Download NAME: T/F 2.12 2.13 MC

NAME:
M339W/389W Financial Mathematics for Actuarial Applications
University of Texas at Austin
In-Term Exam II
Instructor: Milica Čudina
Notes: This is a closed book and closed notes exam. The maximum number of points on this
exam is 100.
Time: 50 minutes
TRUE/FALSE
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2.14 (5)
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3 (2)
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2.15 (5)
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2.16 (5)
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2.17 (5)
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2.18 (5)
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MULTIPLE CHOICE
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2.19 (5)
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2.20 (5)
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9 (2)
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2.21 (5)
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10 (2)
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2.22 (5)
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FOR GRADER’S USE ONLY:
T/F
2.12
2.13
M.C.
Σ
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2.1. TRUE/FALSE QUESTIONS. Please note your answers on the front page.
Problem 2.1. (2 pts) In the setting of the Black-Scholes stock-price model, let S = {S(t), t ≥ 0}
denote the stock price whose drift is α and volatility is σ. Define the new stochastic process
X(t) = ln(S(t)), for every t ≥ 0.
Then,
E[X(t + h) | X(t)] = X(t), for every t ≥ 0 and h > 0.
Solution: FALSE
Since {X(t)} is an (arithmetic) Brownian motion, its increment X(t + h) − X(t) is independent
from X(t) and it is normally distributed with mean αh and variance σ 2 h. So,
E[X(t + h) | X(t)] = E[X(t + h) − X(t) | X(t)] + X(t) = αh + X(t), for every t ≥ 0 and h > 0.
Problem 2.2. (2 pts) Let the movement of a stock price follow a geometric Brownian motion.
The, the realized (rate of) return on that stock follows the normal distribution.
Solution: TRUE
Problem 2.3. (2 pts) The total variation of a standard Brownian motion on the interval from 0
to T is infinite.
Solution: TRUE
Problem 2.4. (2 pts) In the setting of the Black-Scholes stock-price model, let {S(t), t ≥ 0}
denote the price of a non-dividend-paying stock. Then,
1
S(t) = S(0) exp{(α − σ 2 )t + σZ(t)}
2
for every t ≥ 0 and where Z denotes the “driving” standard Brownian motion under the “true”
probability measure P (consistent with the mean rate of return α).
Solution: TRUE
Problem 2.5. (2 points) Asset S pays no dividends and satisfies the following SDE
dS(t) = S(t)(µS dt + σ dZ̃t ), S(0) = s0
where Z̃ is a standard Brownian motion under the risk-neutral probability measure P∗ and with
µS , σ, s0 > 0 constant. Assume that the continuously compouned, risk-free interest rate equals
r = 0.05. Then, µS < 0.05. True or false?
Solution: FALSE
In fact, µS = r.
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Problem 2.6. (2 points) In the Black-Scholes stock-price model, the value function of the
K−strike European put option must satisfy the following terminal condition:
v(T, s) = max(s, K).
Solution: FALSE
It should be
v(T, s) = max(0, K − s).
Problem 2.7. (2 pts) The Black-Scholes option pricing formula can always be used for pricing
both European and American type options.
Solution: FALSE
Problem 2.8. (2 pts) Let the exchange rate for GBP to USD be modeled by a geometric Brownian
motion. Then, the exchange rate from the USD to GBP is also a geometric Brownian motion.
Solution: TRUE
Problem 2.9. (2 pts) An investor uses a proportional investment strategy to build his
portfolio. More precisely, he always invests a fixed proportion of his wealth W in the risky asset
and the remainder at the risk-free interest rate.
Assume that the price of the risky asset is modeled by a geometric Brownian motion. Then,
the investor’s wealth process is a geometric Brownian motion itself.
Solution: TRUE
Problem 2.10. (2 pts) Let Z = {Z(t), t ≥ 0} be a standard Brownian motion. Define the process
Y as
Z t
3
Y (t) = (Z(t)) − 3
Z(s) ds.
0
Then, the process Y has zero drift.
