ft (t, Rt , T ) + 2 ? 2 (t, Rt )frr (t, Rt , T ) dt. We 1 2 choose ? (t) = sign ? Rt f (t, Rt , T ) + ft (t, Rt , T ) + 2 ? (t, Rt )frr (t, Rt , T ) . To avoid arbitrage in this case, we must have ft (t, Rt , T ) + 1 ? 2 (t, Rt )frr (t, Rt , T ) = Rt f (t, Rt , T ), or equivalently, for any r in the 2 range of Rt , ft (t, r, T ) + 1 ? (t, r)frr (t, r, T ) = rf (t, r, T ). 2 56 6. 3. Proof. We note d ? e ds s 0 bv dv C(s, T ) = e? s 0 bv dv [C(s, T )(? bs ) + bs C(s, T ) ? 1] = ? e? s 0 bv dv . So integrate on both sides of the equation from t to T, we obtain e? T 0 bv dv C(T, T ) ? e? t 0 t 0 T bv dv C(t, T ) = ? t s 0 e? T t s 0 bv dv ds. Since C(T, T ) = 0, we have C(t, T ) = e 1 ? a(s)C(s, T ) + 2 ? 2 (s)C 2 (s, T ), we get A(T, T ) ? A(t, T ) = ? bv dv T t e? bv dv ds = T e t s bv dv ds. Finally, by A (s, T ) = T a(s)C(s, T )ds + t 1 2 ? 2 (s)C 2 (s, T )ds. t

Book Report on the Blind Side Essay

The Blind side The Blind Side directed by John Lee Hancock was a visual text about a teenage boy named Michael. Based on a true story Michael, nicknamed Big Mike has grown up in a poor and broken family and goes to a public school where no one really cares about him. Growing up this way has left Big Mike emotionally deprived and lonely. Thanks to his Friend’s dad Michael gets the opportunity to go to a private school on a scholarship. Suddenly he has teachers that care about him and while his life seems to be slowly improving, Big Mike still uses other people’s washing machines in the Laundromat, does not sleep at home and stays at the gym at school because it was warm. The biggest turning point in this movie was when Leigh Anne Touhy sees Big Mike on the side of the road and lets Big Mike have a place to stay. Leigh is a woman that gets what she wants and it is clear from the moment she meets Mike that she would care for him. From this point Big Mike and Leigh build a strong relation ship, he ends up moving in with them and becomes a part of the family. Michael is a big man and one of the reasons he was accepted into the school was because he had the perfect build to play American football and by the end of the movie Big Mike was one of the top players and went on to have a career in it. My favorite character throughout the book was definitely Big Mike, I love how complex of a character the director made him. Appearance wise, Big Mike looked big and tough, he was the type of boy that you would walk the opposite direction when you see him on the street, but when you see him in the movie you discover how kind hearted he is. A classic example of this is when Big Mike is first playing football and he doesn’t tackle anyone because he doesn’t want to hurt anyone. He reminds me of a giant teddy bear that wouldn’t hurt anyone. But then as well as his kind heart if you dig deeper you can see the inner strength he possesses. From a very young age Michael is forced to look after himself and to get through it the way he did could only be done with utter strength. When most people are left with no one they give up but I admire Big Mike so much for turning to himself and doing all he could to make his life work. Not only is this but his loyalty unbreakable. To get Big Mike to tackle at the practice Leigh said to him to pretend that these guys were going to hurt her and his family and that’s when he finally tackled with aggression. You could see then and there that Big Mike would take on anyone that tried to hurt his family and those he cared about. To me it Big Mike is, for lack of a better expression, a total cutie. My favorite part of the movie is when Leigh asks Big Mike if he would like to be part of their family and Big Mike looks at her dead seriously and replies ‘I thought I already was.’ Throughout the whole movie it was this character that constantly impressed and engaged me and the fact that it was based on a true story just adds to this. One of the main ideas in the visual text The Blind Side is courage. Leigh Touhy shows courage when she takes Mike Oher from off the streets and gives him a roof over his head and someone to lean on. This showed courage in the best of ways and it couldn’t off been done to anyone else. Although Leigh’s family did not agree with having Mike there in the first place, then soon got to like the guy and really enjoyed having him there as another brother or son. It also took real courage for Leigh and Michael to legally adopt Big Mike as a child, get his driver license and help him get into the best school available for Mike to go to on a football scholarship. Leigh also had courage for going to her own friends and telling them about Mike. A conversation held with her friends is. Friend: â€Å"You’re changing that boy’s life† Leigh: â€Å"No He’s changing mine†. I thought that Leigh had courage to say this to her friends as Mike is not her own son yet she is taking care of him as he is the only child in the family. A quote from the film The Blind Side to represent Courage is â€Å"that’s why courage is tricky. Should you always do what others tell you to do? Sometimes you might not even know why you’re doing something. I mean, any fool can have courage†. This is saying that courage is important but it’s hard. You should do what you want to do and not what others tell you to do,  but the main point is that anyone can have courage; weather their big, small, tall or short. The text that I immediately thought of was The Dead Poets society and in particular Neil. The reason for this is in my mind I started to immediately compare the difference between Neil and Michael. As I previously stated Big Mike had to deal with the loneliness in his life and the way he did this was by turning to himself for support. I think the reason that Neil committed suicide was loneliness, by this I don’t mean that he had no friends, as it was obvious from the start he did, but that he felt like he had no support from those that mattered. If his dad had shown a slight interest in his acting career then Neil would have been satisfied but because his dad seemed not to care Neil felt like he had no one to turn to and no options. This is the difference that I see between Neil and Michael, inner strength. While Michael appeared soft throughout the visual text, to carry on living especially in certain points in his life took incredible inner strength and as much as I liked Ne il’s character I don’t think he possessed the same strength. Neil is kind of the polar opposite, on the outside he appeared to be strong and content with life but on the inside the way his father was acting slowly killed him inside. Both of these characters had parents that weren’t at all good at their job but it was the difference between the two personalities that ended with one dead and the other a professional football player.

