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 0 X 40?

    Solution: 0 X 40 = |X 20| 20 = |X μ| 20 = 1 |X μ| > 20 1 20202 = 1920.

    1. Look up the distribution of a Poisson random variable with parameter λ, state it and use that to calculate X 1, and X 2 where X is a Poisson random variable with parameter 1.
    2. Given that the m.g.f. ϕX(t) of a Poisson random variable with parameter λ is eλ(et1), 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.
    3. 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 X1 + + X10 > 15. Take λi = 1 where each Xi is a Poisson random variable with parameter 1.
    4. Use the Markov Inequality to get a bound on X1 + + X10 > 15 where each Xi is a Poisson random variable with parameter 1.

    Solution:

    1. X 1 = 1 X = 0 = 1 e1(1)0 0! = 1 e1

      and

      X 2 = 1 X = 0, 1 = 1 e1(1)0 0! e1(1)1 1! = 1 2e1

    2. The mgf of X1 + X2 is the product of the respective mgfs of X1 and X2 that is eλ1(et1) eλ2(et1) = e(λ1+λ2)(et1) which is the mgf of a Poisson random variable with parameter λ1 + λ2.
    3. By part (b), X1 + + X10 is Poisson with parameter 10, so the probability is
      X1 + X10 > 15 0.0487404033.

    4. X1,X2,,X10 are all positive (and integer-valued) random variables and E X1 + + X10 = 10. Then by the Markov inequality bound, X1 + + X10 > 15 = X1 + + X10 16 = 1016 = 0.625. The Markov bound is not sharp, although it is true.
  2. 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 σ2. That is, if Sn represents the price of the stock on day n with S0 given, then
    Sn = Sn1 + Xn,n 1

    where X1,X2, are independent, identically distributed continuous random variables with mean 0 and variance σ2. (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 σ2 = 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?

    Solution: Let X = i=110X i, so E X = 0 and Var X = 10 because of the independence. Then by Chebyshev’s inequality

    5 < X1 + X10 < 5 = |X| < 5 = 1 |X| 5 1 1052 = 1 1025 = 35.

    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.

  3. Find the moment generating function ϕX(t) = E exp(tX) of the random variable X which takes values 1 with probability 12 and 1 with probability 12. Show directly (that is, without using Taylor polynomial approximations) that ϕX(tn)n exp(t22). (Hint: Use L’Hopital’s Theorem to evaluate the limit, after taking logarithms of both sides.)

    Solutions: The m.g.f. is

    ϕX(t) = et + et 2 = sinh(t).

    Then

    ϕX(tn)n = etn + etn 2 n

    Then let

    L(t,n) = log etn + etn 2 n = n log etn + etn 2 = log etn + etn 2 (1n)

    Now apply L’Hopital’s Rule twice to evaluate lim nL(t,n). (Details are omitted.) The result is:

    lim nL(t,n) = t2 2 .

    Therefore ϕX(tn) exp(t22) as n .