MATH424 Midterm 2 sample problem solutions..

MATH424 Midterm 2 sample problem solutions..
(1) Let X have density equal to 4x3 if 0 ≤ x ≤ 1 and equal to 0 otherwise. Let Y = X 2 .
(a) Find density of Y.
(b) Compute EY and V Y.
3
∂y
Solution. (a) We have ∂x
= 2x, hence fY (y) = 4x
= 2x2 = 2y.
2x
R
R1
1
(b) EY = 0 2y × ydy = 23 , EY 2 = 0 2y × y 2 dy = 21 . Hence V Y = E(Y 2 ) − (EY )2 =
1
2
−
4
9
=
1
.
18
(2) Let S = X1 + X2 + · · · + X10000 where Xj are independent identically distributed so that each Xj has
exponential distribution with parameter 2.
(a) Compute ES and V S.
(b) Find approximately P (S > 5100).
Solution. (a) ES = 10000EX = 10000 × 21 = 5000. V S = 10000V X = 10000 × 41 = 2500. Thus
√
σS = 2500 + 50.
(b) By Central Limit Theorem
P (S > 5100) ≈ P (5000 + 50Z > 5100) = P (Z > 2) = 1 − P (Z < 2) ≈ 1 − 0.98 = 0.02.
(3) Let (X, Y ) have density equal to x2 y + 5xy 2 if 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1 and equal to zero otherwise.
(a) Comupte the marginal distribution of X.
(b) Find the conditional distribution of X given that Y = y.
R1
R1 2
2
2
(a) fX (x) = 0 (x2 y + 5xy 2 )dy = x2 + 5x
.
Likewise
f
(y)
=
(x y + 5xy 2 )dx = y3 + 5y2 .
Y
3
0
x2 y + 5xy 2
6(x2 y + 5xy 2 )
f (x, y)
=
=
.
(b) f (x|Y = y) =
fY (y)
(y/3) + (5y 2 )/2
2y + 15y 2
(4) Let (X1 , X2 ) be a normal random vector with mean vector (1, 1) and covariance matrix
4 2
2 3
.
(a) Find the conditional distribution of X1 given that X2 = x2 .
(b) Compute E(X12 X2 ).
Solution. Let Y = X1 − aX2 where a is chosen so that Cov(Y, X2 ) = 0, that is 0 = Cov(X1 , X2 ) −
aV X2 = 2 − 3a and so a = 32 . Note that Y is independent of X2 and
1
2
EY = EX1 − EX2 = ,
3
3
4
4
8
V Y = V X1 + V X2 − Cov(X1 , X2 ) = .
9
3
3
2 8
Thus conditioned on X2 = x2 we have X1 = Y + 23 x2 ∼ N ( 1+2x
, 3 ).
3
(b) By part (a) conditioned on X2 = x2 we have
8 1 + 4x2 + 4x22
25 + 4x2 + 4x22
= V X1 + (EX1 ) = +
=
.
3
9
9
By the formula of total expectation we have
E(X12 )
2
1
25 + 16 + 40
E(X12 X2 ) = E(X2 E(X12 |X2 )) = E(25X2 + 4X22 + 4X23 ) =
=9
9
9
√
where we have used that X2 = 1 + 3Z where Z is a standard normal and so
√
√
E(X23 ) = E(1 + 3 3Z + 9Z 2 + 9 3Z 3 ) = 1 + 0 + 9 + 0 = 10
(the second and the fourth term are zero by symmetry).
(5) Consider Brownian Motion with parameters (1, 2) such that X(0) = 0. Let S = X(1) + X(2) + X(3) +
X(4).
(a) Compute ES and V S.
(b) Find P (S > 12).
1
2
Solution (a)
ES = EX(1) + EX(2) + EX(3) + EX(4) = 1 + 2 + 3 + 4 = 10,
V S = V X(1) + V X(2) + V X(3) + V X(4) + 2Cov(X1 , X2 ) + 2Cov(X1 , X3 ) + 2Cov(X1 , X4 )
+2Cov(X2 , X3 ) + 2Cov(X2 , X4 ) + 2Cov(X3 , X4 ) = 2 + 4 + 6 + 8 + 4 + 4 + 4 + 8 + 8 + 12 = 60.
(b) By part (a)
√
2
P (S > 12) = P (10 + 60Z > 12) = P (Z > √ ) ≈ P (Z > 0.26) = 1 − P (Z < 0.26) ≈ 0.4.
60
(6) Consider Geometric Brownian Motion with parameters (1, 2) such that X(0) = 10. Let M = supt∈[0,1] X(t).
(a) Find the density of M.
(b) Compute P (M > 14).
Solution. (a) Recall that X(t) = 10eW (t) where W (t) is the Brownian Motion with parameters
(1, 2) started from 0. Thus M = 10eR where R(t) = maxt∈[0,1] W (t). By Corollary 3.4.1
r
−
µt
r + µt
2µr/σ 2 ¯
¯
√
√
P (R > r) = e
Φ
+Φ
σ t
σ t
Hence R has density
2µ 2µr/σ2 ¯ r + µt
1
2µr (r + µt)2
(r − µt)2
√
e
Φ
−
−√
exp
+ exp −
σ2
σ2
σ2t
σ2t
σ t
2πtσ
2
2µ 2µr/σ2 ¯ r + µt
1
(r − µt)2
r + µ2 t2
√
= 2e
Φ
−√
+ exp −
.
exp −
σ
σ2t
σ2t
σ t
2πtσ
√
Next dm
= m and r = ln(m/10). Using t = 1, µ = 1 and σ = σ the density of M is
dr
1 m ¯ ln(m/10) + 1
1
(ln(m/10) − 1)2
1
(ln(m/10))2 + 1
√
f (r) =
Φ
+ exp −
.
