Is this equality true for independent random variables $X$ and $Y$? Or maybe we should also assume the same distribution?
2026-04-04 12:10:59.1775304659
$\operatorname{Var}(X+Y\mid X+Y)=\operatorname{Var}(X\mid X+Y)+\operatorname{Var}(Y\mid X+Y)$?
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The conditional distribution of a random variable $Z$ given $Z=z$ is just the unit mass at $z$. This has variance $0$. Therefore $\text{Var}(X+Y \mid X+Y) = 0$. On the other hand, $\text{Var}(X \mid X+Y) \ge 0$; equality means $X+Y$ determines $X$ almost surely. That can happen, but usually is not the case. So your equality is not true in general.