For a random variable $X$ with mean $\mu$ and variance $\sigma^2 < \infty$, define the function $V(x) = \mathbb{E}\left((X − x)^2\right)$. Express the random variable $V(X)$ in terms of $\mu, \sigma^2$ and $X$.
I am not really understanding what $V(X)$ means , are the two big $X$ supposed to be the same? What does the expression mean?
$V(x)$ is a function of a real number $x$. So for example $V(2) = E[(X-2)^2]$ and $V(4) = E[(X-4)^2]$.
Now that $V$ is just another function, you can plug in a random variable $X$ to get a random variable $V(X)$. But simply writing $V(X)=E[(X-X)^2]$ is not quite right, since in the definition of $V$ the expectation is only over the first "$X$" and not over the second "$X$."
It might be helpful to instead write $$V(x) = E[(X-x)^2] = E[(X-\mu + \mu - x)^2] \overset{*}{=} E[(X-\mu)^2] + (\mu-x)^2$$ (make sure you can justify why the starred equality holds) and then plug in $X$.