# Joint Probability question

• July 21st 2011, 09:07 PM
downthesun01
Joint Probability question
Let X and Y each take on either the value 1 or -1. Let

pr(1,1)=pr(X=1, Y=1)
pr(1,-1)=pr(X=1,Y=-1)
pr(-1,1)=pr(X=-1,Y=1)
pr(-1,-1)=pr(X=-1,Y=-1)

Suppose that E[X]=E[Y]=0. Show that

(a)pr(1,1)=pr(-1,-1)
(b)pr(1,-1)=pr(-1,1)

Let p=2p(1,1), Find
(c) Var[X]
(d) Var[Y]
(e) Cov[X,Y]

Ok, I started by making this joint distribution table

http://i54.photobucket.com/albums/g1...un01/table.png

But I don't understand parts C, D, and E. What does p=2p(1,1) mean in relation to C, D, and E?

If I ignore the part about p=2p(1,1) I can see that

var[X]=var[Y]=0.5 and that cov[X,Y]=0 because X and Y are independent
• July 21st 2011, 09:23 PM
mr fantastic
Re: Joint Probability question
Quote:

Originally Posted by downthesun01
Let X and Y each take on either the value 1 or -1. Let

pr(1,1)=pr(X=1, Y=1)
pr(1,-1)=pr(X=1,Y=-1)
pr(-1,1)=pr(X=-1,Y=1)
pr(-1,-1)=pr(X=-1,Y=-1)

Suppose that E[X]=E[Y]=0. Show that

(a)pr(1,1)=pr(-1,-1)
(b)pr(1,-1)=pr(-1,1)

Let p=2p(1,1), Find
(c) Var[X]
(d) Var[Y]
(e) Cov[X,Y]

Ok, I started by making this joint distribution table

http://i54.photobucket.com/albums/g1...un01/table.png

But I don't understand parts C, D, and E. What does p=2p(1,1) mean in relation to C, D, and E?

If I ignore the part about p=2p(1,1) I can see that

var[X]=var[Y]=0.5 and that cov[X,Y]=0 because X and Y are independent

Your values 1/4, 1/4, 1/4 and 1/4 in your cells are only a particular solution. In fact, there a an infinite number of solutions: the general solution is p(-1,-1) = t, p(1, -1) = 0.5 - t, p(-1, 1) = 0.5 - t, p(1, 1) = t where t is a parameter and can have any value $0 \leq t \leq 0.5$. Your particular solution corresponds to t = 0.25.

So what "p=2p(1,1)" does is defines an alternative parameter to express the general solution. My t corresponds p/2.
• July 21st 2011, 09:33 PM
downthesun01
Re: Joint Probability question
Ok. That makes more sense. Thanks