Answer to Question #53234 in Statistics and Probability for Anwer

Question

This is not homework. I am just bothered about question 2.2.1 of Introduction to Mathematical Statistics (Sixth or seventh edition) of Hogg,McKean and Craig. Question for ready reference is If p(x1,x2)=(2/3)^x1+x2 (1/3)^2-x1-x2 for (x1,x2)=(0,0), (1,0), (1,0), (1,1), zero elsewhere is the joint pmf of X1 and X2, find the joint pmf of y1=X1-X2 and Y2=X1+X2.

First I checked that p(x1,x2) is a joint pmf ie p(0,0)=1/9, p(0,1)=2/9, p(1,0)=2/9 and p(1,1)=4/9 and the sum of these probabilities will be 1. Hence p(x1,x2) is a joint pmf.

Starting with y1=X1-X2 and Y2=X1+X2 will give us x1=(y1+y2)/2 and x2=(y2-y1)/2.

By putting values of x1 and x2 in the given equation ie p(x1,x2)=(2/3)^x1+x2 (1/3)^2-x1-x2 will give us p(y1,y2)=(2/3)^[(y1+y2)/2+(y2-y1)/2] (1/3)^[2-(y1+y2)/2-(y2-y1)/2].

Simplifying we obtain p(y1,y2)=(2/3)^y2 (1/3)^2-y2

What is bothering me that last expression that is joint pmf of y1 and y2 does not have y1. so how it can be proved to joint pmf.

Any comments.