Promoting the Go Green Policy Research Paper Example | Topics and Well Written Essays - 2500 words

Promoting the Go Green Policy - Research Paper Example The go green policy was implemented with the objective of saving the environment, by restricting the use of inorganic products. Â  Some scholars have attempted to clear this confusion by defining green or environmental – friendly products as goods that attempt to improve or protect the natural environment. Such preservation or protection is to be achieved by conserving resources and energy while mitigating or eliminating the utilization of pollution, toxic agents and waste (Dangelico & Pujari, 2010, p. 472). The recent natural calamities that afflicted the American continent compelled governments to focus on environmental issues. One of the most important outcomes of environmental damage is climate change. Scholars like Glave have contended that governments and people should act immediately to save the environment for the benefit of future generations (McRae, 2007, p. 133). In the absence of such measures, the future generations will inherit a planet that is beset with defore station, water and air pollution, and unpredictable weather patterns. Going green or making decisions that are friendly towards the environment require the government and the people to play an active part in protecting the earth from degradation, due to human activities (McRae, 2007, p. 133). In order to promote the go green policy, the US government has enacted the Energy Policy Act of 2005. Under the provisions of this act, the Internal Revenue Service provides a number of tax benefits to people who purchase energy-efficient appliances (McRae, 2007, p. 133).

Michael Jordan's Reign In Basketball Research Paper

Michael Jordan's Reign In Basketball - Research Paper Example In 1982 Michael Jordan made the winning shot in NCAA Championship game making his North Carolina Tarheals champions and beginning a legacy that would turn him into a living legend. It wasn’t long after that Jordan was drafted by the Chicago Bulls and in his first season became a fan favorite, and went on to claim the rookie of the year award. It wasn’t long before Michael Jordan became a star in the league. In the 1987 season he averaged 37.2 points per game and scored over 3,000 points; both numbers only matched by Wilt Chamberlain. The next season he again won the league in scoring and won his first league Most Valuable Player Award. Despite Jordan’s unheralded individual success, at this point he had still failed to achieve the allusive NBA Championship victory. In the 1990 season, finally after years of disappointment, Jordan and his Chicago Bulls achieved their first championship. Jordan and the Chicago Bulls would go on to claim the next two championship vi ctories as well. While Larry Bird and Magic Johnson helped greatly popularize the game, it was during this period that Michael Jordan took it to new heights. His heroic style of play brought a widespread audience to the game of basketball, bringing it and himself international acclaim. Still at the height of his accomplishments Jordan announced that he would retire to pursue a career in baseball. Three years later, by the influence and criticism of others, Michael returned to the game. Many people believed Michael would not get back in his groove yet defied odds and created a new atmosphere. As Michael was a true committed person he relied very much so on his self creating independence for himself and influenced the way he live and played. This was displayed by the way he operated, how he portrayed himself, how media coverage portrayed him etc. His motivation

A Essay Example | Topics and Well Written Essays - 500 words - 3

A - Essay Example Slight mistakes in any of the field listed would render the whole project being undertaken futile. The computer is the core of every operation in the current generation of trade and general life. That is why it is taken first as the core qualification for any other do. Continuous development on it brings experience especially when applied to the other fields of concern as listed. I am also a philanthropist. Based on acquisition of all these qualifications, the main aim is profitability and objectivity to the society. The computer society degree has a major aim of easing problem solving skills in the society; the business operations degree will be endeared toward helping profitable engagements in the society. The degree in youth and women empowerment is meant to develop a strong team of women and youth in the society. This is a show if value and giving back to the society in spite of having achieved much in the long run as an individual. One aspect of me that is very predominant therefore is that of bringing change to my immediate environment. This change started with me. The first thing is to get the valuable education required. It is also systematic in that it starts with a foundation on computers. The next change applies to the immediate society; the women and children. This is my society. Developing it will be developing me as a person and the world at large. This equally creates a trend of change in me. This knowledge to the society will create independence in their minds and actions which at the far end makes a society that knows itself and the environment around it. Conservation, business knowledge to marketers and health issues are just but a piece of the society that will be created by my founded knowledge. On top of the above, I am a problem identifier and solver. Though the solving ability will be dependent on the eventual qualifications, the foundation of this lies in the willingness to take up the challenge. The society is marred by

Embracing Globalization while Maintaining Identity Essay

Embracing Globalization while Maintaining Identity - Essay Example As the paper discusses while globalization is not the answer to all the world’s problems, a negative attitude towards it is one that is capable of breeding suspicion and misunderstanding between different cultures and nations. Also, economically speaking, globalization can be far more complicated than merely showing tolerance and acceptance towards other cultures. This essay will first illustrate some of the negative attitudes toward globalization. Next, globalization as it relates to education in Israel will be discussed. Through these examples one can likely see that many people are afraid that globalization will weaken their particular nation and take away from their common identities. In response to these fears, the paper will suggest that it is indeed possible to maintain one’s identity while at the same time accepting globalization. One instance of a negative view towards globalization is â€Å"anti-soccer† Americans as described in the work of Franklin Foer. According to Foer, globalization has actually failed some of its expectations, both economically and culturally, given that a majority of the world’s nations remain poor, and that some cultures actually fear that â€Å"globalized culture† will eventually erode their own unique cultural identities. In order to illustrate the failures of globalization, particularly in its failure to promote the homogenization of culture (and even attracted the reversion to old cultural identities and hostility to other cultures), Foer uses the ever famous worldwide sport of soccer.  