Solution: TRUE
dY (t) = 3(Z(t))2 dZ(t) + 3Z(t)dt − 3Z(t) dt = 3(Z(t))2 dZ(t).
Problem 2.11. (2 pts) Let the process S = {S(t), t ≥ 0} satisfy the following SDE
dS(t) = S(t)[µ dt + σ dZ(t)]
where Z stands for a standard Brownian motion while µ and σ are constant. Then the process
S(t)e−µt has zero drift.
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Solution: TRUE
d[S(t)e−µt ] = −µe−µt S(t) dt + e−µt dS(t)
= −µe−µt S(t) dt + e−µt (S(t)[µ dt + σ dZ(t)]) = e−µt S(t)σ dZ(t).
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2.2. FREE-RESPONSE PROBLEMS. Please, explain carefully all your statements and assumptions. Numerical results or single-word answers without an explanation (even if they’re
correct) are worth 0 points.
Problem 2.12. (15 points) Let S = {S(t), t ≥ 0} denote the stock-price process following a
geometric Brownian motion. The mean stock price at time−1 equals 120 and the median stock
price 115. What is the probability that the time−1 stock price exceeds 100?
Solution: Since S = {S(t), t ≥ 0} is the stock-price process modeled by a geometric Brownian
motion, the stock price at time−1 is lognormally distributed. In fact, using our usual parameters,
we can rewrite it as
1
S(1) = S(0)e(α−δ− 2 σ
(α−δ− 21 σ 2 )
Recall that the median of S(1) equals S(0)e
as
P[S(1) > 100] = P[115e
σZ(1)
2 )+σZ(1)
.
. So, the required probability can be expressed
100
1
> 100] = P Z(1) > ln
σ
115
1
115
115
1
= P Z(1) < ln
ln
=N
.
σ
100
σ
100
Since the mean of S(1) equals S(0)e(α−δ) , we have
p
1 2
120
⇒ σ = 2 ln(1.04348) = 0.2918.
e2σ =
115
So, our final answer is
P[S(1) > 100] = N (0.48) = 0.6844.
The required probability is:
P[S(1) > 100] =
6
Problem 2.13. (20 points) Let Z = {Z(t), t ≥ 0} be a standard Brownian motion. According to
Itô’s lemma, it is possible to express the product Y (t) = et sin(Z(t)) as a sum of a stochastic and
a classical integral. Please, provide this representation.
More precisely, according to Itô’s lemma, we can write
Z t
Z t
σ(s) dZ(s),
µ(s) ds +
Y (t) = Y (0) +
0
0
for every t ≥ 0, and for appropriate stochastic processes µ and σ. Please, determine µ and σ as
functions of the time variables t and the standard Brownian motion Z.
We get:
µ(s) =
and
σ(s) =
Solution: Set F (t, x) = et sin(x). Then Ft (t, x) = et sin(x), Fx (t, x) = et cos(x), and Fxx (t, x) =
−et sin(x). So, by Itô’s lemma,
Z t
Z t
1 s
s
Y (t) =
(e sin(Z(s)) − e sin(Z(s))) ds +
es cos(Z(s)) dZ(s)
2
0
0
Z t
Z t
1
=
es sin(Z(s)) ds +
es cos(Z(s)) dZ(s) .
2 0
0
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2.3. MULTIPLE CHOICE QUESTIONS. Please note your asnwers on the front page.
Problem 2.14. (5 points) Let {Z(t), t ≥ 0} be a standard Brownian motion. Define the stochastic
process
Y (t) = tZ(t).
Then,
(a)
(b)
(c)
(d)
(e)
Rt
Rt
Y (t) = 0 Z(s) ds + 0 Z(s) dZ(s)
Rt
Rt
Y (t) = 0 Z(s) ds + 0 s dZ(s)
Rt
Y (t) = 0 sZ(s) ds
Rt
Y (t) = 0 sZ(s) dZ(s)
None of the above.
