hw4soln - ORIE 3500/5500, Fall 10 HW 4 Solutions Homework 4...

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ORIE 3500/5500, Fall ’10 HW 4 Solutions Homework 4 Solutions Problem 1 (a) Since n 4, it follows that 3 n - 2 10 and hence, 0 p X,Y ( i,j ) 1 for all i and j . We want to show that i,j p X,Y ( i,j ) = 1. Let’s look at the joint pmf table of X and Y . i j 1 2 3 ··· n 1 1 3 n - 2 1 3 n - 2 0 ··· 0 2 1 3 n - 2 1 3 n - 2 1 3 n - 2 ··· 0 3 0 1 3 n - 2 1 3 n - 2 1 3 n - 2 0 . . . . . . . . . . . . . . . . . . n 0 0 0 1 3 n - 2 1 3 n - 2 We see that X i,j p X,Y ( i,j ) = 1 X i =1 2 X j =1 1 3 n - 2 + n - 1 X i =2 i +1 X j = i - 1 1 3 n - 2 + n X i = n n X j = n - 1 1 3 n - 2 = 2 3 n - 2 + n - 1 X i =2 3 3 n - 2 + 2 3 n - 2 = 1 3 n - 2 · (3( n - 2) + 4) = 1 . Therefore, p X,Y is a legitimate joint pmf. (b) The marginal pmf’s are p X ( i ) = ( i +1 j = i - 1 1 3 n - 2 = 3 3 n - 2 , if 2 i n - 1 , min { n,i +1 } j =max { 1 ,i - 1 } 1 3 n - 2 = 2 3 n - 2 , if i = 1 ,n. p Y ( j ) = ( j +1 i = j - 1 1 3 n - 2 = 3 3 n - 2 , if 2 j n - 1 , min { n,j +1 } i =max { 1 ,j - 1 } 1 3 n - 2 = 2 3 n - 2 , if j = 1 ,n. The random variables X and Y are not independent since p X (1) p Y (2) = P ( X = 1) P ( Y = 2) = 2 · 3 (3 n - 2) 2 6 = 1 3 n - 2 = P ( X
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This note was uploaded on 11/11/2010 for the course ORIE 3500 at Cornell.

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hw4soln - ORIE 3500/5500, Fall 10 HW 4 Solutions Homework 4...

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