# Given the variance of sum and difference of two identically distributed random variables, how can I calculate the correlation of the variables?

Suppose $X$ and $Y$ are two identically distributed random variables such that:

\begin{eqnarray*} \sigma^2_{X+Y} = a \, (a \in \mathbb{R}) \\ \sigma^2_{X-Y} = b \, (b \in \mathbb{R}) \end{eqnarray*} I wonder is there a way to find the correlation between the two variables?

Here's what I have tried so far:

\begin{eqnarray*} \sigma^2_{X+Y} = \sigma^2_X + 2\operatorname{cov}(X, Y) + \sigma^2_Y \tag{1}\label{1}\\ \sigma^2_{X-Y} = \sigma^2_X - 2\operatorname{cov}(X, Y) + \sigma^2_Y \tag{2}\label{2}\\ \eqref{1},\eqref{2}\implies \sigma^2_{X+Y} - \sigma^2_{X-Y} = 4\operatorname{cov}(X, Y) \\ \implies \operatorname{cov}(X, Y) = \frac{a-b}{4} \tag{3}\label{3} \end{eqnarray*}

I also know that:

\begin{eqnarray*} \operatorname{corr}(X, Y) = \frac{\operatorname{cov}(X, Y)}{\sigma_X \sigma_Y}\tag{4}\label{4} \end{eqnarray*}

But I have no idea how to calculate the standard deviation ($\sigma$) of $X$ and $Y$. Any help would be appreciated.

You're almost there! The only piece of information that you haven't used yet is the fact that $X$ and $Y$ are identically distributed, which means $\sigma_X^2 = \sigma_Y^2$. So we can drop the subscripts and let $\sigma^2$ denote the common variance of $X$ and $Y$. Adding equations (1) and (2) then gives us $$a + b = 4 \sigma^2$$ or $$\sigma^2 = \frac{a+b}{4}.$$ Since $\sigma_X\sigma_Y = \sigma^2$, the final expression for the correlation is $$\text{Cor}(X,Y) = \frac{a-b}{a+b}.$$
• Hi; +1 for a good answer, but something to keep in mind for next time -- on what look like homework style questions we try to give hints and guidance rather than give the solution. I think most of your approach is good, but it would have been better to have ended leaving something for the OP to do. Please see the mention of homework in the help center and the guidance on answers in the self-study tag wiki -- ideally OPs will mark with that tag but new users don't always know to. – Glen_b Nov 10 '17 at 5:09