For a random variable $X$ with mean $\mu$ and variance $\sigma^2$ define $V(x)=\mathbb{E}(X-x)^2$. By expressing $V(X)$ in terms of $\mu,\;\sigma^2,X$ show that $\sigma^2=\frac{1}{2}\mathbb{E}(V(X))$.
My question is: isn't $V(X)=0?$ I've rewritten it as $V(X)=\sigma^2-(\mu-X)^2$ but this leads nowhere. I think I am misunderstanding something, or maybe there is a mistake in the question.
My question is: isn't $V(X)=0?$
No, actually $V(X)$ is a random variable. Let me suggest to ponder the following proof, keeping track of the parts that are random variables and of those that are deterministic.
By definition, for every real number $x$, $V(x)=E(X^2-2xX+x^2)$, that is, $V$ is polynomial function of degree $2$ defined by $$V(x)=E(X^2)-2E(X)x+x^2.$$ Thus, $V(X)=E(X^2)-2E(X)X+X^2$ and finally, using the linearity of the expectation, $$E(V(X))=E(X^2)-2E(X)E(X)+E(X^2)=2\sigma^2.$$ Exercise: Show that $E(V(X))=E((X-Y)^2)$, where $Y$ is independent of $X$ and has the same distribution.