Contemporary Developments in Business and Management Case Study

Contemporary Developments in Business and Management - Case Study Example For that reason, since its origin in Nov. 1999, the corporation has gathered above VND 10 billion in its premium, out of which 7 billion has been invested again into the nationwide financial system. The hard work of this corporation in developing its company acts, improving the excellence of its clients care programs, and causative to the nationwide social and economic growths have been acknowledged by administrative organizations, clients and the people of Vietnam. This paper will discuss the internal and external analysis of Prudential Vietnam and some factors involved in the decision-making process of the said company (online). Prudential in Vietnam is a global retail pecuniary services group that intends to facilitate people protected and augment their own and their dependents' monetary comfort by providing investments, security and other products and services that are appropriate for their requirements. The only strategy of Prudential Vietnam is to construct flourishing and more and more beneficial businesses in each of these markets and in that way make the most of profits to their shareholders eventually. This report assesses the impact of external and internal factors on Prudential Vietnam and evaluates the industry's responses to such factors. Established in the UK in 1848, Prudential plc is one of the principal life and annuity giver in the UK and a top worldwide monetary services corporation with more than US$430 billion (June 2006) in finances under supervision, more than 21m clients and approximately 23,000 staff members globally. For the sake of meeting its mounting requirements of its clients, this corporation has passed to marketplace incorporated variety of monetary services productions that now consists of life the assurance, retirement funds, mutual funds, banking, asset board and all-purpose insurance (online). In 1995, the Prudential plc has established its first and foremost representative agency in Vietnam. From that time Prudential Vietnam has turned out to be a most dependable organization for myriads of clients through Vietnam - providing them appropriate support to meet their investments, security and their investment wants. Prudential plc has established its markets in Malaysia, Singapore, and Hong Kong, and following value-creation openings and predictions in the district's countless highly-potential and competitive marketplaces (online).

The Role of the United Nations in a Collective Security Essay

The Role of the United Nations in a Collective Security - Essay Example Recent international developments, however, have proven that collective security as envisaged by the UN Charter of 1945 is inadequate to meet the exigencies of the times. In the past, the collective security function of the United Nations had often failed because it had become a battleground of the two superpowers which emerged after the 2nd World War. The UN, especially its, however, security functions, was held hostage to the power play of these two countries. Even before the dust of the war had settled, the intense competition for global supremacy between the United States (US) and the Union of the Soviet Socialist Republic (USSR) began to take root into what has been called the Cold War, so termed because despite the intense fighting between the two sides on all fronts it did not entail the use of weapons. This intense conflict between the two superpowers affected the UN and its collective security functions because of its inherent structural defect. It would seem that the name United Nations is not the same as ‘equal nations’ because five of its members are not only given permanent standing but a commanding veto vote. When the organization was established in 1945, the countries which fought together with the Axis powers namely, the US, USSR, the United Kingdom (UK), China, and France were accorded permanent seats in the Security Council (SC) (Krasno 2004). As members of the P5, both the US and the USSR, together with the other three countries, have the sole veto power over any draft resolution of the UN (Zhu & Hearn 1999).

Sheltering the Deep Essay Example | Topics and Well Written Essays - 250 words

Sheltering the Deep - Essay Example the use of a phrase or word with application of humour in order to empathise different meaning or different uses of the words that sound the same or are a like (James 4). In the article, the author uses pun in different paragraphs like, â€Å"underwater canyon, bottlenose whales". In this phrase, she uses â€Å"bottlenose whales† to make sure that the readers understand the meaning of the word whales. Another pun used is the â€Å"underwater communities†. Another example of a pun in the article is the illustration of "rich waters". She is trying to explain about the birds that uses water to find food and applies pun for the reader to understand. Assonance is also applied in the report in different parts. In many sentences, the author repeats vowel â€Å"e† and â€Å"i† in different sentences. She has used narrowed and swallowed in the same sentences stating that after the lagoon narrowed then even become shallower. Alliteration is used where by she says, â€Å"race

How Did the Indsutrial Development Unite or Divide the North and the South Essay Example for Free

