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Let $x\sim \mathcal{N(\mu_x,\Sigma_x \succ 0)}$ and $y\sim \mathcal{N(\mu_y,\Sigma_y \succ 0)}$ be multivariate Gaussian distributions, with $\mu_x \in \mathbb{R}^n$ and $\mu_y \in \mathbb{R}^m$ being the expected value vectors of $x$ and $y$ respectively. Let $M\in \mathbb{R}^{n \times m}$ be any matrix. Suppose that $x$ and $y$ are independent, that is $$\begin{bmatrix}x\\y\end{bmatrix}\sim \mathcal{N}\left(\begin{bmatrix}\mu_x\\\mu_y\end{bmatrix},\begin{bmatrix}\Sigma_x&0_{n \times m}\\0_{m \times n}&\Sigma_y \end{bmatrix}\right)\,.$$

What is the expression of $\mathbb{E}(x^T M y)$ in terms of $\mu_x,\mu_y,\Sigma_x,\Sigma_y$?

My guess would be $\mathbb{E}(x^T M y)=\mu_x^T M \mu_y$, but I'm confused how to prove this.

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$x$ and $y$ are independent so $x$ and $My$ are independent. Therefore, $$ \mathbb E[x^TMy]=\mathbb E[x^T]\mathbb E[My]=\mathbb E[x]^TM\mathbb E[y]=\mu_x^TM\mu_y. $$

$x$ and $y$ do not need to be Gaussian.

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Consider the general case

$$\begin{bmatrix}x\\y\end{bmatrix}\sim \mathcal{N}\left(\begin{bmatrix}\mu_x\\\mu_y\end{bmatrix},\begin{bmatrix}\Sigma_{xx}&\Sigma_{xy}\\\Sigma_{yx}&\Sigma_{yy} \end{bmatrix}\right)\,.$$

Recall that the Cross-covariance_matrix between $x$ and $y$ is

$$\Sigma_{xy} = \Sigma_{yx}^T = \text{Cov}(x,y) = E[(x-\mu_x)(y-\mu_y)^T] = E[xy^T] - \mu_x \mu_y^T $$

Consequently $E[yx^T] = \Sigma_{yx} + \mu_y\mu_x^T$. Using the linearity of the expectated value and the trace trick we obtain

\begin{align} \def\tr{\operatorname{tr}} E[xMy^T] &= E[\tr(x^T M y)]\\ &=E[\tr(Myx^T)]\\ &= \tr(E[Myx^T])\\ &= \tr(ME[yx^T]) \\ &= \tr(M(\Sigma_{yx}+\mu_y\mu_x^T)) \\ &= \mu_x^T M\mu_y + \tr(M\Sigma_{yx}) \end{align}

In the special case when $\Sigma_{yx}=0$, the formula simplifies to $E[xMy^T]=\mu_x^T M\mu_y$

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