This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: UCSD ECE153 Handout #17 Prof. YoungHan Kim Thursday, April 28, 2011 Solutions to Homework Set #4 (Prepared by TA Lele Wang) 1. Two independent uniform random variables. Let X and Y be independently and uniformly drawn from the interval [0 , 1]. (a) Find the pdf of U = max( X,Y ) . (b) Find the pdf of V = min( X,Y ). (c) Find the pdf of W = U V . (d) Find the probability P { X Y  1 / 2 } . Solution: (a) We have F U ( u ) = P { U u } = P { min( X,Y ) u } = P { X u,Y u } = P { X u } P { Y u } = u 2 for 0 u 1. Hence, f U ( u ) = braceleftBigg 2 u, u 1 , , otherwise. (b) Similarly, 1 F V ( v ) = P { V > v } = P { max( X,Y ) > v } = P { X > v,Y > v } = P { X > v } P { Y > v } = (1 v ) 2 , or equivalently, F V ( v ) = 1 (1 v ) 2 , for 0 v 1. Hence, f V ( v ) = braceleftBigg 2(1 v ) , v 1 , , otherwise. 1 (c) First note that W = U V =  X Y  . (Why?) Hence, P { W w } = P { X Y  w } = P ( w X Y w ) . Since X and Y are uniformly distributed over [0 , 1], the above integral is equal to the area of the shaded region in the following figure: x y 1 1 w w The area can be easily calculated as 1 (1 w ) 2 for 0 w 1. Hence F W ( w ) = 1 (1 w ) 2 and f W ( w ) = braceleftBigg 2(1 w ) , w 1 , , otherwise. (d) From the figure above, P { X Y  1 / 2 } = P { W 1 / 2 } = 1 / 4 . 2. Waiting time at the bank. Consider a bank with two tellers. The service times for the tellers are independent exponentially distributed random variables X 1 Exp( 1 ) and X 2 Exp( 2 ), respectively. You arrive at the bank and find that both tellers are busy but that nobody else is waiting to be served. You are served by the first available teller once he/she becomes free. Let the random variable Y denote your waiting time. Find the pdf of Y . Solution: First observe that Y = min( X 1 ,X 2 ). Since P { Y > y } = P { X 1 > y,X 2 > y } = P { X 1 > y } P { X 2 > y } = e 1 y e 2 y = e ( 1 + 2 ) y for y 0, Y is an exponential random variable with pdf f Y ( y ) = braceleftBigg ( 1 + 2 ) e ( 1 + 2 ) y , y , , otherwise. 2 3. Two envelopes. An amount A is placed in one envelope and the amount 2 A is placed in another envelope. The amount A is fixed but unknown to you. The envelopes are shuffled and you are given one of the envelopes at random. Let X denote the amount you observe in this envelope. Designate by Y the amount in the other envelope. Thus ( X,Y ) = braceleftBigg ( A, 2 A ) , with probability 1 2 , (2 A,A ) , with probability 1 2 . You may keep the envelope you are given, or you can switch envelopes and receive the amount in the other envelope....
View
Full
Document
This note was uploaded on 09/16/2011 for the course ECE ECE153 taught by Professor Younghankim during the Spring '11 term at UCSD.
 Spring '11
 YoungHanKim

Click to edit the document details