How Did the Indsutrial Development Unite or Divide the North and the South Essay During the Civil War, the advances of the Industrial Revolution introduced great changes in the industrial and technological development. Both the North and the South created many advances in railroad and water transportation. The Union, however, was far more advanced technologically than the Confederate states . Consequently, the North made greater and more effective use of progress in weapons, communication, transportation and medicine than South . Although the industrial development made the nation very widely known, both the south and the north were divided because their differences. The Civil War was the first modern war that helped strengthen the technology and industrial system. But their industry and technology distinguished the two sides, which represented different economic conditions. The North had developed a strong economy that was becoming day-by-day more industrialized. By the nineteenth century, large factories and organizations sprang up throughout the north. Also, the population of the country was increasing and immigrants from all over Europe came along. The North was becoming a huge success but the South was falling behind. The North was rising in a higher success rate than the South. The Union flourished more factories and more transportation. Canals were being handmade, there was an increase of labor force and there it was becoming more adequate to transport product through trains . Inventions were also becoming to life. For example, the Telegraph was becoming a extremely useful. Invented by Samuel F. B. Morse, the Telegraph was inexpensive to make and was ideal for long distance communication. The north had more advantages in growing the economy because it had twice as big as the population from the south. It had much greater man power and it had a better work force. Many factories from the north built war material to supply to the Union. However, slavery was decreasing around the 1860’s and factories were pouring in by the immigrants from Europe. In fact, seven out of every eight immigrants that traveled to the U. S. settled in the North rather than the South. The economy in the North was also increasing therefore immigrants settled there to establish their own business. Northerners were far more likely to have careers in business, medicine, or education . Also, children were slightly more prone to attend school than Southern children. As for the South, the warm climate and the fertile soil made it ideal for farmers to grow significant amounts of crops. There were more abundant natural resources in the south and because agriculture was so profitable few Southerners saw a need for industrial development . There were no large cities aside from a few known places. Most of the known cities existed near shipping ports to send agricultural produce to Northern destinations. However, the South had difficulty with transportation and most products were sent by water. Only a few train tracks were located in the South. In the other hand, Southern children tended to spend less time in school and most Southern families based their teachings in gravitating toward military careers as well as agriculture . The first half of the nineteenth century was a time of expansion and improvement of transportation systems. States in the North and the Midwest chartered and built overland roads and turnpikes. The Turnpike Era† (1790-1820) consisted of Americans relying on roads for internal transportation. Canals, such as the Erie Canal, tied New York City to the Great Lakes. Steamboats and railroads improved the movement of goods and people, forging ties that served both sides well during the Civil War. The first federal charter corporation that created the dream of the transcontinental railroad was the Union Pacific Railroad Company and the Central Pacific Company . Both of these companies gathered many immigrants, at low pay, to work massive hours to construct the railroad. However, better transportation fostered an upgrade on trade within the country and dispersed new civilization to the west. The industrial revolution created many social problems. Poverty became a growing concern, especially the fact that factory wages were scarcely adequate for family survival . Most residents experienced hunger and destitution. Among the poor, child labor was very common. Most parents forced their children to look for jobs instead of going to school for survival. Southerners often cited these factors as crimes whenever the North challenged its institution of slavery. The Industrial Revolution brought Southern landowners an invention that they adopted and embraced: The Cotton Gin. Invented by Eli Whitney, the cotton gin made slavery profitable and made cotton the nations number one export . The South also adopted the steam engine, mainly to aid the cotton gin and to use on steamships to transport cotton. Ironically, the success of the cotton gin, by fostering slavery, helped to separate the two sides of the country and bring about the Civil War. The pace of immigration also stimulated economic growth while increasing differences between North nd South. Immigrants, mostly from Europe at this time, were supplied with low-cost labor. Most immigrants lived in the North where jobs were constantly available but had no respect to the workers. The use of standard, interchangeable parts, especially the manufacturing of guns, clocks, and sewing machines , allowed the nation to advance technologically by using unskilled workers. During the Civil War, with Southern representatives of Congress gone and the Republican Party controlling the house of Congress and the presidency, â€Å"the government set about to aid business and technology†. In 1862, the Department of Agriculture was founded. It provided a national center to coordinate agricultural development and promote scientific farming. â€Å"A house divided against itself cannot stand. I believe this government cannot endure permanently half-slave and half-free. † This quotation was from Abraham Lincoln in 1858. Abraham Lincoln did not want the North and South to separate but for the Industry to grow bigger . In the first part of the quotation, â€Å"A house divided against itself cannot stand†, portrays that the United States needs to be UNITED not divided. A house needs to stand tall and not let anything else break it down. It is true that the â€Å"government cannot endue permanently half-slave and half-free† because this needs to be a united country not a haft this haft that country. Lincoln convinced others that the United States could not be this way. It had to bet glued together again and it had to abolish anything that was not right. However, throughout time, The Divided States of America was soon becoming the United States of America. After Lincoln’s death, three amendments were ratified that help America put back to place. The 13th amendment concluded that slavery was officially abolished . The 14th amendment granted â€Å"all persons born or naturalized in the United States, â€Å"to be citizens which included former slaves that were freed . The 15th amendment granted African Americans the right to vote . These three amendment helped bit by bit to repair the United States. Even though today there is still a difference in the North and South, our nation will always be together. The United States grew tremendously during the Industrial Revolution. Inventions were made, transportation was spread out, new jobs were increasing and more knowledge was diffusing. Throughout time, our population was growing and our nation got to spread out to the west to expand our land and culture. Even though our presidents may have made mistakes, we get to learn what we have done wrong and use that in our future. Our nation may have been divided for awhile but we can always retain it back. Back where it always was, united.

