Washington University in St. Foundation of Finance (FIN 538) Master of Finance Program
Summer 2021
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FINAL EXAM (Duration: 120 min)
1. (30 points) Let Xi, i = 1, 2 be independent random variables that have the following distri- bution: Xi = 1 with probability 43 and Xi = 0 with probability 14 :
B2 = X1 + X2
Denote an upward move of Xi by H and a downward move by L. Consider the following sample space for the above random variables:
Ω = {HH,HL,LH,LL}
Let Fi, i = 1, 2 be the natural filtration associated with the process Bi.
(a) (15 points) Describe Z1 = E[B2|B1] as a map from Ω into the set of real numbers R. Is the process Bi, i = 1, 2 a martingale? Why or why not?
Answer. We have
Z1(HH) = E[B2|B1](HH) = E[B2|B1 = 1] = E[B2|{HH, HL}]
= 1 + 43 ∗ 1 + 14 ∗ 0
Z1(HL) = Z1(HH) = 74
Z1(LH) = E[B2|B1](LH) = E[B2|B1 = 0] = E[B2|{LH, LL}] = 0 + 34 ∗ 1 + 14 ∗ 0
Z1(LL) = Z1(LH) = 34
Bi, i ∈ {1, 2} is not a martingale because B1 ̸= E[B2|B1].
(b) (15 points) Suppose you observed that B2 = 1 but did not observe either X1 or X2. What are the expected values of X1 and X2 given this information?
Answer. We have
E[X1|B2 = 1] = E[X1|{HL, LH}]
= P rob(HL|{HL, LH}X1(HL) + P rob(LH|{HL, LH}X1(LH)
= 21 ∗ 1 + 12 ∗ 0 = 12
Similarly,
E[X2|B2 = 1] = E[X2|{HL, LH}]
= P rob(HL|{HL, LH}X2(HL) + P rob(LH|{HL, LH}X2(LH) = 21 ∗ 0 + 12 ∗ 1
= 12 2. (30 points in total):
Similarly,
(a) (10 points) Find a such that the process Xt = e−t+aBt is a martingale where Bt is a standard Brownian motion. Answer. By Ito’s lemma:
a2 dXt =Xt (−1+ 2)dt+adBt
√√ Xt isamartingaleifandonlyifthedrifttermiszero: a= 2ora=− 2.
(b) (10 points) Calculate the (unconditional) expectation E[(Bs + Bt)Bu], for s ≤ t ≤ u where Bt is a standard Brownian motion.
Answer. We have E[(Bs + Bt)Bu] = E[BsBu] + E[BtBu], where
Similarly,
E[BsBu] = E[Bs(Bu − Bs) + Bs2]
= E[Bs(Bu − Bs)] + E[Bs2]
= E[Bs]E[(Bu − Bs] + s =s
E[BtBu] = E[Bt(Bu − Bt) + Bt2]
= E[Bt(Bu − Bt)] + E[Bt2]
= E[Bt]E[(Bu − Bt] + t =t
E[(Bs +Bt)Bu]=s+t
(c) (10 points) Apply Ito’s lemma to calculate dX for X = B2e t B−2ds where B is a stan-
dard Brownian motion. Calculate the instantaneous expected rate of return dt . Answer. By the product rule:
tt0st0st0s
dX = d(B2)e t B−2ds + B2d(e t B−2ds) + d(B2)d(e t B−2ds) (1)
d(Bt2) = 2BtdBt + dt SetY =tB−2ds,thendY =B−2dt.Thus
d(e t B−2ds) = deYt = B−2eYt dt = B−2e t B−2dsdt
Plug these into (1) we get
dX = (2B dB + dt)e t B−2dsdt + B2B−2e t B−2dsdt + (2B dB + dt)B−2e t B−2dsdt
ttt0stt0sttt0s = (2B dB + 2dt)e t B−2dsdt
tt 0s which implies
dXt = 2dBt+ 2dt Bt2
1dXt 2 2
dtEt X =Et[BdBt+B2dt]=
3. (30 points) Consider the price of a security that follows the process in the figure below. At each date t = 0 and t = 1, the price jumps up with a risk-neutral probability of q and jumps down with a risk-neutral probability of 1 − q. The risk-free rate is the same across the two periods and the states.
(a) (15 points) Find the risk-neutral probability q, the risk free rate and the security price at t = 0.
Answer. Let r denote the risk free rate. We have from the second period prices: 4 =6q+1−q=5q+1
Solving the above yields
r=0 q = 53
The security’s price at t = 0 is
The date-0 price of this call option
S0 = 3 ∗ 4 + 2 ∗ 3 = 18 555
(b) (15 points) Find the date-0 price of a call option that matures at the end of the second period and that has a strike price of 4. The call option’s payoffs are:
If the state is
upup:6−4=2
updown : max(1 − 4, 0) = 0 downup:5−4=1 downdown : max(0 − 4, 0) = 0
is therefore:
∗ 2 + 2 ∗ 3 ∗ 1 = 24 55 25
4. (10 points) Calculate V ar t eBs dB where B is a standard Brownian motion. 0st
Answer. We have
t t 2 t 2 t
V ar eBs dBs = E[ eBs dBs ] − E[ eBs dBs] = E[ 0000
Consider a partition {t0 = 0,t1,··· ,tn = t} of the interval [0,t]. We have
]= lim E[ eBti−1(Bti −Bti−1) ] n→∞ i=1
t2n 2
lim E[ e2Bti−1 (Bti − Bti−1 )2 + 2 eBti−1 (Bti − Bti−1 )eBtj−1 (Btj − Btj−1 )]
i=1 1≤i