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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 ]