The Challenges Of The Twenty First Century Education Essay

The Challenges Of The Twenty First Century Education Essay Introduction This chapter reviews the relevant literature used in this study in a thematic and systematic manner. It begins with reviewing texts related to the contemporary local and global challenges. It then discusses the national goals of education in Kenya relating them to the demands of the 21st century. Teacher education in Kenya is then reviewed and finally the system of education in Kenya is examined. 2.2.0 The challenges of the twenty first century The 21st Century has been conceived as: The age of Globalization, the age of Knowledge Economy, and the Information age. Globalization refers to the contemporary social reality, which is characterized by change, uncertainty, unpredictability, complexity, interdependence and diversity. According to Giddens (1990) and Albrow (1994), it refers to the process by which human relations are increasingly being intensified. As a result, economic, political, cultural and social distinctions are becoming less and less inhibitive. Advancement, especially in the information and telecommunication sectors has compressed time and space and the world is gradually becoming a borderless forum for human interaction popularly known as the global village. The educational challenges in a globalized world include: ensuring that learners acquire the technological skills that enable them to access the global information and telecommunication networks, transforming learner attitudes and dispositions to enable them adapt to change and uncertainty, fostering character traits in learners that make them functional in a cross-cultural and democratic setting. Such traits include open-mindedness, tolerance and intellectual autonomy, enhancing learners capacity and attitude to think critically and creatively. Knowledge has become the most important factor in economic development in a globalized world, hence the use of the term Knowledge economy to refer to the contemporary global economy. Consequently, the ability of a society to produce, select, adapt, commercialise, and use knowledge is critical for sustained economic growth and improved living standards (World Bank, 2002). Education needs to go beyond merely informing learners. Learners have to be enabled to learn on their own, make sense of and apply knowledge innovatively. Learners, therefore, need to take charge of their thinking and direct it towards solving problems as wells as formulating and pursuing desired goals. Information and telecommunication technologies that facilitate and support knowledge-based activities have become extremely useful. Information has become easily accessible with the use of the internet hence the use of the term information age to refer to the 21st century. Merely consuming information uncritically is dangerous in a globalized age. Education should therefore enable learners to select, interrogate, evaluate and utilize information efficiently. 2.2.1 The global challenges According to Haag (1982) education systems in the world have expanded quantitatively rather than qualitatively making them unable to address current social problems. Although a lot of resources have been devoted to education, it appears like the systems of education have been ineffective in addressing social ills like inequality, intolerance, resistance to change, crime and violation of human rights among others. These social ills manifest themselves in rising levels of poverty, ethnic and racial conflicts and violation of human rights in many parts of the world. Ward and McCotter (2004) argue that developing thinking skills in educational institutions, though necessary, is inhibited by inappropriate teaching methods used by teachers, ineffective evaluation procedures and rigorous overload of the curriculum. Other negative factors include ineffective educational policies that emphasize content and structure of education while neglecting quality and process of education. Jelinek (1978 ) refers to the dominant expository methods of instruction as didacticism in which it is assumed that education is an act of depositing knowledge into learners who are mere depositories. The process of instruction is lifeless, petrified, motionless, static and compartmentalised and thus strange to the experience of the learners. The approach is irrelevant to reconstruction of the experience of the learners since they are considered to know nothing. Ultimately the approach tends to turn humans into automatons and therefore negates their dignity and abilities as human beings. Without development of thinking skills, graduates from educational institutions are observed to be limited in analytical, creative and innovative abilities that are essential in the modern knowledge based world of work. Leat (1999) blames inability of education systems to address contemporary challenges on faulty academic cultures and traditions. For instance, attainment of good grades and certificates using whatever means is valued as opposed to transformation of learners into effective agents of change in society. Unhealthy competition among learners is tolerated instead of developing the culture of cooperation and mutual responsibility. Rigid adherence to existing academic cultures inhibits ability to respond to change. Unhealthy competition among learners promotes antagonism and undermines team spirit. According to Perkins (1990) the culture in most educational institutions is characterized by minimal informative feedback and emphasis on traditional ways of doing things. This diminishes the exercise of individual initiative and choice. Consequently learners are denied an opportunity to develop essential characteristic of democratic citizenship. Such characteristics include respect for dialogue, freedom of expression and self-determination through individual choice. Barrow et al. (2006) reports that a study done in India revealed that the main challenges facing teacher education include meeting childrens specific learning needs, respecting students cultural and socio-economic context and involving parents and communities in school activities. The teachers therefore require interpersonal and counseling skills that can enable them to empathetically discern the learning needs of pupils. The teachers also need to be able to deal with diversity of learners and forge school-community collaboration. Douglass (2006) perceives preparation of employable graduates as the main challenge of education in the 21st century. According to him, emphasis on employability has led to other challenges. For instance, there are sharp divisions between scientific-technological academic disciplines on the one hand and social science and humanities on the other. He argues that there is an overemphasis on the value of scientific-technological disciplines leading to bifurcat ion of knowledge. According to NCCPPE(2008), the main challenges that education must confront in the world today include: conservation of the environment and sustainability of natural resources; the provision of health care; renewal of economic vitality; coping with change through learning; promoting core human values like justice, peace and equity and protecting human rights. This study endeavoured to find out the extent to which global challenges mentioned above impact on Kenya. It further sought to establish whether and how the system of PTE in Kenya was responsive to them. The study also designed a framework by which the challenges can be classified for effective examination. For instance, the following categories of challenges among others were be included in the framework: cognitive, cultural, political, social, emotional, economic, ethical, technological, and religious. Cognitive challenges include obstacles that hinder effective thinking. Such obstacles render learners incapable of analytical, evaluative and creative thinking. Such learners merely copy and reproduce the ideas of other people without being