Q:

Given Y = g(X), why is Var(Y|X) = 0?

Given $Y = g(X)$, why is $Var(Y|X) = 0$? 
After replacing $Y$ with $g(X)$...
$Var(g(X)|X) = 0$ is not intuitive to me. 
Intuitively, I feel like the $Var(g(X)|X)$ should be the whole function $g(X)$.
Any insights would be helpful.  Thanks.

A:

$\newcommand{\Var}{\operatorname{Var}}\newcommand{\E}{\mathbb E}\newcommand{\F}{\mathcal F}$This is easily done with the measure theoretic definition. One defines for a square integrable random variable $X$ and a $\sigma$-algebra $\F$:
\begin{align}
\Var(X|\F):=\E[X^2|\F]-\E[X|\F]^2
\end{align}
Now we want $\Var(Y|X)=\Var(Y|\sigma (X))$, and that is:
\begin{align}
\Var(Y|\sigma(X))=\E[Y^2|\sigma(X)]-(\E[Y|\sigma(X)])^2
\end{align}
Since $Y=g(X)$ we have $Y$ is $\sigma(X)$-measurable (as long as $g$ is nice enough $(\star)$). That implies $\E[Y^2|\sigma(X)]=Y^2$ and $\E[Y|\sigma(X)]=Y$. Hence:
\begin{align}
\Var(Y|\sigma(X))=Y^2-Y^2=0
\end{align} 
$(\star)$: Surely one needs to require $g$ to be measurable and such that $\E[g(X)^2]$ exists, etc. 

