I would like to understand the following expectation of a PDF:
Let $\theta\sim\mathcal{N}(\mu, \Sigma_1)$, $x\sim\mathcal{N}(\theta, \Sigma_2)$. Let $f(x)$ be an arbitrary function of $x$. Then how can I get rid of the expectation in $\mathbb{E}_\theta[\mathbb{P}_x(f(x))]$?
I have seen Normal distribution with mean coming from normal distribution, and understood that $x\sim\mathcal{N}(\mu, \Sigma_1+\Sigma_2)$. My guess for the answer is $\mathbb{P}_{x\sim\mathcal{N}(\mu, \Sigma_1+\Sigma_2)}(f(x))$, but I'm not sure how to formalize it.
Thanks a lot for any hint.
Assuming your notation is intended to denote "the expectation over random variable $\theta $ of the marginal probability density ${p_x}\left( x \right)$ (I apologize, I have trouble creating your specific notation) subsequently evaluated at value $f\left( x \right)$", then you are correct, since the marginal density ${p_x}\left( x \right)$ does not depend on $\theta $ (this is called a nuisance parameter, and the operation of marginalizing it away is what leads to the marginal variance/covariance ${\Sigma _1} + {\Sigma _2}$), so that the expectation operation is effectively a no op, and just integrates out to unity.
There are, however, other possible ways to interpret your notation, depending upon the level of precision you employed. In particular, if it is intended to denote "the expectation over random variable $\theta $ of the conditional probability density ${p_x}\left( {\left. x \right|\theta } \right)$ subsequently evaluated at value $f\left( x \right)$", then the expectation does actually play, although I believe it leads to the same result here, since $${E_\theta }\left[ {{p_x}\left( {\left. x \right|\theta } \right)} \right] = {p_x}\left( x \right)$$ which is how you obtain the marginal density from the conditional density in the first place.
If your use of ${p_x}\left( {f\left( x \right)} \right)$ is actually intended to denote marginal density ${p_y}\left( y \right)$ (or conditional density ${p_y}\left( {\left. y \right|\theta } \right)$) for random variable transformation $y = f\left( x \right)$, rather than just subsequent evaluation of the $x$ density at value $f\left( x \right)$, then things would get more complicated. But my guess is this is not what you mean.