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SOLUTIONS MOCK EXAM 2021
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 (wA 2 sigA 2+wB 2 s igB 2+(1 wA wB) 2 s igC 2+ + 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 (wA 2 sigA 2+wB 2 s igB 2+(1 wA wB) 2 s igC 2+ + 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 u ( j 1) d ( ( 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 ) ) + } > Q

M1< function ( r , m, de l t a t , u , d){ + return (((1+ r/m) (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) ( 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) (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) ( 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.