Question

Checking my probability answer. Could someone please take a look (expectation)

Answer 1

Let X & Y be independent random variables with some distribution (taking values of 0, 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.

Answer 2

So what I did was:
\[\Omega_{X+Y} = \{0+0, 1+0, 0+1, 1+1\}\]\[\Omega_{|X-Y|}=\{|0-0|, |0-1|, |1-0|, |1-1|\}\]

Answer 3

\[E(X+Y) = \sum_{x+y}(x+y)*P(x+y)=0*1/4+1*2/4+2*1/4=1\]\[E(|X-Y|)=\sum(|x-y|)*P(|x-y|)=0*2/4+1*2/4=1/2\]

Answer 4

\[E((X+Y)(|X-Y|)) = 0*10/16+1*4/16+2*2/16=4/16+4/16=1/2\]

Answer 5

The values (10/16, etc.) came from:
{0*0, 0*1, 0*1, 0*0,
1*0, 1*1, 1*1, 1*0,
1*0, 1*1, 1*1, 1*0,
2*0, 2*1, 2*1, 2*0}

Answer 6

So:
\[1/2=E((X+Y)(|X-Y|)=E(X+Y)*E(|X-Y|)=1*1/2=1/2\]

Answer 7

Was this approach correct?

Answer 8

I am stumped on proving that the events are not independent.. Do I just need to come up with two events, say A is the event that X+Y = 2 and B is the event that |X-Y| = 0
and show that P(A intersect B) != P(A)*P(B) or something? I can't seem to come up with an example where that's not true.

Answer 9

@phi any ideas?

Answer 10

No, I have done little with random variables

Answer 11

This is just an idea:
Two events A and B are independent if and only if their joint probability equals the product of their probabilities.

You showed E(product) = product of E
if and only if means product of E must imply E (product)
if that fails, they are not independent.