# Lower bound for reduced variance after conditioning

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 $$\mathbb{E}[\operatorname{Var}(X|Y)]\leq\operatorname{Var}(X)$$. Under normality, it is known that $$\operatorname{Var}(X|Y)=\frac{\sigma^2\tau^2}{\sigma^2+\tau^2}=\tau^2\left(1-\frac{\tau^2}{\sigma^2+\tau^2}\right)$$ almost surely, and this inequality is stronger than $$\mathbb{E}[\operatorname{Var}(X|Y)]\leq\operatorname{Var}(X)$$ 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?

The LHS of the expression you wrote is the minimum mean square error (i.e., the expectation $$\inf E(X-\hat X)^2$$ where $$\hat X$$ is measurable on $$Y$$). On the other hand, the expression you wrote ($$\hat \sigma^2:=\sigma^2 \tau^2/(\sigma^2+\tau^2)$$) is the error of the optimal linear estimator, so it always bounds from above the optimal error, that is $$\hat \sigma^2\geq E Var(X|Y)$$; equality is achieved in the Gaussian case.