For mixture of multivariate Bernoulli distribution we have that,
$$p(x|\mu,\pi) =\sum_{k=1}^{K}\pi_kp(x|\mu_k)$$ where $$p(x|\mu_k) = \prod_{i=1}^{D}\mu_{ki}^{x_i}(1-\mu_{ki})^{1-x_i}$$
I read it from the book that
$$E[x] = \sum_{i=1}^{K}\pi_k\mu_k$$ $$\operatorname{cov}[x] = \sum_{k=1}^{K}\pi_k(\Sigma_k+\mu_k\mu_k^T) - E[x]E[x]^T$$
The mean is trivial to prove, however I can't find proof for the covariance and I don't know how to prove it.
Can anyone help?
You know that, in general, $\operatorname{cov}[x] = E[x x^T] - E[x]E[x]^T$ and you want to compute the first term. We can use the property: $E[g(X)] = E [ E(g(X)|k]]$
But $E(x x^T | k)$ (i.e., fixing the population 'k' index of the mixing), is $\Sigma_k+\mu_k\mu_k^T$ (don't let the $k$ subindexes confuse you: they are the first and seconds moments of $x$, for a fixed population index $k$).
Now, the above is a function ok $k$, and its expectation is $\sum_{k=1}^K \pi_k (\Sigma_k+\mu_k\mu_k^T)$
Notice that this result is not really connected to the Bernoulli distribution, it's valid in general.