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Math489/889 Stochastic Processes and Advanced Mathematical Finance Solutions for Homework 7 Steve Dunbar Due Mon, November 2 , 2009 1. Time to review all of the information we have about coin-tossing fortunes and random walks. You have a record of 100 tosses of a coin made in class on September 21, 2009. The essence of all the following questions in Problem 1 is to estimate in various ways an answer to the following question: How ”typical” was your coin flip sequence? All 2100 coin flip sequences are equally likely of course, so yours is neither more nor less typical than any other in that way. However, some sets or events of coin flip sequences are more or less “typical” as measured by the probability of a corresponding event. How “typical” as measured by the probability of events containing your sequence is your sequence in one or more of these categories? (a) What is your value of T100 , and the number of heads and the number of tails in your record? Let ² = |T100 | and use Chebyshev’s inequality as in the proof of the Weak Law of Large Numbers to provide an upper bound on the probability that for all possible records |T100 | > ². (b) For your record let a be the absolute value of the difference of the number of Heads you obtained and the expected number of heads. Use the Central Limit Theorem to estimate the probability P [|Tn − nµ| ≤ a] 1 (c) Use a full blank 8.5 × 11 piece of paper and draw the record of the random walk you recorded as a graph of Tn versus n. Choose the time ( or step scale) so that the full time axis for your graph is 10 inches. What is the appropriate scale for the space (or Tn ) axis? (d) What is the excess of Heads over Tails in your record? What is the probability that among all possible records, one would observe an excess of heads over tails at least as large as what you observed? Use the half-integer correction to provide a good estimate. (e) What is the amount of time your random walk was on the positive side? What is the probability of spending at least this much time on the positive side, among all possible walks? 2. If you buy a lottery ticket in 50 independent lotteries, and in each lottery your chance of winning a prize is 1/100, write down and evaluate the probability of winning and also approximate the probability using the Central Limit Theorem. (a) exactly one prize, (b) at least one prize, (c) at least two prizes. Explain with a reason whether or not you expect the approximation to be a good approximation. Solution: The exact probabilities are easy: (a) (b) µ ¶µ ¶1 µ ¶49 50 1 99 ≈ 0.3055586198 1 100 100 ¶0 µ ¶50 µ ¶µ 1 99 50 ≈ 0.3949939329 1− 0 100 100 (c) µ ¶µ ¶0 µ ¶50 µ ¶ µ ¶2 µ ¶49 50 1 50 1 99 99 − ≈ 0.08943531310 1− 1 0 100 100 100 100 2 The normal approximations (from the Central Limit Theorem using the half-integer, or histogram area corrections) are h i 3/2−50·(1/100) √ √ (a) P 1/2−50·(1/100) < Z < ≈ 0.4223907551, 99·50/100 99·50/100 h i √ (b) P 1/2−50·(1/100) < Z] = 1/2 = 0.5, 99·50/100 h i √ (c) P 3/2−50·(1/100) < Z ≈ 0.0776092449. 99·50/100 The normal approximations are not good since the rule of thumb npq > 18 is not satisfied, in fact npq ≈ 1/2. 3. Find a number k such that the probability is about 0.6 that the number of heads obtained in 1000 tossings of a fair coin will be between 440 and k. Solution: Since we wish to count the number of heads, let Xi = 1 with probability 1 if the fair coin comes up heads on the ith flip, and Xi = 0 with probability P 1/2 if it comes up tails. Then µ = 1/2 and σ 2 = 1/4, The Sn = ni=1 Xi counts the number of heads. We seek k so that P [440 ≤ S1000 ≤ k] = 0.6 This is equivalent to · ¸ 440 − (1/2)1000 Sn − (1/2)1000 k − (1/2)1000 √ √ √ P ≤ ≤ = 0.6 (1/2) 1000 (1/2) 1000 (1/2) 1000 By the CLT, this can approximated with √ Φ((k − 500)/(5 2)) − Φ(−3.795) = 0.6 where Φ(x) is the c.d.f. function for the N (0, 1) random variable. Evaluating with a table √ Φ((k − 500)/(5 2)) − 0.000075 = 0.6 so and √ Φ((k − 500)/(5 2)) = 0.600075 √ (k − 500)/(5 10) = 0.25354 so k = 504. 3 4. Suppose you bought a stock at a price b + c, where c > 0 and the present price is b. (Too bad!) You have decided to sell the stock after 30 more trading days have passed. Assume that the daily change of the 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 . Write an expression for the probability that you do not recover your purchase price. Justify with reference to a mathematical theorem why you can write this expression. Solution: Since we explicitly assume that the daily changes are independent, identically distributed continuous random variables with mean 0 and variance σ 2 , the hypotheses of the Central Limit Theorem are satisfied. Then " # 30 X P S0 + Xi < b + c = P [S3 0 < c] i=0 · ¸ c ≈P Z< √ σ 30 µ ¶ c √ =Φ σ 30 5. A bank has $1,000,000 available to make for car loans. The loans are in random amounts uniformly distributed from $5,000 to $20,000. Make a model for the total amount that the bank loans out. How many loans can the bank make with 99% confidence that it will have enough money available? Let X1 , X2 , X3 . . . be a sequence of random variables representing the individual loan amounts. These random variables may reasonably be assumed to be independent, and of course are identically distributed random variables on the interval [5000, 20000]. Then E [Xi] = 12500 and Var [Xi ] = 18,750,000 so σ = 4330.127020. Then the total loan amount is Sn = X1 + · · · + Xn . We seek P [Sn > 1000000] ≤ 0.01. 4 h This is approximately the probability P Z > example from tables) this requires 1000000−12500n √ 4330.127020 n i . Note (for 1000000 − 12500n √ > 2.326347874 4330.127020 n or n < 73.10947801 so the bank can expect to make about 73 loans. 6. Evaluate p2k,2n and graph the probability mass function for 2n = 30 and all admissible values. Use this to show that p2k,2n is a probability mass function for 2n = 30. (Some computer software will make this easy and pleasant, but is not necessary.) Solution: 2k p2k,30−2k 0 0.1444644481 2 0.0747229903 4 0.0581178814 6 0.0503688305 8 0.0459889322 10 0.0433609932 12 0.0418395549 14 0.0411363691 16 0.0411363691 18 0.0418395549 20 0.0433609932 22 0.0459889322 24 0.0503688305 26 0.0581178814 28 0.0747229903 30 0.1444644481 7. Required for Mathematics Graduate Students, Extra Credit for anyone else By actually evaluating the integral, show that Z √ 1 2 1 α p = arcsin( α) π 0 π x(1 − x) 5 p Use this to show that 1/(π x(1 − x)) is a probability density function for 0 < x < 1. √ √ Solution: Let u = x, so that du = 1/(2 √x) dx. Changing the limits of integration u = 0 when x = 0 and u = α when x = α. After the substitution, the integral becomes 1 π Define Z √ α √ 0 √ 2 2 du = arcsin( α). π 1 − u2 0 α≤0 F (α) = arcsin( α) 0 < α < 1 1 1≤α √ 2 π Then F (α) ≥ 0, limα→−∞ F (α) = 0 and limα→+∞ F (α) = 1 and F (α) is non-decreasing. Hence F (α) is a valid c.d.f. and the given function is a valid p.d.f. 6