sensitive to context. Cultural challenges include stereotypes and prejudices that are embodied in traditions and norms of particular social groups. These traditions are passed on from one generation to another without critical scrutiny. In additio n cultural beliefs and practices tend to be ethnocentric. As such they promote conflicts and misunderstanding among social groups. Political challenges include ideologies and policies that are designed to facilitate acquisition and maintenance of political power. Often such ideologies are propagated in a competitive way without regard to sufficient analysis, evaluation and fair-mindedness. Social challenges include the need to accommodate diversity in the contemporary setting that is increasingly becoming multicultural. There is need for tolerance, open-mindedness and humility in interpersonal interactions. Emotional challenges have to do with inability to understand and deal with ones own feeling as well as the feelings of others. This is especially so in a social context that is dynamic, uncertain and stressful. Economic challenges include scarcity of resources and controversial methods of distributing the resources. These challenges are compounded by the increasing needs and wants characterised by the consumerist lifestyles. Ethical challenges arise from disagreements on principles of right and wrong leading to ethical relativism. Such relativism makes moral values difficult to apply across board leading to ethical confusion. Technological challenges include inability to control and manage the use of technology in a way that benefits society without endangering wellbeing. Such challenges include checking abuse of the internet, regulating the mass media and ensuring that nuclear technology does not get into the hands of terrorists. Religious challenges include animosity among different faiths that sometimes expresses itself in overt violent practices. Religious fanatics often engage in breach of human rights and criminal activities in the name of God. All these challenges cannot be effectively addressed unless education empowers the learners and society at large to think for themselves, analyse and evaluate issues, question beliefs and claims as well as develop the ability to creatively solve problems. 2.2.2 The challenges in Africa Assie-Lumumba (2006) perceives the debt burden, ethnic violence, armed conflicts and the scourge of HIV and AIDS as the most visible challenges that impact education in Africa. What is needed is an education system that can empower Africans to participate in the production and application of knowledge relevant in addressing these challenges and promote broad societal advancement. In Ghana, the need to develop thinking skills among learners has been recognized as a viable way of addressing contemporary challenges facing Africa. However in practice, the development of such skills has not been given adequate attention (Acheampong, 2001; Hill, 2000). There exists a mismatch between the professed value of thinking skills in education and actual efforts to develop such skills in teacher training institutions. According to Owu-Ewie (2007), classroom environment in many educational institutions in Ghana inhibit thinking in students. The teachers have been observed to be autocratic and rigid in imposing their views on students. The opinions of students are disrespected and discarded thus discouraging learner participation, curiosity and creativity. Teachers make poor use of questioning and motivation and use the lecture method predominantly. The education system fosters rote learning, drilling and exam orientation. Such a system does not facilitate a healthy teacher-learner interaction. According to Barrow et al (2006), studies done in Namibia and Nigeria reveal additional challenges facing education in Africa. In Namibia, although educational policies are strongly based on active learning and learner-centered theoretical foundation, in practice, these lofty theories are not effectively implemented. The educational policies are poorly understood, interpreted and executed. The study findings support school-based teacher professional development programs, associated with whole-school improvement programs, as very promising ways of increasing understanding and effective implementation of active-learning policies. In Nigeria, religious tensions and economic empowerment are the challenges that education needs to address. Teachers have therefore to be empowered to facilitate inter-faith harmony, creativity and self-reliance among learners. In Ethiopia, poor quality of education, insufficient financing, lack of equity and poor management are the key challenges facing educa tion (Ethiopian National Agency for UNESCO, 2001). Consequently, the ongoing educational reform encompasses every aspect of the educational system- the curricula, teacher training, educational inputs, educational finance, organization and management, structure of education, career structure of teachers, and evaluation. The reform is aimed at total restructuring of the educational system. This study examined the challenges above and related them to those in Kenya and the rest of the world. It also compared the strategies employed to address the challenges with a view to reconstructing a more responsive approach to the challenges relevant to PTE in Kenya. 2.2.3 The challenges in Kenya In Kenya, some of the 21st century challenges are HIV AIDS, gender awareness, and sensitivity to human rights (K.I.E., 2004a and 2004b). Others include poverty, crime, drug abuse, and unemployment. These challenges require that education empowers the learner to reflect and respond to them pro-actively. While informing the learners about these problems is important, empowerment of the learner demands a transformation that transcends the cognitive dimension to include all other faculties of the human person. This study explored these non-cognitive dimensions such as the creative, cultural, ethical and social among others. According to RCE (2007), sustainable development in Kenya (as in any other country) is complex since it encompasses social issues such as peace and security, human rights, gender equality, cultural diversity and intercultural understanding. Other issues include poor governance, corruption, increased incidences of diseases, erosion of cultural values and morals, among others. The economic issues include corporate social responsibility and accountability, ethical marketing, increasing levels of poverty and the widening gap between rich and poor. The other issues include trends of unsustainable production and consumption leading to inefficiency and wastefulness, poor enforcement of policies and regulations governing production and marketing. Environmental challenges include the energy, nutritional and other domestic needs of an expanding population, unsustainable use of natural resources (water, land), rural/urban migration, climate change, rural development, urbanization, disaster prev ention and mitigation concerns. This study examines the extent to which these concerns are addressed in PTE with specific reference to pedagogical approaches employed. Abagi and Odipo (1997) argue that the operation of primary education system in Kenya faces the problem of inefficiency. Low completion rates and national pupil-teacher ratio make inefficiency evident. In addition, teaching-learning time was found not to be utilized efficiently in primary schools. The factors that may be responsible for inefficiencies include: ineffective education policies and management processes, misallocation of resources to various educational levels; school based factors such as teachers attitudes, time utilization, school environment; and household based factors such as poverty. The inefficiencies identified above lead one to question the quality and relevance of education in Kenya. While Abagi and Odipo (1997) discuss inefficiency from the point of view of mismatch between resource inputs and desired output in terms of qualified graduates able to contribute to national development, this study explored another dimension of inefficiency which involves discrepanc y between pedagogical approaches and the achievement of educational goals and objectives. The focus therefore was on the process of teaching and learning in addition to other material resources employed to pursue the achievement of educational goals and objectives. 