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Two random variables $Z$ and $W$ are uncorrelated if $E(ZW)= E(Z)E(W)$.

Let $X$ and $Y$ be independent random variables receiving 1 with probability $\frac{1}{2}$ and $0$ otherwise.

Prove that $X+Y$ and $|X-Y|$ are uncorrelated random variables but are not independent random variables.

So far what I did is $E(X+Y) = E(X) + E(Y) = \frac{1}{2} + \frac{1}{2} = 1$.

But how do I calculate $E(|X-Y|)$??

user59036
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1 Answers1

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There are four possible cases: $$ \begin{cases} X=1\text{ and }Y=1 \\ X=1\text{ and }Y=0 \\ X=0\text{ and }Y=1 \\ X=0\text{ and }Y=0 \end{cases} $$ In each cases, find $X+Y$ and $|X-Y|$. That will make it possible to find $E(X+Y)$ and $E(|X-Y|)$. You will also need $E((X+Y)|X-Y|)$. So in each of the four cases, find $(X+Y)|X-Y|$.

Then you will be able to show that $X+Y$ and $|X-Y|$ are uncorrelated.

To show that they are not independent, find $\Pr(X+Y=0\ \&\ |X-Y|=0)$ and $\Pr(X+Y=1)\cdot\Pr(|X-Y|)=0$ and similarly for the other possible values of $X+Y$ and $|X-Y|$.

Summary: For a thing like this, enumerate the possible cases.

Michael Hardy
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