Solution: (b)
Problem 2.15. (5 pts) The current price of a share of stock is given to be S(0) = 45. The stock
is assumed to follow a geometric Brownian motion. Let σ = 0.4. The stock is projected to pay a
single $5.00 dividend in 1 month.
Assume that r = 0.08.
Consider an at-the-money call on that stock with expiry in 3 months. Find the price VC (0) of
this call.
(a) 1.233
(b) 1.255
(c) 1.697
(d) 3.013
(e) None of the above.
Solution: (c)
The prepaid forward price on this stock is
P
F0,T
(S) = S(0) − De−rTD = 45 − 5e−
0.08
12
= 40.0332. ≈
P
VC (0) = F0,T
(S)N (d1 ) − Ke−rT N (d2 )
with
1
1
P
d1 = √ [ln(F0,T
(S)/K) + (r + σ 2 )T ] = · · · = −0.38,
2
σ T
√
d2 = d1 − σ T = −0.58.
We get that the price is
VC (0) = 40.0332N (−0.38) − 45e−0.08/4 N (−0.58) = 40.0332(1 − 0.6480) − 45e−0.02 (1 − 0.719) = 1.697.
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Problem 2.16. (5 points) The current price of a continuous-dividend-paying stock is given to be
$92. The stock’s volatility is 0.35 and its dividend yield is 0.02.
The continuously-compounded, risk-free interest rate is 0.05.
Consider a $90-strike European call option on the above stock with exercise date in a quarteryear.
What is the Black-Scholes price of this call option?
(a) 5.05
(b) 7.66
(c) 7.71
(d) 7.89
(e) None of the above.
Solution: (d)
d1 = 0.26, d2 = 0.08.
So,
VC (0) = 92e−0.02/4 × 0.6026 − 90e−0.05/4 × 0.5319 ≈ 7.89.
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Problem 2.17. (5 points) Consider a non-dividend-paying stock whose price S = {S(t), t ≥ 0}
is modeled using a geometric Brownian motion. Suppose that the current stock price equals $100
and that its volatility is given to be 0.25.
The continuously compounded, risk-free interest rate is assumed to equal 0.04.
Consider a derivative security which entitles its owner to obtain a European call option on the
above stock six months from today, i.e., at time t∗ = 1/2. The call option is to be 3−month
to expiration at time of delivery and have the strike equal to 105% of the time−t∗ price of the
underlying asset. This contract is called a forward start option.
What is the price of the forward start option?
(a) 3.13
(b) 8.65
(c) 10.51
(d) 13.55
(e) None of the above.
Solution: (a) or (e)
At time t∗, the required Black-Scholes price of the call option equals
∗
VC (t∗ ) = S(t∗ )N (d1 ) − 1.05S(t∗ )e−r(T −t ) N (d2 )
= S(t∗ )(N (d1 ) − 1.05e−0.01 N (d2 ))
with
1
0.252
1
d1 =
− ln(1.05) + 0.04 −
×
= −0.2478,
0.125
2
4
√
d2 = d1 − σ T − t∗ = −0.2478 − 0.125 = −0.3728.
So, N (d1 ) = 1 − N (0.25) = 1 − 0.5987 = 0.4013 and N (d2 ) = 1 − N (0.37) = 1 − 0.6443 = 0.3557
and Hence,
VC (t∗ ) = S(t∗ )(0.4013 − 1.05e−0.01 × 0.3557) = S(t∗ )0.31531.
So, one would need to buy exactly 0.031531 shares of stock to be able to buy the call option in
question at time−t∗ . This amount of shares costs $3.1531.
Problem 2.18. (5 points) A pair of stock prices is modeled using the following system of SDE:
dS(t)
= c(0.01 dt + 0.1 dZ(t))
S(t)
dQ(t)
= 0.03 dt + 0.2 dZ(t)
Q(t)
with Z a standard Brownian motion.
The asset S does not pay dividends, while the asset Q is assumed to pay dividends continuously
with the dividend yield equal to 0.02.
The continuously-compounded, risk-free interest rate equals 0.04. What is the value of the
constant c?
