# Probability proof question

• Oct 8th 2007, 12:50 PM
clockingly
Probability proof question
I"m not really sure how to go about this problem, so any help would be greatly appreciated!

Let X and Y be independent random variables with the same distribution, taking values 0 and 1 with equal probability. Show that

E((X+Y)(|X-Y|)) = E(X+Y)E(|X+Y|), but that X+Y and |X-Y| are not independent.

(Here, E is the expectation)
• Oct 8th 2007, 01:51 PM
CaptainBlack
Quote:

Originally Posted by clockingly
I"m not really sure how to go about this problem, so any help would be greatly appreciated!

Let X and Y be independent random variables with the same distribution, taking values 0 and 1 with equal probability. Show that

E((X+Y)(|X-Y|)) = E(X+Y)E(|X+Y|), but that X+Y and |X-Y| are not independent.

(Here, E is the expectation)

X+Y can take the values 0, 1, 2, with probability 1/4, 1/2, 1/4

|X-Y| can take the values 0, 1, with probabilty 1/2, 1/2

Also the RV ((X+Y), |X-Y|) can take values (0,0), (0,1), (1,0), (1,1), (2,0), (2,1),
with probabilities 1/4, 0, 0, 1/2, 1/4, 0.

So now you can just evaluate E((X+Y)(|X-Y|)) and E(X+Y)E(|X+Y|) to show that
they are equal

If X+Y and |X-Y| are independent then their joint distribution:

P((X+Y)=a, |X-Y|=b)=P((X+Y)=a)P(|X-Y|=b)

and you have sufficient information to check this

RonL
• Oct 8th 2007, 04:41 PM
clockingly
Thanks! I just realized that I made a mistake in typing the problem. Sorry about that. It's actually show that E((X+Y)(|X-Y|)) = E(X+Y)E(|X-Y|) but I'm assuming the values would be the same.

Okay, so using this information, I'm getting that:
E(X+Y) = 0*1/4 + 1*1/2 + 2*1/4 = 1
and
E(|X-Y|) = 0*1/2 + 1*1/2 = 1/2

Multiplying these two expectations gives 1/2, which is E(X+Y)E(|X-Y|).

However, how would one find (X+Y) and (|X-Y|) individually and then compute the expectation? I guess I'm a little unclear about what values to use for this because there are coordinates involved.
• Oct 8th 2007, 05:32 PM
clockingly
Sorry, you can disregard my previous question. I just figured it out! :D