Show that $X + Y$ and $|X − Y |$ are uncorrelated

Question

Be $X$ and $Y$ Independent Bernoulli-distributed random variables with parameter $p= \frac{1}{2}$. Show that $X + Y$ and $|X − Y |$ are uncorrelated.

So I have to show $cov(X + Y,|X − Y |)= 0$

$cov(X + Y,|X − Y |)= \mathbb{E}[(X+Y)|X-Y|]-\mathbb{E}[X+Y] \mathbb{E}[|X-Y|] = 0$

My problem here is to calculate the values of $\mathbb{E}[(X+Y)|X-Y|]$ , $\mathbb{E}[X+Y]$ and $\mathbb{E}[|X-Y|]$.

Could you help me to solve this?

Unfortunately there are no solutions in my exercise book.

Answer 1

We can simply do the calculations explicitly. The value $(X,Y)$ is equally likely to be any of $(0,0), (0,1), (1,0), (1,1).$ Then

$$ \begin{align*} \mathbb{E}[ X |X-Y| ] &= \frac{1}{4} ( 0 + 0 + 1 + 0) = \frac{1}{4}\\ \mathbb{E}[|X-Y|] &= \frac{1}{4} ( 0 + 1 + 1 + 0) = \frac{1}{2}\\ \mathbb{E}[X] &= \frac{1}{2} \\ \end{align*} $$

So then $\text{Cov}(X, |X-Y|) = \frac{1}{4} - \frac{1}{2} \cdot \frac{1}{2} = 0.$ Symmetrically, $\text{Cov}(Y, |X-Y|) = 0$ as well, and by linearity we have $\text{Cov}(X+Y, |X-Y|) = 0.$

Answer 2

Hint: One has $(X+Y)(X-Y) = X^2-Y^2$ and for a Bernouilli variable, $X^2=X$.