Random variables expectation

Question

Let $X, X′$ be independent random variables taking values $\pm1$ with probabilities $\frac{1}{2}$. Let $0\le r \le 1$ and set:

$\ Y = \left\{ \begin{array}{ll} X & \mbox{with probability $r$}\\ X' & \mbox{with probability $1-r$}.\end{array} \right. $

Find $\rho(X, Y ) = \frac{E[XY]-E[X]E[Y]}{\sqrt{Var(X)Var(Y)}}$.

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My attempt: $E[X]=E[X']=0$, $Var(X)=Var(X')=1$, $E[Y]=E[X] r + E[X'] (1-r) = 0$, $E[Y^2]= X^2 r + X'^2 (1-r)=r+(1-r)=1$ so $Var[Y]=1$. So all boils down to finding $E[XY]$ which I don't know how to do.

Answer 1

There are several methods. One is to look at the joint distribution

  X  X'   Y  XY    Prob
 +1  +1  +1  +1    1/4
 +1  -1  +1  +1    r/4
 +1  -1  -1  -1  (1-r)/4  
 -1  +1  +1  -1  (1-r)/4
 -1  +1  -1  +1    r/4
 -1  -1  -1  +1    1/4

which gives $E[XY]= r$. As a check, this is clearly correct when $r=0$ or $1$

Answer 2

We can write $Y=BX+(1-B)X'$ where $B,X,X'$ are independent and $B\sim\mathsf{Bernoulli}(r)$.

Then: $$\mathbb EXY=\mathbb E[BX^2+(1-B)XX']$$ Can you take it from here?