Conditional variance - $Var(X + U | X) = Var(U)$? I am wondering if the following equality holds - $Var(X + U | X) = Var(U)$? where $X$ and $U$ are two independent random variables?
It seems can we say $Var(X + U | X) = Var(X|X) + Var(U|X) = Var(X|X) + Var(U)$ as $U, X$ independent? And then $Var(X|X) = 0$? But that doesn't seem right.
Can someone clarify this for me, I'm not very experienced with conditional variance.
 A: Unless there's something missing, that looks right to me, but if you want to argue it fully you may want to insert a step or two.  For example
$\text{Var}(X + U | X) = \text{Var}(X|X) + \text{Var}(U|X)$ 
I think you should explain here that you're using $\text{Cov}(X,U|X)=0$ by expanding the variance into its three components and then arguing one is 0.
A: More remarkably, independence between $X$ and $U$ provides a regular system of conditional distributions of $X+U$ given $X$ by setting  $${\cal L}(X+U \mid X=x) = \text{"law of $x+U$"}.$$ 
Then $Var(X+U \mid X)=Var(U)$ because $Var(x+U)=Var(U)$ for any $x$.
A: $Var(X+U|X)=Var(U|X)$ sounds absolutely logical: if the value of $X$ is known, then $X$ has conditional variance 0 (it is a certain variable), so the conditional variance of $(X+U)$ will be the conditional variance of $U$. Then, if $X$ and $U$ are independent the conditional variance of $U$ is simply the variance of $U$.
Another way to look at it: recall that for any random variable $Z$ (that has a variance) and $(a,b) \in \mathbb{R}^2$, $Var(aZ+b)= a^2Var(Z)$, the additive constant $b$ vanishes from the variance (which is easily understood: $b$ only affects the magnitude of $(aZ+b)$, not its dispersion around the mean). Now in a conditional variance $Var(X+U|X)$ the $X$ is just like an additive constant $b$, and vanishes just the same.
A: I hope I'm adding/complementing sth, if not sorry for the excess of answers.
$\text{Var}(X+U|X)= E((X+U-E(X+U|X))^2|X)=E((X+U-X-E(U))^2|X)$
$=E((U-E(U))^2|X)=E((U-E(U))^2)$, where the last equality holds because $X$ and $U$ are independent.
