Math489/889 Stochastic Processes and Advanced Mathematical Finance Homework 5

Transcription

1 Math489/889 Stochastic Processes and Advanced Mathematical Finance Homework 5 Steve Dunbar Due Fri, October 9, 7. Calculate the m.g.f. of the random variable with uniform distribution on [, ] and then obtain E[X] and Var[X]. Solution: φ U (t) = exp(tx) dx = exp t t Note that the expression for this moment generating function is not defined at t = although certainly the integral converges at t = and in fact is. This is an example of a removable singularity. Although it is possible to differentiate this function and then evaluate by taking limits (perhaps using L Hopital s Rule) it is easier to express this function as the analytic function that it is by means of its power series expression: φ U (t) = + t/ + t /6 + = i= t i (i + )! Then it is easy to see that E[U] = φ U () = / and E[U ] = φ U () = /3. Hence Var[U] = /3 /4 = /. The differentiability of the function is guaranteed by the analyticity.. If you buy a lottery ticket in 5 independent lotteries, and in each lottery your chance of winning a prize is /, write down and evaluate the probability of winning and also approximate the probability using the Central Limit Theorem.

2 (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) (c) ( ) ( ) ( ) ( ) ( ) ( ) ( 5 ) ( ) ( ) 5 99 ( 5 ) ( ) ( ) The normal approximations (from the Central Limit Theorem using the half-integer, or histogram area corrections) are (a) P[ / 5 (/) 99 5/ < Z < 3/ 5 (/) 99 5/ ] , (b) P[ / 5 (/) 99 5/ < Z] = / =.5, (c) P[ 3/ 5 (/) 99 5/ < Z The normal approximations are not good since the rule of thumb npq > 8 is not satisfied, in fact npq /. 3. Find a number k such that the probability is about.6 that the number of heads obtained in tossings of a fair coin will be between 44 and k. Solution: Since we wish to count the number of heads, let X i = with probability if the fair coin comes up heads on the ith flip, and X i = with probability / if it comes up tails. Then µ = / and

3 σ = /4, The S n = n i= X i counts the number of heads. We seek k so that P[44 S k] =.6 This is equivalent to 44 (/) P[ (/) S n (/) (/) k (/) (/) ] =.6 By the CLT, this can approximated with Φ((k 5)/(5 )) Φ( 3.795) =.6 where Φ(x) is the c.d.f. function for the N(, ) random variable. Evaluating with a table Φ((k 5)/(5 )).75 =.6 so and so k = 54. Φ((k 5)/(5 )) =.675 (k 5)/(5 ) = Suppose you bought a stock at a price b + c, where c > and the present price is b. (Too bad!) You have decided to sell the stock after 3 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 and variance σ. That is, if S n represents the price of the stock on day n with S given, then S n = S n + X n, n where X, X,... are independent, identically distributed continuous random variables with mean and variance σ. Write an expression for the probability that you do not recover your purchase price. 3

4 Solution: 3 P[S + X i < b + c] = P[S 3 < c] i= [ P Z < c ] σ 3 ( ) c = Φ σ 3 5. A bank has $,, available to make for car loans. The loans are in random amounts uniformly distributed from $5, to $,. How many loans can the bank make with 99% confidence that it will have enough money available? Solution: Let X, X, X 3... 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 uniform random variables on the interval [5, ]. Then E[Xi] = 5 and Var[X i ] = 8,75, so σ = Then the total loan amount is S N = X + +X n. We seek P[S n > ].. This is approximately the probability P[Z > 5n ]. Note (for n example from tables) this requires 5n n > or n < so the bank can expect to make about 73 loans. 6. Evaluate p k,n and graph the probability mass function for n = 3 and all admissible values. Use this to show that p k,n is a probability mass function for n = 3. (Some computer software will make this easy and pleasant, but is not necessary.) 4

5 Solution: k p k,3 k Required for Mathematics Graduate Students, Extra Credit for anyone else By actually evaluating the integral, show that π α x( x) = π arcsin( α) Use this to show that /(π x( x)) is a probability density function for < x <. Solution: Let u = x, so that du = /( x) dx. Changing the limits of integration u = when x = and u = α when x = α. After the substitution, the integral becomes Define π α u du = π arcsin( α). α F (α) = π arcsin( α) < α < α Then F (α), lim α F (α) = and lim α + F (α) = 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. 5