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If given a known covariance, \begin{equation} cov(X,Y), \end{equation} what would the covariance, \begin{equation} cov(RX,RY) \end{equation} be, if R is an independant random variable with a variance $R_v$ and an expectation $R_e$?

I believe the covariance if all variables are mutually independant can be given by: \begin{equation} cov(RX,RY) = R_vX_eY_e \end{equation} where $X_e$ and $Y_e$ are the expectations of X and Y, but is it possible to solve this if there is a covariance between X and Y?

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If you also know the means of $X,Y$, you can use the definition of covariance: $$\begin{align}\operatorname{cov}(RX,RY)&=E[R^2XY]-E[RX]E[RY]\\&=E[R^2]E[XY]-E[R]^2E[X]E[Y]\\&=(\sigma_r^2+\mu_r^2)(c_{xy}+\mu_x\mu_y)-\mu_r^2\mu_x\mu_y\end{align}$$

where $c_{xy}=\operatorname{cov}(X,Y)$.

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  • $\begingroup$ Awesome, thank you that answers it perfectly! $\endgroup$
    – JEK
    Commented May 14, 2020 at 18:51
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    $\begingroup$ Actually, is it definitely $E[R^2] = \sigma_r^2 + \mu_r^2$? I was under the impression that is was $E[R^2] = \sigma_r + \mu_r^2$. $\endgroup$
    – JEK
    Commented May 14, 2020 at 19:11
  • $\begingroup$ $\sigma_r^2$ is the variance of $R$ in common notation $\endgroup$
    – gunes
    Commented May 14, 2020 at 19:14
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    $\begingroup$ Oh yeah, sorry I am just learning statistics after doing physics my whole life lol. $\endgroup$
    – JEK
    Commented May 14, 2020 at 19:15

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