Calculate covariance and correlation

Question

$X$ is to $Ber(1/2)$ and $Y$ is to $N(0,1)$. Assume $X$ is indep to $Y$.

We define $Z=X+Y, W=X-Y$. Find $Cov(Z,W)$ and $Corr(Z,W)$.

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Firstly, we know some information:

$E(X)=1/2, E(Y)=0$

$Var(X)=1/4, Var(Y)=1$

$E(Z) = E(X+Y)=E(X)+E(Y)=1/2$

$E(W)=E(X-Y)=E(X)-E(Y)=1/2$

Then, $Cov(Z,W)=E(ZW)-E(Z)E(W)=E(ZW)-(1/2)(1/2)=E(ZW)-1/4$

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$E(ZW)=E([X+Y][X-Y])=E(X^2-Y^2)=E(X^2)-E(Y^2)$

We can use variances here to help us.

$Var(X)=E(X^2)-(E(X))^2=E(X^2)-(1/2)^2=1/4$

Therefore, $E(X^2)=1/4+(1/2)^2=1/2$

Similarly, $Var(Y)=1=E(Y^2)-(E(Y))^2=E(Y^2)-0$

Therefore, $E(Y^2)=1$

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Then, $E(ZW)=E(X^2)-E(Y^2)=1/2-1=-1/2$, and $Cov(Z,W)=-1/2-1/4=-3/4$

But my textbook says it is $-1/2$.

Have I done a step wrong?

Answer 1

The book is wrong. It contains an error in its calculation of $\mathsf{Cov}(ZW)$.

$$\begin{align}\mathsf {Cov}(ZW)&=\mathsf E(X^2)-\mathsf E(Y^2)-\mathsf E(X)\mathsf E(W)\\ &= \tfrac 12-1-(\tfrac 12)\require{cancel}\color{red}{\cancelto{\tfrac 12}{\color{black}{(0)}}}\end{align}$$

So, therefore the correct answer is indeed: $\mathsf {Cov}(ZW) =-\tfrac 34$.

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To check: Covariance is Bilinear.$$\begin{align}\mathsf{Cov}(Z,W)&=\mathsf {Cov}(X+Y,X-Y)\\&=\mathsf{Cov}(X,X)+\mathsf{Cov}(X,-Y)+\mathsf{Cov}(Y,X)+\mathsf{Cov}(Y,-Y)\\ &= \mathsf{Var}(X)-\mathsf{Var}(Y)\\&=\tfrac 14-1\\&=-\tfrac 34\end{align}$$