IEOR 4703: Homework 2 Solutions

Transcription

1 IEOR 4703: Homework 2 Solutions Exercises for which no programming is required Let U be uniformly distributed on the interval (0, 1); P (U x) = x, x (0, 1). We assume that your computer can sequentially generate such U indepently upon demand. 1. Let X = (b a)u + a. Show that X is uniformly distributed on the interval (a, b); that is, show that P (X x) = (x a)/(b a), x (a, b). Thus, if your computer can generate a U, then it can in fact generate any rv X that is uniformly distributed on the interval (a, b) for any desired a and b. SOLUTION: If x (a, b), then for y = (x a)/(b a), it follows that y (0, 1) and thus P (U y) = y, that is, P (U (x a)/(b a)) = (x a)/(b a). We now use this fact: 2. Continuation: X has cdf given by P ((b a)u + a x) = P (U (x a)/(b a)) = (x a)/(b a). x a b a, if x (a, b); F (x) = 1, if x b; 0, if x a. Show that in fact setting X = (b a)u + a is just the inverse transform method. SOLUTION: It is easily sen that F 1 (y) = (b a)y + a, y (0, 1). 3. Let V = 1 U. Show that V is also uniformly distributed on the interval (0, 1). SOLUTION: We must show that P (V x) = x, x (0, 1). To this : P (V x) = P (1 U x) = P (U 1 x) = 1 (1 x) = x. 4. Recall that conditional probabilities are defined as: P (A B) = P (A B)/P (B), where A B denotes the intersection of the two events. Conditional on the event B = {U.5}, show that U is uniformly distributed on the smaller interval (0, 0.50). That is, P (U x U.50) = x/(.50), x (0,.50). More generally, conditional on the event B = {U (a, b)}, where 0 < a < b < 1, show that U is uniformly distributed on the smaller interval (a, b). P (U x U (a, b)) = (x a)/(b a), x (a, b). SOLUTION: With A = {U x} and B = {U.5}, and x (0,.50), we have A B = {U x}. Thus P (A B)/P (B) = P (U x)/(p (U.50)) = x/(.50). More generally, when B = {U (a, b)}, and A = {U x} where 0 < a < x < b < 1, we have A B = {U (a, x]} and thus P (A B)/P (B) = P (U (a, x])/(p (U (a, b))) = (x a)/(b a). 5. Give an algorithm that uses two uniforms U 1, U 2 for generating the following distribution: F (x) = 0.40(1 e 2x ) (1 e 2 x ), x 0. 1

