Expectation of an unbiased estimator (under variational inference setting) I am reading this ICML2016 paper, and am puzzled with the first inequality (converted to equality) on section 2.2. 
Assume the model is $P(x,h)$ where $h$ are the hidden variables. Also assume $\hat{I}$ is an unbiased estimator of the likelihood term, $P(x)$. From this we can conclude that $$E_{P(x)}[\hat{I}]-P(x)=0\Rightarrow E_{P(x)}[\hat{I}] = P(x)$$ now assume we want to establish a lower bound on $P(x)$ (similar to EM approach) and plug in the estimator $\hat{I}$, instead of $P_{\theta}(x)$ in the lower bound formulation. 
For this imagine the posterior distribution over the latent variables $h$, to be estimated using  $Q(h|x)$ (i.e., $Q$ is a variational posterior). So if we want to write 
\begin{align}
&\log P(x) =\log \sum P(x,h)\\&\Rightarrow \log P(x) = \log \sum P(x,h)\frac{Q(h|x)}{Q(h|x)}\\&\Rightarrow \log E_{Q(h|x)}[P(x,h)] \ge E_{Q(h|x)}[\log P(x,h)]\\ &\Rightarrow \log \hat{I} \ge E_{Q(h|x)}[\log P(x,h)]
\end{align}
Here are two the puzzling parts:


*

*I don't understand how they could drive


$$E_{Q(h|x)}[\log \hat{I}]\leq \log E_{Q(h|x)}[\hat{I}] = \log P(x)$$
 given all mentioned in the above.


*They also say since $\hat{I}$ is an unbiased, it can be written


$$E_{Q(h|x)}[ \hat{I}] = P(x)$$
  which is not clear why, given the unbiased estimator definition.
 A: Let's start with the second question first: The fact that $\hat{I}$ is unbiased can be seen as follows. First, in the paper it is also stated that the samples $h^i$ of the latent variables are independent. Thus, we have 
\begin{align}
E_{Q(h|x)}(\hat{I}) 
 & = E_{Q(h|x)}\left( \frac{1}{K} \sum_{i=1,\ldots,K} \frac{P(x,h)}{Q(h|x)}\right)\\
 & =  \frac{1}{K} \sum_{i=1,\ldots,K} 
      E_{Q(h|x)}\left(  \frac{P(x,h)}{Q(h|x)}\right)\\
 & =  \frac{1}{K} \sum_{i=1,\ldots,K} 
      \int  \frac{P(x,h)}{Q(h|x)} \cdot Q(h|x)\,\text{d}h\\
 &= P(x)
\end{align}
The reason why this expectation is taken over $Q(h|x)$ rather than $P(x)$ is that we actually want to estimate the function $P(x)$. By taking the expectation $E_{P(x)}$, we would integrate out $x$ and obtain an expression that only depends on $h$ rather than $x$.
From this we can now easily obtain the answer to your first question. As $\log$ is concave, the inequality follows from Jensen's inequality which states that for concave functions $\phi(\cdot)$ we have $E(\phi(x)) \leq \phi(E(x))$. The equality follows from the proof above.
