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I have a $K$-dimensional vector $z$ of Bernoulli random variables with success probabilities vector $p$ and a $K \times D$ dimensional matrix $A$ such that each row $A_k$ is a $D$-dimensional Gaussian random variable with mean vector $\mu_k$ and covariance matrix $\Sigma_k$. I would like to compute the following expected value:

$$ \mathbb{E}_{z, \{A_k\}}[\operatorname{diag}(z)\, A A^T \, \operatorname{diag}(z)]$$

where $\operatorname{diag}(z)$ is a square $K \times K$ matrix with $z$ as its diagonal elements.

How can I do this?

Also, if anyone can suggest a better title, please let me know :)

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  • $\begingroup$ If $z_i$ is independent from both $z_j $ and $A_j$, then the expected value is the matrix $\operatorname{diag}(p)\, \bigl(\Sigma + \mu\mu^T\bigr)\, \operatorname{diag}(p)$. Can we assume that these variables are independent? $\endgroup$
    – framago
    Oct 13 at 20:45
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    $\begingroup$ Yes, they are independent! $\endgroup$ Oct 13 at 21:30
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Define $B:=\mathrm{diag}(z)A = \pmatrix{&z_1 A_1& \\ &\vdots& \\ &z_K A_K}$. Note that $M :=\mathbb{E}[\mathrm{diag}(z)\, A A^T \, \mathrm{diag}(z)] = \mathbb{E}[BB^T]$.

Let's do some calculations: $$M_{ij} = \mathbb{E}[(BB^T)_{ij}]=\mathbb{E}[z_i A_i (z_j A_j)^T]=\mathbb{E}[z_iz_jA_i{A_j}^T]=\mathbb{E}[z_iz_j]\,\mathbb{E}[A_i{A_j}^T]=\mathbb{E}[z_i]\,\mathbb{E}[z_j]\,\mathbb{E}[A_i{A_j}^T]=p_ip_j\bigl(\mathrm{Cov}(A_i, A_j) + \mathbb{E}[A_i]\mathbb{E}[{A_j}^T]\bigr) = p_i\bigl(\mathrm{Cov}(A_i, A_j)+\mu_i{\mu_j}^T\bigr)p_j$$

Hence the result is $$M = \mathrm{diag}(p)\bigr(\Sigma + \mu{\mu}^T\bigr)\mathrm{diag}(p)$$ where $\mu=\pmatrix{&\mu_1&\\&\vdots&\\&\mu_K&}$ and $\Sigma_{ij}=\mathrm{Cov}(A_i,A_j)$

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    $\begingroup$ I think this is almost correct. The one problem is that you can't just replace $z$ with $p$ in the expectation. To see why, suppose E[A A^T] = I and z is 1 dimensional. Then we have $E[z_1^2]= E[z_1] = p_1$. But your answer gives $p_1^2$. $\endgroup$ Oct 14 at 1:34
  • $\begingroup$ You are right! The calculation of $M_{ii}$ must be treated separately $\endgroup$
    – framago
    Oct 14 at 8:09

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