2 SOLUTION: This is an example of a mixture, F (x) = pf 1 (x) + (1 p)f 2 (x) where p = 0.40, F 1 is exponential at rate 2 and F 2 is Weibull, with an easily found inverse F2 1 (y) = (1/4)[ln(1 y)] 2, y (0, 1). So we can use the composition method: (a) Generate U 1 and U 2. (b) If U 1 0.4, then set X = ( 1/2) ln(u 2 ); otherwise set X = (1/4)[ln(U 2 )] Let X 1,..., X n be iid rvs distributed as F (x) = P (X x), and assume that F 1 (y) is known in closed form. (a) Give an algorithm for generating a rv m = min{x 1, X 2,..., X n } that uses only ONE uniform U. SOLUTION: This is analogous to simulating the maximum, which we went thru in class lecture, same idea using the inverse transform method. Let G(x) = P (m x). Then P (m > x) = P (X 1 > x,..., X n > x) = (1 F (x)) n. Thus G(x) = 1 (1 F (x)) n. Thus solving y = G(x) for x in terms of y yields G 1 (y) = F 1 (1 (1 y) 1/n ). Since 1 U is itself uniform we obtain m = F 1 (1 (U) 1/n ) as our algorithm. (b) Give the algorithm in the special case when F (x) = 1 e λx, x 0; what kind of distribution does m have here? SOLUTION: In this case m exp(nλ), so m = (1/(nλ)) ln U. (c) Give the algorithm in the special case when F (x) = x, x (0, 1) is the uniform (0, 1) distribution; what kind of distribution does m have here? SOLUTION: F y = y, so we obtain X = (1 (U) 1/n ). G(x) = 1 (1 x) n, so taking the derivative yields the Beta density g(x) = n(1 x) n 1, x (0, 1). 7. Suppose that X 1 and X 2 are iid exponentials; P (X i x) = 1 e λx, x 0. Prove that U def = X 1 /(X 1 + X 2 ) is uniform on (0, 1) and is indepent of the sum X 1 + X 2. (Hint: consider the change of variables u = x 1 + x 2, v = x 1 /(x 1 + x 2 ); now find the probability density function g(u, v) using the fact that the joint density of X 1, X 2 is given by f(x 1, x 2 ) = λ 2 e λ(x 1+x 2 ). Jacobian, etc.) SOLUTION: Noting that the sum V = X 1 + X 2 is an Erlang distribution with density g(v) = λ 2 ve λv, v > 0, and recalling that the uniform density on (0, 1) is given by g(u) = 1, u (0, 1), it must be shown that the joint density of U = X 1 /(X 1 + X 2 ) and V = X 1 + X 2 is given by g(u, v) = 1 λ 2 ve λv, u (0, 1), v > 0. (The product of the two because of indepence.) We consider the function h(x 1, x 2 ) = (h 1 (x 1, x 2 ), h 2 (x 1, x 2 )), where h 1 (x 1, x 2 ) = u = x 1 /(x 1 + x 2 ) and h 2 (x 1, x 2 ) = v = x 1 + x 2 ). From general multidimensional calculus, changing variables in this manner involves computing the determinant of the 2 by 2 Jacobian matrix of partial derivatives ( ( h1 x 1 h 2 h 1 x 2 h 2 x 1 x ) = x 2 (x 1 +x 2 ) 2 x 1 (x 1 +x 2 ) 2 ). 2

3 The determinant is thus J(x 1, x 2 ) = (x 1 + x 2 )/(x 1 + x 2 ) 2 = 1/(x 1 + x 2 ) = v 1. The general formula for computing g is g(u, v) = J(x 1, x 2 ) 1 f(x 1, x 2 ) = vf(x 1, x 2 ) = yielding the result. vλ 2 e λ(x 1+x 2 ) = λ 2 e λv, Exercises for which programming is required 1. Program up the code (in MATLAB) for generating a Poisson rv X with mean α (below); then using α = 50, simulate via Monte Carlo simulation to estimate P (X = 50), the probability that X takes on its own mean as its value. (This is the Mode.) Generate, n = 1000 iid samples, X 1,..., X n and use the estimate 1 n n X i. i=1 Repeat for n = 5, 000 and then n = 10, 000.(The exact answer is of course e α α 50 /50!.) %This algorithm generates a Poisson rv with inputed mean alpha %User inputs alpha alpha = input( Value of alpha: ); X = 0; P = 1; u = rand; P = P*u; while (P >= exp(-alpha)) X u = rand; P = P*u; X= X+1; &This code estimates P(X=50) for a Poisson rv with mean alpha=50 alpha = input( Value of alpha: ); count = 0; %this variable keeps track of the number of times that X equals 50 over the ite 3

4 n = 10000; for i=1:n X = 0; P = 1; u = rand; P = P*u; while (P >= exp(-alpha)) u=rand; P=P*u; X=X+1; if ( X == 50) count = count+1; count / n 2. Simulate a compound Poisson process up to time T = 10, when λ = 5 and the iid jumps J are distributed as P (J = 1) = 0.30, P (J = 1) = 0.10, P (J = 4) = Use the following (but write up in MATLAB): (a) t = 0, N = 0, X = 0. (b) Generate a U (c) t = t + [ (1/λ) ln (U)]. If t > T, then stop. (d) Generate J distributed as G. (e) Set N = N + 1 and set X = X + J (f) Go back to 2. t=0; N=0; X=0; lambda=5; T=10; while (t<=t) t=t+(-1/lambda*log(rand)); Y=rand; 4

5 J=1*(Y<=0.3)-1*(Y>0.3)*(Y<=0.4)+4*(Y>0.4); N=N+1; X=X+J; X N 5