Let $X$ be a random variable with variance $tau^2$ and $Y$ be another random variable such that $Y-X$ is independent of $X$ and has mean zero and variance $sigma^2$. (One can think of $Y$ as a noisy observation of $X$.) It follows from the law of total variance that $mathbb{E}(operatorname{Var}(X|Y))leqoperatorname{Var}(X)$. Under normality, it is known that $operatorname{Var}(X|Y)=frac{sigma^2tau^2}{sigma^2+tau^2}=tau^2left(1-frac{tau^2}{sigma^2+tau^2}right)$ almost surely, and this inequality is stronger than $mathbb{E}(operatorname{Var}(X|Y))leqoperatorname{Var}(X)$ as it quantifies *how much* the expected variance is reduced.

I wonder if one could prove something similar in general. If this does not hold in general, would it help to assume the random variables are sub-Gaussian?