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SOLUTIONS MOCK EXAM
Question1: (a) See Exercise 2 Tutorial 1. (b) > v1 <−runif (6 ,−5 ,5) > v1 [ 1 ] −3.534955 −2.125690 −3.197416 −2.690970 1.171644 4.615299 > v2 <−runif (14 ,−5 ,5) > v2 [ 1 ] −1.19118968 0.99891338 1.94085672 −4.69556951 3.59148792 0.02149171 [ 7 ] −2.65920813 4.52136889 −0.60572579 1.77306420 −0.71442411 4.73246855 [ 1 3 ] −1.71294960 −1.27886442 > Marray<−array (c ( v1 , v2 ) ,dim = c ( 3 , 6 , 3 ) ) > Marray , , 1 [ , 1 ] [ , 2 ] [ , 3 ] [ , 4 ] [ , 5 ] [ , 6 ] [ 1 , ] −3.534955 −2.690970 −1.1911897 −4.69556951 −2.6592081 1.7730642 [ 2 , ] −2.125690 1.171644 0.9989134 3.59148792 4.5213689 −0.7144241 [ 3 , ] −3.197416 4.615299 1.9408567 0.02149171 −0.6057258 4.7324685 , , 2 [ , 1 ] [ , 2 ] [ , 3 ] [ , 4 ] [ , 5 ] [ , 6 ] [ 1 , ] −1.712950 −2.125690 1.171644 0.9989134 3.59148792 4.5213689 [ 2 , ] −1.278864 −3.197416 4.615299 1.9408567 0.02149171 −0.6057258 1 2[ 3 , ] −3.534955 −2.690970 −1.191190 −4.6955695 −2.65920813 1.7730642 , , 3 [ , 1 ] [ , 2 ] [ , 3 ] [ , 4 ] [ , 5 ] [ , 6 ] [ 1 , ] −0.7144241 −1.278864 −3.197416 4.6152986 1.940857 0.02149171 [ 2 , ] 4 .7324685 −3.534955 −2.690970 −1.1911897 −4.695570 −2.65920813 [ 3 , ] −1.7129496 −2.125690 1.171644 0.9989134 3.591488 4.52136889 > Marray [ , 2 , 1 ] [ 1 ] −2.690970 1.171644 4.615299 > mean( v1 ) [ 1 ] −0.9603482 > print ( Marray [ , Marray [3 , ,3 ] >mean( v1 ) , 3 ] ) [ , 1 ] [ , 2 ] [ , 3 ] [ , 4 ] [ 1 , ] −3.197416 4.6152986 1.940857 0.02149171 [ 2 , ] −2.690970 −1.1911897 −4.695570 −2.65920813 [ 3 , ] 1 .171644 0.9989134 3.591488 4.52136889 Question2: (a) > rA<−0 .11 # expec ted re turn o f s t o c k A > rB<−0 .14# expec ted re turn o f s t o c k B > rC<−0 .085# expec ted re turn o f s t o c k C > sigA<−0 .30 # standard d e v i a t i on o f A > s igB<−0 .45 # standard d e v i a t i on o f B > s igC<−0 .30 # standard d e v i a t i on o f C > sigmaAB<− sigA∗s igB∗0 .3 # > sigmaAC<− sigA∗s igC∗0 .15 # covar iance between a s s e t s A and C > sigmaBC<− s igB∗s igC∗0 .45 # covar iance between a s s e t s B and C > rp<−function (wA,wB){ + return (wA∗rA+wB∗rB+(1−wA−wB)∗rC)} > rp (1/3 ,1/3) [ 1 ] 0 .1116667 > varp<−function (wA,wB){ 3+ return (wAˆ2∗sigAˆ2+wBˆ2∗s igBˆ2+(1−wA−wB)ˆ2∗s igCˆ2+ + 2∗wA∗wB∗sigmaAB+2∗wA∗(1−wA−wB)∗sigmaAC+2∗wB∗(1−wA−wB)∗sigmaBC) + } > sqrt ( varp (1/3 ,1/3) ) [ 1 ] 0 .2607681 > varp (1/3 ,1/3) [ 1 ] 0 .068 > sigmap<−sqrt ( varp ) > plot ( sigmap , rp , pch=” . ” ) (b) > r f<−0 .02 > amax<−max( ( rp−r f )/sigmap )# the p o s i t i v e tangency o f sharpe r a t i o > amax [ 1 ] 0 .3675796 > x<−seq (0 , 3 ,by=0.01) > l ines (x , amax∗x+r f ) 0.3 0.4 0.5 0.6 0.7 0.8 0. 05 0. 10 0. 15 0. 20 sigmap rp (c) 4> varp<−function (wA,wB){ + return (wAˆ2∗sigAˆ2+wBˆ2∗s igBˆ2+(1−wA−wB)ˆ2∗s igCˆ2+ + 2∗wA∗wB∗sigmaAB+2∗wA∗(1−wA−wB)∗sigmaAC+2∗wB∗(1−wA−wB)∗sigmaBC) + } > wA<−seq (0 , 1 , by=0.001) > WBn<−(1/ ( rB−rC ) )∗(0.12−rC−wA∗ ( rA−rC ) ) > Wamin<−match(min( varp (wA,WBn) ) , varp (wA,WBn) ) > wA[Wamin ] [ 1 ] 0 .56 > WB1<−(1/ ( rB−rC ) )∗(0.12−rC−wA[Wamin ] ∗ ( rA−rC ) ) > WB1 [ 1 ] 0 .3818182 > WC1<−1−wA[Wamin]−WB1 > WC1 [ 1 ] 0 .05818182 Question3: > bu i ld s tock t r e e<−function (S , u , d ,N){ + t r e e=matrix (0 ,nrow=N+1, ncol=N+1) + for ( i in 1 : (N+1)){ + for ( j in 1 : i ){ + t r e e [ i , j ]=S∗uˆ( j −1)∗d ˆ( ( i −1)−( j −1))}} + return ( t r e e )} > QM2<−function ( r , d e l t a t , u , d){ + return ( (exp( r∗de l t a t)−d)/ (u−d ) ) + } > QM1<−function ( r , m, de l t a t , u , d){ + return (((1+ r/m)ˆ(m∗de l t a t)−d)/ (u−d ) ) + } > Bin Op Pri2<−function ( t ree , d e l t a t , r1 , r2 , m, u , d , K){ + D1<−(1+r1/m)ˆ(−m∗de l t a t ) + D2<−exp(−r2∗de l t a t ) + Q1<−QM1( r1 ,m, de l t a t , u , d ) + Q2<−QM2( r2 , d e l t a t , u , d ) + Q<−c (Q1,Q2) + D<−c (D1 , D2) + opt ion1 t r e e=matrix (0 , nrow=nrow( t r e e ) , ncol=ncol ( t r e e ) ) + opt ion1 t r e e [nrow( opt ion1 t r e e ) , ]=pmax(2∗ (K−sqrt ( t r e e [nrow( t r e e ) , ] ) ) , 0) 5+ for ( i in (nrow( t r e e )−1):1){ + for ( j in 1 : i ){ + option1 t r e e [ i , j ]=max(max(2∗ (K−sqrt ( t r e e [ i , j ] ) ) , 0 ) , + (Q[ i ] ∗opt ion1 t r e e [ i +1, j +1]+(1−Q[ i ] ) ∗opt ion1 t r e e [ i +1, j ] ) ∗D[ i ] ) + } + } + return ( opt ion1 t r e e ) + } > Bin Op Pri va l<−function (S , T, r1 , r2 ,m, K, u , d ,N){ + Q<−c (QM1( r1 ,m, de l t a t=T/N, u , d ) ,QM2( r2 , d e l t a t=T/N, u , d ) ) + t r e e<−bu i ld s tock t r e e (S=S , u=u , d=d ,N=N) + opt ion<−Bin Op Pri2 ( t ree , d e l t a t=T/N, r1=r1 , r2=r2 , u=u ,m=m, d=d , K=K) + return ( l i s t (Q=Q, s tock=tree , opt ion=option , p r i c e=opt ion [ 1 , 1 ] ) ) + } > r e s u l t s 1<−Bin Op Pri va l (S=30, T=8/12 , r1 =0.05 , r2 =0.06 ,m=52, + u=1.15 ,d=0.8 ,K=6, N=2) > r e s u l t s 1 $Q [ 1 ] 0 .6194234 0.6291467 $ s tock [ , 1 ] [ , 2 ] [ , 3 ] [ 1 , ] 30 .0 0 .0 0 .000 [ 2 , ] 24 .0 34 .5 0 .000 [ 3 , ] 19 .2 27 .6 39 .675 $opt ion [ , 1 ] [ , 2 ] [ , 3 ] [ 1 , ] 1 .154789 0.0000000 0 [ 2 , ] 2 .202041 0.5426693 0 [ 3 , ] 3 .236439 1.4928596 0 $p r i c e [ 1 ] 1 .154789 Question4: (a) > S<−c (50 ,50∗1 .08 ,50∗ 0 . 88 ) #(S(0) , uS (0) , dS (0) ) 6> V<−c (max(S [2 ] −48 ,0) ,max(S [3 ] −48 ,0) ) > V [ 1 ] 6 0 > QM<−function ( r , m, de l t a t , u , d){ + return (((1+ r/m)ˆ(m∗de l t a t)−d)/ (u−d ) ) + } > Q<−QM( 0 . 1 , 1 2 , 3/ 1 2 , 1 . 0 8 , 0 . 8 8 ) > Q [ 1 ] 0 .7260446 > D1<−function ( r ,m, de l t a t ){(1+ r/m)ˆ(−m∗de l t a t )} > D<−D1( 0 . 1 , 12 , 3/12) > D [ 1 ] 0 .975411 > BinOpP<−(Q∗V[1]+(1−Q)∗V[ 2 ] ) ∗D > BinOpP [ 1 ] 4 .249151 > Del<−(V[1]−V[ 2 ] ) / (S [2]−S [ 3 ] ) # compute the d e l t a > Del [ 1 ] 0 . 6 (b) For the Hedging strategy, check similar argument as in Homework 5 question 3 (ii). (c) Assume that C(0) = £4 ¡ £4.2. Then, the value of the option is too cheap • At time t = 0, short sell 0.6 share and use the proceeds (0.6× 50 = £20) to buy a call option at £4 and invest the rest 30-4=£26 at the risk free interest rate 10%. • At time t = T = 3 months, there are two possible outcomes: : S1(T ) = uS(0) = £48. The call option is exercised. The option buyer collects 26(1+ 0.1 12 )3 ' £26.655 representing the future value of £26 invested at the risk free interest rate. She additionally short sale 0.4 stock at 54 × 0.4 = £21.6 and use the proceeds 26.655+21.6 = £48.255 to purchase one share at £48 and close the short position in one share by returning it to the owner. By following this strategy, there is a risk free profit of pounds 0.255. : S2(T ) = dS(0) = £44. The option is not exercised. Collect 26(1 + 0.1 12 )3 ' £26.655 representing the future value of £16.5 invested at the risk free interest rate. Buy 0.6 stock for 0.6 × 44 = £26.4 and close the short position in 0.6 stock by returning it to the owner. By following this strategy, there is a risk free profit of pounds 26.655− 26.4 ' £0.255.