− √
exp −
m
m 10
2
2
2 π
2
ln
1.2
+
1
ln
1.2
−
1
¯
¯
√
√
(b) P (M > 12) = P (R > ln 1.2) = 1.2Φ
+Φ
2
2
¯
¯
≈ 1.2Φ(0.84)
+ Φ(−0.57)
= 1.2 × 0.2 + 0.72 = 0.96.
(7) Suppose that the stock price at time t is described by the geometric Brownian Motion with drift
parameter µ = 0.2 and volatility parameter σ = 0.15. Assume that the interest rate is 10%. Let
D(S, K) be the no-arbitrage cost of the digital call option which at time 1 pays you $ 1 if S(1) > K
where S = S(0).
(a) Find D(100, 105).
(b) Compute ∂D
(100, 105).
∂S
2
to
e
Solution. Under the risk neutral probability S(t) ∼ GBM (r − σ2 , σ 2 ) Thus no arbitrage cost equals
−rt
−rt
E(I(S(t) > K)) = e
−rt
P (S(t) > K) = e
where
ω
¯=
√
σ2
P (exp r − t + σ tZ > K) = e−rt P (Z > ω
¯)
2
(r − σ 2 /2)t + ln s − ln K
√
.
σ t
Accordingly
∂D
e−rt
∂ω
¯
e−rt −¯ω2 /2
2
= √ e−¯ω /2
=√
e
.
∂s
∂s
2π
2πtσs
Substituting the numerical data we get ω
¯ ≈ 0.27 and so D(100, 105) ≈ e−0.1 × 0.4 ≈ 0.36 and
∂D
≈ 0.023.
∂s
3
(8) Suppose that the stock price at time t is described by the geometric Brownian Motion with drift
parameter µ = 0.2 and volatility parameter σ = 0.15. Assume that the interest rate is 5%. Consider
an option which after 4 years pays you $ 10 for every year where the price of the stock at the end of the
year was higher that the price of the stock at the beginning of the year. For example if S(0) < S(1),
S(1) > S(2), S(2) < S(3) and S(3) < S(4) then the payoff is $ 30 and if S(0) < S(1), S(1) < S(2),
S(2) > S(3) and S(3) > S(4) then the payoff is $ 20. Find no-arbitrage cost of this option.
Solution. Under risk neutral probability the stock price follows Geometric Brownian Motion with
2
parameters (0.05 − 0.15
, 0.152 ) = (0.03875, 0.0225). Let U be the number of times the stock goes up.
2
Then U = U1 +U2 +U3 +U4 where Uj is the indicator of the event that the stock goes up during the year
j. Thus expected payoff is =10E(U ) = 40P (E) where E is the event that the stock will goes up during
the year. We have P (E) = P (S(j + 1)/S(j) > 1) = P (0.03875 + 0.15Z > 0) ≈ P (Z > −0.26) ≈ 0.6.
Thus the cost is 40 × 0.6e−0.05×4 = $ 19.65.
(9) The present price of a certain stock is $ 100 and the price of the call option with experation time 1
year and strike price $ 105 costs $ 12. The interest rate is 5%.
(a) Find the implied volatility of the stock.
(b) If you want to replicate the above call option using delta hedging strategy, how many shares of
the stock should you own at time 0?
Solution. Using monotonicity of option price in σ we find by trial and error that σ ≈ 0.3. Then
2
ω ≈ 0.05+0.30.3/2−0.05 ≈ 0.15. Hence Φ(ω) ≈ 0.56 and so one has to buy approximately 0.56 shares of the
stock.
(10) A certain stock can either gain 10% or loose 5% during each period. The interest rate is 5%. Describe
delta hedging strategy to replicate the option which after 2 periods pays you the price of the stock
times the number of time the stock went up during those two periods.
Solution. Suppose that the initial stock price is S. The payoff of the above option equals 2.42S if
stock goes up twice, 1.045S if stock goes up once, and 0 if the stock goes down both times. Under risk
neutral probabilities the stock goes up with probability 1.05−0.95
= 23 and it goes down with probability
1.1−0.95
1.1−1.05
= 31 .
1.1−0.95
Thus if the stock goes up (and its price becomes 1.1S) during the first period then the expected
payoff is 23 2.42S + 13 × 1.045S ≈ 1.95S and so the option is worth 1.95/1.05 ≈ 1.86S. To be able to
= 8.33 shares of
meet the payment of either a = 2.42S or b = 1.045S we need to have 2.42S−1.045S
0.151.1S
stock which are worth 9.16S and to have −7.3S in the bank.
Thus if the stock goes up (and its price becomes 0.95S) during the first period then the expected
payoff is 23 1.045S + 13 × 0 ≈ 0.69S and so the option is worth 0.69/1.05 ≈ 0.66S. To be able to meet
the payment of either a = 1.045S or b = 0 we need to have 1.045S−0
= 7.33 shares of stock which are
0.150.95S
worth 6.96SS and to have −6.3S in the bank.
Therefore the cost of the option is 23 1.86S + 31 0.66S = 1.39S. To be able to meet the payment of
either a = 1.86S or b = 0.66S we need to have 1.86S−0.66S
= 8 shares of stock which are worth 8SS
0.15S
and to have −6.61S in the bank.
Thus to replicate the option you need 1.39S. The strategy is the following:
(1) Initially borrow 6.61S and buy 8 shares of the security;
(2) If the stock goes up borrow extra money to but 0.33 more shares.
(3) If the stock goes down sell 8 − 7.33 = 0.67 shares and payoff part of your dept.