Download Math489/889 Stochastic Processes and Advanced

Math489/889
Stochastic Processes and
Advanced Mathematical Finance
Homework 4
Steve Dunbar
Due Wednesday, October 10, 2007
1. Suppose X is an exponentially distributed random variable with mean
E[X] = 1. For x = 0.5, 1, and 2, compare P[X ≥ x] with the Markov
inequality bound.
Solution:
P[X ≥ 0.5] = e−0.5 ≈ 0.6065306597 ≤ 1/0.5 = 2
P[X ≥ 1] = e−1 ≈ 0.3678794412 ≤ 1/1 = 1
P[X ≥ 2] = e−2 ≈ 0.1353352832 ≤ x1/2 = 0.5
The Markov inequality bound is not sharp.
2. Suppose X is a Bernoulli random variable with P[X = 1] = p and
P[X = 0] = 1 − p = q. Compare P[X ≥ 1] with the Markov inequality
bound.
Solution: E[X] = 1 · p + 0 · q = p so P[X ≥ 1] = p ≤ p/1 = p. The
Markov inequality bound is sharp.
3. Suppose X is a continuous random variable with mean and variance
both equal to 20. What can be said about P[0 ≤ X ≤ 40]?
Solution: P[0 ≤ X ≤ 40] = P[|X − 20| ≤ 20] = P[|X − µ| ≤ 20] =
1 − P[|X − µ| > 20] ≥ 1 − 20/202 = 19/20.
1
4. Let X1 , X2 , . . . , X10 be independent Poisson random variables with
mean 1. First use the Markov Inequality to get a bound on P[X1 +
· · · + X10 > 15]. Next use the Central Limit theorem to get an estimate
of P[X1 + · · · + X10 > 15].
Solution: X1 , X2 , . . . , X10 are all positive (and integer-valued) random variables and E[X1 + · · · + X10 ] = 10. Then by the Markov inequality bound, P[X1 + · · · + X10 > 15] = P[X1 + · · · + X10 ≥ 16] =
10/16 = 0.625. The variance of Xi is also 1, so by the Central Limit
Theorem,
15 − 10 · 1
S10 − 10 · 1
√
√
>
]
1 · 10
1 · 10
S10 − 10 · 1
√
]
≈ P[Z >
1 · 10
= 0.0569231491.
P[X1 + . . . X10 > 15] = P[
The Markov Inequality Bound is not sharp. By results from probability
theory, X1 + · · · + X10 is Poisson with parameter 10, so the probability
is
P[X1 + . . . X10 > 15] ≈ 0.0487404033.
5. A first simple assumption is that the daily change of a company’s stock
on the stock market is a random variable with mean 0 and variance σ 2 .
That is, if Sn represents the price of the stock on day n with S0 given,
then
Sn = Sn−1 + Xn , n ≥ 1
where X1 , X2 , . . . are independent, identically distributed continuous
random variables with mean 0 and variance σ 2 . (Note that this is an
additive assumption about the change in a stock price. In the binomial
tree models, we assumed that a stock’s price changes by a multiplicative
factor up or down. We will have more to say about these two distinct
models later.) Suppose that a stock’s price today is 100. If σ 2 = 1,
what can you say about the probability that after 10 days, the stock’s
price will be between 95 and 105 on the tenth day?
2
Solution:
P[−5 < X1 + . . . X10
P10
Xi
−5
5
√
√ ]
> 5] = P[ √ < i=1
<
1 · 10
1 · 10
1 · 10
−5
5
√ ]
≈ P[ √ < Z <
1 · 10
1 · 10
= 0.8861537018.
6. Find the moment generating function φX (t) = E[exp(tX)] of the random variable X which takes values 1 with probability 1/2 and −1
with probability 1/2. Show directly (that
is, without using Taylor
√
polynomial approximations) that φX (t/ n)n → exp(t2 /2). (Hint: Use
L’Hopital’s Theorem to evaluate the limit, after taking logarithms of
both sides.)
Solutions: The m.g.f. is
e−t + et
= sinh(t).
φX (t) =
2
Then
√
φX (t/ n)n =
√
e−t/
n
√
+ et/
2
n
n
Then let
L(t, n) = log
√
e−t/
n
√
e−t/
√
+ et/
2
n
n
n !
√
+ et/
2
n
= n log
√ −t/√n
e
+ et/ n
= log
/(1/n)
2
Now apply L’Hopital’s Rule twice to evaluate limn→∞ L(t, n). (Details
are omitted.) The result is:
lim L(t, n) =
n→∞
t2
.
2
√
Therefore φX (t/ n) → exp(t2 /2) as n → ∞.
3
7. (For Math Graduate students only) Prove that if E[eaX ] exists, then
P[X ≥ ] ≤ e−a E[eaX ]
for a > 0.
Solution: Let F (x) be the c.d.f of X. Then
Z ∞
dF (x)
P[X ≥ ] =
Z ∞
−a
ea dF (x)
=e
Z ∞
≤ e−a
ea dF (x)
Z ∞
−a
eaX dF (x)
≤e
−a
=e
4
−∞
aX
E[e
]