Intuitive Difference between posterior variance $V(Y|X)$ and variance of conditional expectation $V(E(Y|X))$

Question

Intuitive Difference between posterior variance $V(Y|X)$ and variance of conditional expectation $V(E(Y|X))$

From Eves theorem we have as expectation is non-negative,

$V(Y) \geq V(E(Y|X))$

It means after knowing information X, my prior variance V(Y) has reduced

Is posterior variance $V(Y|X)$ or $V(E(Y|X))$. Both look similar to me. Please elaborate intuitive difference /relation between these 2. I am a beginner. Request your patience.

Answer 1

You can prove that if $Y\in L^2(\Omega, \mathcal{F}, \mathbb{P})$, then

$Var(Y)=Var(\mathbb{E}(Y\mid X))+\mathbb{E}(Var(Y\mid X))$.

Note that $Var(Y\mid X)$ is a random variable and $Var(\mathbb{E}(Y\mid X))$ is a number.

Answer 2

$V(Y|X)$ is the posterior variance of $Y$ given $X$. It is a random variable which is a function of $X$; for each possible value $x$ of $X$, $V(Y|X=x)$ is the variance of the random variable $Y$ conditioned on $X=x$.

$V(E(Y|X))$ is harder to give an intuitive explanation for. Just as $V(Y|X)$ is a random variable, $E(Y|X)$ is a random variable, where $E(Y|X=x)$ is the expected value of $Y$ given $X=x$. Since $E(Y|X)$ is a random variable, it has a variance, and $V(E(Y|X))$ is that variance. But I do not know what that quantity is really useful for. As mentioned in the other answer, $$V(E(Y|X))=V(Y)-E(V(Y|X))=E[V(Y)-V(Y|X)],$$ so you can interpret $V(E(Y|X))$ as the "average reduction in variance in $Y$ you get by learning the value of $X$".