Finding covariance of $X+Y$ and $X-Y$, where $X$ and $Y$ normal RV's

Question

How would I find the covariance of $X+Y$ and $X-Y$, given that $X$ and $Y$ are independent normal random variables, both with mean $0$ and variance $1$?

My attempt: $$Cov(X+Y,X-Y)=\mathbb{E}[(X+Y)(X-Y)]-\mathbb{E}[X+Y]\mathbb{E}[X-Y]$$ $$=\mathbb{E}[(X+Y)(X-Y)]-(\mathbb{E}[X]+\mathbb{E}[Y])\mathbb{E}[X-Y])$$ $$=\mathbb{E}[(X+Y)(X-Y)]-(0)\mathbb{E}[X-Y])$$ $$=\mathbb{E}[(X+Y)(X-Y)]$$

I know that due to the independence of $X$ and $Y$ we have that $\mathbb{E}[XY]=\mathbb{E}[X]\mathbb{E}[Y]=0$, however I am not sure how to implement this into my answer to finish it off. Thanks for help

Answer 1

You can also use linearity: $$ \text{Cov}(X+Y,X-Y)=\text{Cov}(X,X)-\text{Cov}(X,Y)+\text{Cov}(Y,X)-\text{Cov}(Y,Y)\\ =1-0+0-1=0. $$

Answer 2

The completion to your attempt is as follows: $$\begin{align*} \mathbb E [(X+Y)(X-Y)] &= \mathbb E [X^2 - Y^2] \\ &= \mathbb E [X^2] - \mathbb E[Y^2] \\ &= \operatorname{Var}[X] + \mathbb E[X]^2 - (\operatorname{Var}[Y] + \mathbb[Y]^2) \\ &= 1 + 0 - 1 - 0 \\ &= 0, \end{align*}$$ where we have used the fact that $$\operatorname{Var}[X] = \mathbb E[X^2] - \mathbb E[X]^2.$$ One advantage to this approach is that you can now compute the covariance when $X$ and $Y$ have arbitrary means and variances; e.g., $$\begin{align*} \operatorname{Cov}[X+Y, X-Y] &\overset{\rm ind}{=} \sigma_X^2 + \mu_X^2 - \sigma_Y^2 - \mu_Y^2 - (\mu_X + \mu_Y)(\mu_X - \mu_Y) \\ &= \sigma_X^2 - \sigma_Y^2. \end{align*}$$ Note the result is independent of the means.

Answer 3

To finish off your calculation, note that $EX^2 = EY^2$.