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### Solution to Homework Three

Course: MATH A, Spring 2011
School: HKU
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Word Count: 473

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5030 Quantitative MAFS Modeling of Derivatives Securities Solution to Homework Three Course Instructor: Prof. Y.K. Kwok 1. F is generated by the partition P = {{3, 2}, {1, 1}, {2, 3}}. (i) Since {2, 3} P and X (2) = 4 = X (3) = 9, X is not F -measurable. (ii) Since {2, 3} P and X (2) = 2 = X (3) = 3, X is not F -measurable. Dene the random variable X ( ) = max(, 3). Now, X ( ) = 3 for all , hence X is F...

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5030 Quantitative MAFS Modeling of Derivatives Securities Solution to Homework Three Course Instructor: Prof. Y.K. Kwok 1. F is generated by the partition P = {{3, 2}, {1, 1}, {2, 3}}. (i) Since {2, 3} P and X (2) = 4 = X (3) = 9, X is not F -measurable. (ii) Since {2, 3} P and X (2) = 2 = X (3) = 3, X is not F -measurable. Dene the random variable X ( ) = max(, 3). Now, X ( ) = 3 for all , hence X is F -measurable. 2. (a) Suppose F is generated by a partition P . It suces to show that this property is valid for every B P . Consider E [IB E [X |F ]] = E [ X | B ] P ( ) B = E [ X | B ] P (B ) = X ( )(P ( )/P (B ))P (B ) B = X ( )P ( ) B = E [XIB ]. J E [X |Bj ]1Bj , and consider (b) Recall E [X |F ] = j =1 E [max(X1 , , Xn )|F ] J E [max(X1 , , Xn )|Bj ]1Bj = j =1 J Kj max(X1 (k,j ), , Xn (k,j ))P (k,j )1Bj = j =1 k=1 while max(E [X1 |F ], , E [Xn |F ]) J = max Kj J Kj X1 (k,j )P (k,j )1Bj , , j =1 k=1 j =1 k=1 1 Xn (k,j )P (k,j )1Bj . It is obviously that the maximum value among the various sums of X (k,j ), = 1, , n, cannot be greater than the value obtained by taking the maximum value among X1 (k,j ), , Xn (k,j ) and performing the summation afterward. Hence, we obtain the desired The result. 3. property E [Xt+1 Xt |Ft] = 0, t = 0, 1, . . . , T 1 does imply that X is a martingale because T 1 E [XT |Ft] = E [Xs+1 Xs |Ft ] + E [Xt |Ft] s=t T 1 = E [E [Xs+1 Xs |Fs ]|Ft] + Xt s=t = Xt , t = 0, 1, . . . , T 1. 4. Note that Nk+1 Nk is independent of Fk since the successive binomial trials are independent and p = probability of success in the (k + 1)th trial, we have E [Nk+1 (k + 1)p (Nk kp)|Fk ] = E [Nk+1 Nk p|Fk ] = E [Nk+1 Nk p] = 0. Then from Problem 3, we deduce that Yk is a martingale. 5. Consider a portfolio consisting of 4 units of money market account with interest rate r = 0.3 and shorting one unit of asset, then we have V (0) = 4 S (0) = 0, V (2; i ) = 4(1 + r)2 S (2; i ) 0.76, i = 1, 2, 3, 4. Hence this is an arbitrage opportunity. 6. We dene k to be the largest non-negative integer such that uk dnk S X , that is, ln X k lnSdn . It is seen that u d X uj dnj S when j k 0 when j > k . max(X uj dnj S, 0) = The price formula for a European put option with terminal payo max(X S, 0) for the n-period binomial model is given by p= n j =0 n Cj pj (1 p)nj max(X uj dnj S, 0) Rn k n k n Cj pj (1 = XR p) nj j =0 where p = n Cj pj (1 p)nj S j =0 up . R 2 uj dnj , Rn
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