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Suppose I have two independent random vectors, $X$ and $Y$, both of size $n\times 1$. We have that $\mathbb{E}[X] = \mu_X$ and $\mathbb{E}[Y] = \mu_Y$. We also have that $\mathrm{Cov}(X) = \Sigma_X$ and $\mathrm{Cov}(Y) = \Sigma_Y$. I am seeking the properties of the random vector obtained by the elementwise product of these random vectors, $Z = X\odot Y$. I know that the expectation is $\mathbb{E}[Z] = \mathbb{E}[X]\mathbb{E}[Y]$, but I don't know how to obtain the covariance $\mathrm{Cov}(Z)$.

As this is intended to be the first step toward the inner product, it may be simpler to discuss the expectation and variance of the random variable $z = X^\intercal Y$.

Thank you for your help!

trebledawson
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  • Have you checked https://stats.stackexchange.com/questions/76961/variance-of-product-of-2-independent-random-vector as covafriance of single RV is equal to variance of a singel RV! – Emir Ceyani May 08 '20 at 21:28
  • I did look at that answer, but I wasn't sure how to apply it. Here, we are assuming that $\Sigma_X$ is a fixed but unknown covariance matrix, corresponding to the covariance of each of the elements ($X_i, X_j)$ (and similarly for $\Sigma_Y$). It appears that that answer assumes that each _element_ of each random vector is also independent, with a diagonal covariance matrix, which I cannot assume here. – trebledawson May 08 '20 at 21:34
  • No, this claim "that each element of each random vector is also independent, with a diagonal covariance matrix" comes from the fact that you have two mutually independent random vectors. – Emir Ceyani May 08 '20 at 22:13
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    Ah, then in that answer, is $\mathrm{Var}(Y_1)$ (for example) referring strictly to the first diagonal element of $\Sigma_Y$? – trebledawson May 08 '20 at 22:36
  • Yes, as you said it's the first diagonal element – Emir Ceyani May 08 '20 at 22:38
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    Thank you very much! – trebledawson May 08 '20 at 22:45

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