Correlation of sum of independent variables with its parts. if Z=X+Y, what is Cor(Z,X)?

Question

If $Z = X + Y$, where $X$ & $Y$ are independent random variables,
is there some formula to work out $\rho(Z,X)$, based on $\sigma_X$, $\sigma_Y$?

For example, I've noticed that for $\sigma_X$ = $\sigma_Y$, over many samples, $\rho(Z,X)$ ~= $\rho(Z,Y)$ ~= $0.7065$

I expect $Z$ to be correlated to $X$ & $Y$, but what is the maths?
$X$ & $Y$ are roughly normally distributed if that makes a difference?

Sorry if it's a stupid question, help much appreciated!

Answer 1

Have you tried to work with the definition of the correlation? \begin{align} \rho(Z,X) &= \rho(X+Y,X) \\ &= \frac{\mbox{cov}[X+Y,X]}{\sqrt{\mbox{var}[X+Y]\mbox{var}[X]}} \\ &=\frac{\mbox{var}[X]}{\sqrt{(\mbox{var}[X]+\mbox{var}[Y])\mbox{var}[X])}} \\ &= \frac{\sqrt{\mbox{var}[X]}}{\sqrt{\mbox{var}[X]+\mbox{var}[Y]}} \\ &= \frac{\sigma_X}{\sqrt{\sigma_X^2+\sigma_Y^2}}, \end{align} where I used the independence of $X$ and $Y$.

Answer 2

By definition, $\rho_{XY} = \frac{cov(X,Y)}{\sigma_X \sigma_Y} $, so

\begin{align} \rho_{Z,X+Y} &= \frac{cov(Z, X+Y)}{\sigma^2_Z}& Var(Z)=Var(X+Y)=\sigma^2_Z\\ &= \frac{cov(Z,X)}{\sigma^2_Z}+\frac{cov(Z,Y)}{\sigma^2_Z}\\ &= \frac{cov(Z,X)\sigma_X}{\sigma_Z\sigma_Z \sigma_X} + \frac{cov(Z,Y)\sigma_Y}{\sigma_Z\sigma_Z \sigma_Y} \\ &= \frac{\sigma_X}{\sqrt{2}\sigma_X}\rho_{X,Z} + \frac{\sigma_Y}{\sqrt{2}\sigma_Y}\rho_{Y,Z} & \sigma_Z = (\sigma_X^2+\sigma_Y^2)^{1/2}=\sqrt{2}\sigma_X \\ &=\frac{1}{\sqrt{2}}(\rho_{X,Z} + \rho_{Y,Z}) = \frac{0.7065}{\sqrt{2}} \end{align}