Download Math489/889 Stochastic Processes and Advanced

Math489/889
Stochastic Processes and
Advanced Mathematical Finance
Homework 7
Steve Dunbar
Due Wed, October 20, 2010
1. 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.
2. (a) Look up the distribution of a Poisson random variable with parameter λ, state it and use that to calculate P [X ≥ 1], and P [X ≥ 2]
where X is a Poisson random variable with parameter 1.
(b) Given that the m.g.f. φX (t) of a Poisson random variable with
t
parameter λ is eλ(e −1) , show that the sum of independent Poisson
random variables X1 with parameter λ1 and X2 with parameter
λ2 is again Poisson with parameter λ1 + λ2 .
(c) Using the fact that the sum of two independent Poisson random
variables with means λ1 and λ2 is again Poisson with mean λ1 +λ2
find the exact probability that P [X1 + · · · + X10 > 15]. Take λi =
1 where each Xi is a Poisson random variable with parameter 1.
(d) Use the Markov Inequality to get a bound on P [X1 + · · · + X10 > 15]
where each Xi is a Poisson random variable with parameter 1.
Solution:
1
(a)
P [X ≥ 1] = 1 − P [X = 0] = 1 −
e−1 (1)0
= 1 − e−1
0!
and
P [X ≥ 2] = 1 − P [X = 0, 1] = 1 −
e−1 (1)0 e−1 (1)1
−
= 1 − 2e−1
0!
1!
(b) The mgf of X1 + X2 is the product of the respective mgfs of X1
t
t
t
and X2 that is eλ1 (e −1) · eλ2 (e −1) = e(λ1 +λ2 )(e −1) which is the mgf
of a Poisson random variable with parameter λ1 + λ2 .
(c) By part (b), X1 + · · · + X10 is Poisson with parameter 10, so the
probability is
P [X1 + . . . X10 > 15] ≈ 0.0487404033.
(d) 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 Markov bound is not sharp, although it is
true.
3. 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?
P
Solution: Let X = 10
i=1 Xi , so E [X] = 0 and Var [X] = 10 because
2
of the independence. Then by Chebyshev’s inequality
P [−5 < X1 + . . . X10 < 5] = P [|X| < 5]
= 1 − P [|X| ≥ 5]
≥ 1 − 10/52 = 1 − 10/25 = 3/5.
Note that the problem does not assume that the daily changes Xi are
normal, so we cannot use the normal distribution for the sum. With
only 10 summands, the Central Limit Theorem cannot be reliably used.
4. 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
using Taylor
√ n is, without
2
polynomial approximations) that φX (t/ n) → exp(t /2). (Hint: Use
L’Hopital’s Theorem to evaluate the limit, after taking logarithms of
both sides.)
Solutions: The m.g.f. is
φX (t) =
Then
e−t + et
= sinh(t).
2
√
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
+ et/ n
e
= 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