2.3.0 Global responsiveness to contemporary issues Shah, (1997) discusses the implications of globalization in the 21st century which include information revolution, dynamic demand of relevant skills, uncertainty of a borderless world economy, and intense competition among others. He suggests that responsiveness to this reality demands a paradigm shift in the management of human relationships in all spheres of life. Although he takes a political perspective and dwells on how a shift in modes of governance needs to be effected, this study adapts Shahs ideas to a globalized educational context. For instance the shift from management to leadership in political governance can be equated to the shift from authoritarian teaching to facilitative teaching in education. In both cases, control (of citizens/learners by politicians/teachers) is discouraged while participation, consultation and involvement by all parties is encouraged. This change is so fundamental that Shah refers to it as a cultural transformation. He describes it as follows: The culture of governance is also slowing changing from a bureaucratic to a participatory mode of operation; from command and control to accountability for results; from being internally dependent to being competitive and innovative; from being closed and slow to being open and quick; and from that of intolerance from risk to allowing freedom to fail or succeed (Shah, 1997) While responsiveness as described above is desirable and even necessary in the world today, it is not easy to accomplish. In developing countries for instance, the reform of the public sector has been attempted in many countries without tangible results. Shah (1997) regards attempts to reform the public sector in many developing countries as an illusion or dream. The command and control orientation is so entrenched that developing a client orientation that emphasizes collaboration and service is difficult to achieve. Consequently human relationships are devoid of a sense of responsibility and mutual respect. Relating these ideas to primary teacher education, responsiveness would mean enabling learners to be self-reliant and self-disciplined as well as actively engaging both teachers and learners as partners in the learning process. Leithwood et al (1994) assert that modern education systems have to address broader and complex goals, use a diversity of forms of instruction and strategies for learning to cater for diverse clientele as opposed to the traditional youthful learners. In addition, educational institutions will need technical resources to facilitate independent decision making and collaboration with other agencies. Future institutions of education will need greater decentralization of authority, empowerment of staff, and increased accountability to the stake holders they serve. Much effort will be employed to review the content of the curriculum and the process of instruction in order to enhance learning as well as forge useful links between the educational institutions and their environment. Specifically the following responses will be needed: provision of higher order thinking skills like analysis, evaluation and creativity. use of flexible client-centred forms of instruction and willingness to collaborate with other educational agencies. Addressing of cultural, religious and ethnic diversity and tensions and Increasing respect for the rights of individual, ability, race, age, sex etc which enhances equity as an education goal as well as equal access to knowledge. Beyer (1997) asserts that the best way to respond to contemporary challenges is to empower learners to learn. This implies facilitation of efficient and effective thinking. Such learning involves careful development of learner inclinations as well as their abilities to think skilfully. Barrow et al. (2006) reports that a research carried out in India suggest that one way of facilitating responsiveness among learners is to create a learning environment in which children are motivated to participate actively and are encouraged to think beyond their own context. This concurs with Beamons (1997) and Beyers (1997) view that the classroom environment should be motivating and provide opportunities for learners to use their cognitive and creative abilities. In order to develop the thinking skills of learners, the teachers need to use cognitive instruction approaches and learner-centred strategies which stimulate thinking. This study explored such approaches and strategies and endeavoured to find out whether they were being used in PTE in Kenya. Kea et al. (2006) recommend multicultural teacher education preparation as a viable way of promoting responsiveness to the challenges of the 21st century. Teachers who have learned culturally responsive pedagogy are believed to be more confident and effective in instructing children from diverse social, cultural, religious and economic backgrounds. Culturally responsive pedagogy involves adapting the content of instruction and teaching styles, curriculum, methodology, and instructional materials responsive to students values and cultural norms. Thus, the ultimate challenge for teacher educators is to prepare reflective practitioners who can connect, commit, and practice an ethos of care with diverse students and their families. According to Douglass (2006), responsiveness to contemporary challenges requires cross-cultural and human rights education. Governments should ensure that their primary and secondary educational systems provide for a balance and integration of national history and identity formation with knowledge of other cultures, religions, and regions. Educational approaches should be suitable for transforming the values of young people, their perceptions and knowledge about other civilizations, cultures and peoples across all regions. Critical thinking should be promoted in order to enhance fair-mindedness and objectivity in a world where information is being generated and disseminated at an amazing rate. Critical thinking is essential for analysing, evaluating and applying information. It is useful in combating misperceptions, prejudices, inaccuracies and outright lies among others. Critical thinking enhances the learners ability to separate fact from opinion, to evaluate information for bias, to construct and deconstruct meaning logically and relevantly. Such skills are important for promoting tolerance, mutual respect and responsible citizenship. This study examines the responsive approaches discussed above in the light of the Kenyan context. It aims at developing and recommending a comprehensive proposal of responsive pedagogies for teacher education that are suitable for addressing contemporary issues in Kenya. According to NCPPHE (2008) the most important educational goal is to facilitate learning by students and the society in general and thus create a learning society. In such a society, learning is a way of life and is therefore life-long. Optimized learning is that which helps strengthen democratic and civic institutions in the nation. This concept of learning extends beyond the education of students in classrooms to include educations impact on societal organizations, businesses, corporations, and cultures. This view is also adopted by Partnership for the 21st century (2004), which is a conglomeration of public and private partnership focusing on improving education in the 21st century. The partnership reached out to hundreds of educators, business leaders and employers to determine a vision for learning in the 21st century, to reach a consensus on