prac.final.sol

Problem 2 (10 points) If U is uniform on (0 , 2 π ) and Z is independent of X and exponential with rate 1, show that the random variables X and Y defined by X = √ 2 Z cos U Y = √ 2 Z sin U are independent standard normal random variables. Answer: The joint density of U and Z is f U,Z ( u, z ) = 1 (0 , 2 π ) ( u ) · 1 2 π e - z . The Jacobian of the map from U, Z to X, Y is given by J = cos U √ 2 Z - √ 2 Z sin U sin U √ 2 Z √ 2 Z cos U = 1 . Finally, solving for U, Z in terms of X, Y , we find that Z = 1 2 ( X 2 + Y 2 ) U = arctan( Y/X ) . We conclude that the joint density of X, Y is given by f X,Y ( x, y ) = 1 2 π e - 1 2 ( x 2 + y 2 ) = 1 √ 2 π e - 1 2 x 2 1 √ 2 π e - 1 2 y 2 , which is the density of i.i.d. standard normal random variables. 3

Problem 5 (20 points) Let T 1 and T 2 be two independent exponential variables with rates λ 1 and λ 2 , respectively. Define T min = min ( T 1 , T 2 ), and let X min be a random variable which equals 1 if T 1 < T 2 and 2 if T 2 < T 1 . (a) (5 points) The distribution of T min is exponential with some parameter λ . Find λ . Answer: λ 1 + λ 2 We find that F T min ( t ) = 1 - P ( T min ≥ t ) = 1 - P ( T 1 ≥ t ) P ( T 2 ≥ t ) = 1 - e - ( λ 1 + λ 2 ) t , which means that T min is exponential with parameter λ 1 + λ 2 . (b) (5 points) Find P ( X min = 1). Answer: λ 1 λ 1 + λ 2 We find that P ( X min = 1) = P ( T 1 < T 2 ) = Z ∞ 0 Z t 2 0 λ 1 λ 2 e - λ 1 t 1 - λ 2 t 2 dt 1 dt 2 = λ 2 Z ∞ 0 e - λ 2 t 2 (1 - e - λ 1 t 2 ) dt 2 = 1 - λ 2 λ 1 + λ 2 = λ 1 λ 1 + λ 2 . (c) (10 points) Show that T min and X min are independent. Answer: We compute that P ( T min ≥ t, X min = 1) = P ( t ≤ T 1 < T 2 ) = Z ∞ t Z t 2 t λ 1 λ 2 e - λ 1 t 1 - λ 2 t 2 dt 1 dt 2 = λ 2 Z ∞ t e - λ 2 t 2 ( e - λ 1 t - e - λ 1 t 2 ) dt 2 = e - ( λ 1 + λ 2 ) t - λ 2 e - ( λ 1 + λ 2 ) t λ 1 + λ 2 = λ 1 λ 1 + λ 2 e - ( λ 1 + λ 2 ) t = P ( T min ≥ t ) P ( X min = 1) , which gives independence. 6