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Let $\vec{b}$ denote a vector of $n$ Bernoulli variables and let $A \in \mathbb R ^{n\times n}$ be a constant matrix. Can we compute the following expectation:

$$\mathbb E[ \mathrm{diag}(\vec{b}) A \,\mathrm{diag}(\vec{b})],$$

taken over the joint distribution of $\vec{b}$, which is:

$$p(\vec{b}) = \prod_i^n a_i^{b_i} (1-a_i)^{1-b_i},$$

where $a_i$ are the parameters of the individual Bernoulli distributions? Note that the expectation is also a matrix $n \times n$ and it should be expressed in terms of the parameters $a_i$.

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  • $\begingroup$ Just to verify my understanding: $\mathrm{diag}(\vec{b})$ is a diagonal $n \times n$ matrix, right? (Its diagonal entries $\in \{0,1\}$.) So $\mathrm{diag}(\vec{b}) A \,\mathrm{diag}(\vec{b})$ is also an $n \times n$ matrix, and you want its expectation, which is also an $n \times n$ matrix, right? $\endgroup$
    – antkam
    Apr 18, 2018 at 14:45
  • $\begingroup$ Correct. I'm looking for the expectation in terms of the parameters $a$ (will update the question) $\endgroup$ Apr 18, 2018 at 16:28

1 Answer 1

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Let $B = \mathrm{diag}(\vec{b})$, which is a diagonal matrix with diagonal entries $\in \{0, 1\}$.

For any matrix $M$, denote by $M_{*j}$ the $j$th column of $M$ (which is an $n \times 1$ column vector). E.g., $B_{*j}$ is either $0$ or $1$ at its $j$th position, but $0$ everywhere else. We have:

  • if $b_j = 0$ then $(AB)_{*j} = A \times B_{*j} = 0$

  • if $b_j = 1$ then $(AB)_{*j} = A \times B_{*j} = A_{*j}= j$th column of $A$

I.e. $AB$ consists of those $j$th columns of $A$ where $b_j=1$. The columns where $b_j = 0$ have been replaced with $0$.

Similarly, Left-multiplying by $B$ picks out those $i$th rows where $b_i = 1$.

Thus, $(BAB)_{ij} = b_i A_{ij} b_j$, i.e. it is the original $A_{ij}$ entry iff both $b_i=b_j=1$, and $0$ otherwise.

If the above is not convincing enough :) go to algebra:

$$(BAB)_{ij} = \sum_k \sum_l B_{ik} A_{kl} B_{lj} = B_{ii} A_{ij} B_{jj}$$

where the last equality is due to $B$ being diagonal and so all $B_{ik}$ vanish except for $k=i$, and all $B_{lj}$ vanish except for $B_{jj}$.

Finally, $E[(BAB)_{ij}] = E[b_i A_{ij} b_j]$. Now using $P(b_i = 1) = a_i$, we have:

  • For the diagonal entries, $b_i = b_j$ is the same random variable, so $E[(BAB)_{ii}] = a_i A_{ii}$

  • For the non-diagonal entries, $b_i, b_j$ are two independent random variables, so $E[(BAB)_{ij}] = a_i A_{ij} a_j$.

Incidentally, the non-diagonal entries are the same as those in $C=\mathrm{diag}(\vec{a}) A\,\mathrm{diag}(\vec{a})$. The diagonal entries are different though, because $C_{ii} = a_i A_{ii} a_i = a^2_i A_{ii} \neq a_i A_{ii}$.

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