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

Question

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.

Answer 1

$\newcommand{\Var}{\operatorname{Var}}\newcommand{\E}{\mathbb E}\newcommand{\F}{\mathcal F}$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.

Answer 2

I am assuming you have a definition for $E[Y \vert X]$. \begin{align*} Var (Y \vert X) &= E [ (Y - E[Y \vert X] )^2 \vert X] \\ &= E[Y^2 \vert X] - (E[Y \vert X])^2 \\ &= (g(X))^2 - (g(X))^2 \\ &= 0. \end{align*}