# Bayesian MMSE estimators from a transformation of the observations

Consider a random variable X whose value we want to estimate using a Bayesian MMSE estimator. Let $O_1(X)$ be a set of observations which depend on $X$ in some complex way (captured by $P(O_1|X)$) then the MMSE estimator is the conditional mean $\hat{X}_1=\mathbb{E}[X|O_1]$. Now consider another set of observation which is a possibly complex transformation of the first $O_2(O_1)$ and the corresponding MMSE estimator $\hat{X}_2=\mathbb{E}[X|O_2]$. Now it is obvious that if the mapping $O_2(O_1)$ is deterministic then $MSE[\hat{X}_1]\le MSE[\hat{X}_2]$ from the minimality of the MSE of $\hat{X}_1$.

My question is if this is true also when the mapping $O_2(O_1)$ is probabilistic, that is defined by a conditional distribution $P(O_2|O_1)$. Intuitively, it should be as any other stochasticity in the mapping just seems to introduce additional noise as it does not depend on $X$. But I wonder if one can show this explicitly.

EDITED: as the MMSE estimator is unbiased, indid $MSE[\hat{X}_i]=V[\hat{X}_i]$

This fact is called the data processing inequality. If the variables $X$, $O_1$ and $O_2$ form a Markov chain ($X \rightarrow O_1\rightarrow O_2$), then $Var [X|O_1] \leq Var[X|O_2]$, with equality iff $E[X|O_1] = E[X|O_2]$.