(a) 3
10
(b)
(c)
(d)
(e)
8
10
13
None of the above.
Solution: (b)
The Sharpe ratios of the two assets needs to be the same since they have a common source of
uncertainty Z. We have
0.01c − 0.04
0.05 − 0.04
c−4
5−4
1
=
⇒
=
=
⇒ 2(c − 4) = c ⇒ c = 8.
0.1c
0.2
c
2
2
Problem 2.19. We use the Black-Scholes model for the evolution of the stock price S. The current
stock price is S(0) = $60, and its volatility given to be σ = 0.20 The stock pays no dividends.
The continuously-compounded, risk-free interest rate equals 0.06.
What is the price of a two-month, $60-strike put option on the above stock?
(a) 1.62
(b) 2.55
(c) 2.84
(d) 3.37
(e) None of the above.
Solution: (a)
In our usual notation, the price is
VP (0) = Ke−r·T N (−d2 ) − S(0)N (−d1 )
with
√
60
(0.2)2
1
ln
+ 0.06 +
= (0.3 + 0.1)/ 6 = 0.16,
60
2
6
√
d2 = d1 − 0.2/ 6 = 0.08.
1
p
d1 =
0.2 1/6
The standard normal table gives us
N (d1 ) = 0.5636
⇒
N (−d1 ) = 0.4364,
N (d2 ) = 0.5319
⇒
N (−d2 ) = 0.4681.
So,
VP (0) = 60 e−0.01 × 0.4681 − 0.4364 = 1.6225.
Problem 2.20. Alice wagers to deliver one share of stock S to Bob in one year if the stock price at
that time exceeds $70.00. Assume that the current price of the said stock equals S(0) = 60.00, and
that its volatility parameter is given to be 0.25. Moreover, the stock pays dividends continuously
with the dividend yield of 0.01. The continuously-compounded, risk-free interest rate equals 0.04.
What is the Black-Scholes price of Alice’s bet?
(a) 16.72
(b) 21.13
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(c) 38.27
(d) 39.51
(e) None of the above.
Solution: (b)
We recognize that Alice’s bet is actually a written asset call. So, its price is the Black-Scholes
price of an asset call, i.e., in our usual notation
VAC (0) = S(0)e−δT N (d1 )
with
1
d1 =
0.25
60
1
2
ln
+ 0.04 − 0.01 + (0.25) = −0.37.
70
2
So, our answer is
VAC (0) = 60e−0.01 N (−0.37) = 60e−0.01 (1 − 0.6443) ≈ 21.13.
Problem 2.21. The stochastic process S = {S(t), t ≥ 0} satisfies the stochastic differential
equation
dS(t) = S(t)(α dt + σ dZ(t))
with α and σ constant and Z = {Z(t), t ≥ 0} a standard Brownian motion. Define the stochastic
process Y as
p
Y (t) = S(t)
for every t ≥ 0. Then, V ar[ln(Y (t)] equals . . .
2
(a) σ4 t
(b) 2σ 2 t
2
(c) σ
√t
(d) σt
(e) None of the above.
Solution: (a)
Problem 2.22. Source: Sample MFE Problem #16.
Assume that the Black-Scholes framework holds. Let S = {S(t), t ≥ 0} denote the price of a nondividend-paying stock. The stocks volatility is given to be 20%. The continuously compounded
risk-free interest rate is 4%.
You are interested in contingent claims with payoff being the stock price raised to some power.
For 0 ≤ t ≤ T, consider the equation
P
F0,T
[Sa ] = [S(t)]a
where the left-hand side stands for the prepaid forward price at time t of a contingent claim that
pays the amount (S(T ))a at time T . An obvious solution to the above equation is a = 1. It
corresponds to the stock itself. The other solution to the above equation is . . .
(a) −4
12
(b) −2
(c) 0
(d) 2
(e) None of the above.
Hint: You might find it handy to know that
P
F0,T
[Sa ] = [S(t)]a e(r(a−1)+a(a−1)σ
Solution: (b)
2 /2)(T −t)