the definition of 21st century skills, knowledge and expertise which will enable learners to thrive in contemporary world. This work endea voured to bridge the gap between the knowledge and skills most students learn today in school and the knowledge and skills required in the 21st century communities and workplaces. The critical knowledge and skills identified by Partnership for the 21st century (2004) are learning and innovation skills (creativity and innovation, Critical thinking and problem solving, communication and collaboration), information, media and technology skills (information literacy, media literacy and ICT literacy) as well as life and career skills (flexibility and adaptability, initiative and self-direction, social and cross-cultural skills, productivity and accountability, leadership and responsibility). The partnership also identified interdisciplinary themes which summarises the content that students should learn in the 21st century. These themes include global awareness, civic literacy, health literacy and financial, economic, business and entrepreneurial literacy. This study used these themes, knowledge and skills to evaluate the success of teacher education in Kenya in equipping primary school teachers to prepare pupils for the demands of the 21st century. 2.3.1 Responsiveness to contemporary issues in Africa According to Barrow et al. (2006) research carried out in Nigeria indicated that responsive approaches in education in Africa should aim at fostering moral values such as patience, tolerance, friendliness, compassion, empathy and fair-mindedness. In addition teaching methods and strategies should create and sustain a stimulating learning environment. Teachers need to use a variety of teaching methods to reach students at different levels of ability. In Namibia, the research recommended teachers reflection on their practice in order to seek way of improving teaching and learning. Learner-centred teaching, variation of teaching methods and strategies and positive teacher attributes like kindness and courtesy were also argued to contribute to responsiveness to contemporary issues in education. Ethiopian National Agency for UNESCO (2001) describes responsive education from the Ethiopian perspective as characterized by access to basic education for all, production of responsible citizens who can solve problems and cooperate with others in productive socio-economic activities. Other characteristics of responsive education include equity, community participation in education, and suitability of academic disciplines to the needs of the country. Quality and quantity need to accompany each other in educational matters; for instance, new educational institutions should be established and the existing ones strengthened in order to produce professionals at a quantity and quality levels that match the requirements of the country. 2.3.2 Responsiveness to contemporary issues in Kenya In Kenya, attempts at making education responsive to contemporary issues and challenges is reflected in the various educational reports that have been generated since independence (Republic of Kenya, 1964, 1976, 1981, 1988 and 1999). This is evident in the review of the said reports below. The Ominde Report (Republic of Kenya, 1964) was the first commission of independent Kenya to address matters of education. It made a lasting contribution by articulating, among other issues, the goals of education in Kenya which are reinforced in the commission reports that followed. Among the recommendations that the report gave included that: partnership should be forged between government and regional and local authorities in the planning and administration of education; educational policy should be consciously directed towards promoting national unity; religious instruction should be handled as an academic subject on educational lines dissociated from the sectarian objectives of any religious group; primary education should provide training in the rudiments of citizenship; education should be responsive to contemporary needs and modern educational practice; education should be child-cantered and child study experts should be included as lecturers in teacher training colleges. The Ominde Report (Republic of Kenya, 1964) identified a number of issues of which the following are relevant to this study and remain relevant to the current situation in Kenya: Education in Kenya should foster as sense of nationhood and promote nationhood. The post 2007 election violence revealed the fact that nationhood and national unity have not yet been successfully achieved in Kenya(Republic of Kenya, 2008a) and education, to be relevant must develop among learners and teachers alike patriotism, tolerance and mutual social responsibility. Education should serve the people and the needs of Kenya without discrimination. It should develop in learners the traits of fair-mindedness, empathy and justice. Education should enable learners at all levels to adapt to change. This requires analytical and evaluative skills that enable one to examine ones context, identify and define problems clearly and accurately. It also requires creative skills to enable one to formulate and implement relevant solutions to problems. The teaching methods prevalent in educational institutions after independence were faulted by the Ominde Report (Republic of Kenya, 1964) because they neglected learner participation, imagination and understanding and emphasised drilling and authoritarian teaching. Little attempt was made to adapt instruction to the needs of learners. In teacher training colleges, few lecturers were well grounded on the crucial question of how children learn or fail to learn. The report appealed for a paradigm shift in teaching and learning as indicated in the following: We do not believe that the students will effectively break loose from the old bookish, rote methods until they have themselves shared in the exhilaration of autonomous learning and have discovered how much more complete is their final mastery (Republic of Kenya, 1964, p.116) Gachathi Report (Republic of Kenya, 1976) expounded on the implications of the ideology of African Socialism on education in Kenya. The Report recommended the following issues which remain relevant to the contemporary Kenyan context: Education needs to continue promoting national unity in order to address social and economic challenges facing the country; education should be a tool for removing social and regional inequalities and creating international consciousness; education should enhance adaptability and management of change; education should foster mutual responsibility and cooperation and education should develop positive attitudes and values that motivate people to serve diligently, honestly and efficiently. The Report recommended that all educational institutions should give increasing emphasis on problem-solving teaching methods that have a bearing on the real life situation of the Kenya environment. This study examines the extent to which PTE has addressed the concerns and recommendation of the Gachathi report. The Mackay Report (Republic of Kenya, 1981) was mainly concerned with the establishment of the second university in Kenya. It recommended the establishment of the 8-4-4 system of education. It reiterated the importance of the following objectives of education in Kenya: fostering national unity based on the adaptations of the rich cultural heritage of the Kenyan people; facilitation of the needs of national development; development of skills, knowledge attitudes, talents and personalities of learners; fostering positive international consciousness and promotion of social justice and morality. The Report observed that formal education had tended to concentrate on imparting knowledge for the sake of passing examinations instead of facilitating problem solving. Kamunge Report (Republic of Kenya, 1988) addressed education and manpower training for the rapidly changing Kenyan society. It recommended