Examples_Functions+of+multiple+random+variables

# Examples_Functions+of+multiple+random+variables - ) ( ) 2 /...

This preview shows page 1. Sign up to view the full content.

Example-1 : Given that two random variables X and Y are statistically independent and identically distributed (i.i.d.) according to an exponential distribution. That is, ) ( ) ( x u e x f x X λ = , and ) ( ) ( y u e y f y Y = . Find the pdf of a new random variable defined as Y X Z + = , a summation of two i.i.d. random variables. Solution : Using the important fact that the pdf of the sum of independent random variable is the convolution of their individual pdfs we have, ) ( ) ( ) ( z f z f z f Y X Z = . For 0 z , 0 ) ( = z f Z , and for 0 > z , . ) ( ) ( ) ( 2 0 2 ) ( 0 0 z z z u z z u Y z X Z ze du e du e e du u z f u f z f = = = = Hence, ). ( ) ( 2 z u ze z f z Z = -- (*) Question : Is this a valid pdf? Method-I: Check for non-negative (Yes) and . 1 ) ( 0 = dz z f Z Method-II: Recall that, for a chi-square distributed r.v. ) ( ~ 2 N T χ , its pdf
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ) ( ) 2 / ( 2 ) ( 2 / 2 / 1 2 / t u e N t t f t N N T = . 1. Fact-1: Exponential distribution is 2 distribution with the d.o.f = 2. When 2 = N , we have ) 2 ( ~ 2 T with pdf ) ( 2 1 ) ( 2 / t u e t f t T = , an exponential pdf with parameter 2 / 1 = . 2. reproducing property of the 2 distribution Z is still 2 distributed. When 4 = N , we have ) 4 ( ~ 2 T with pdf ) ( 4 ) ( 2 / t u e t t f t T = , The same pdf as (*) with parameter 2 / 1 = . Therefore ) ( ) ( 2 z u ze z f z Z = in (*) is a valid pdf....
View Full Document

Ask a homework question - tutors are online