I am working on a problem that asks for Var(Y|X=1)
Is this the same as E(Y$^2$|X=1) - $\big($E(Y|X=1)$\big)$$^2$
2026-03-27 00:01:37.1774569697
Formula for finding Conditional Variance
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As I understand it, the notation $E( \cdot \mid X = 1)$ signifies the integral with respect to the measure $\mu( \cdot ) = \dfrac{\mathbf{P}(\cdot \cap \{X = 1\})}{\mathbf{P}(X = 1)}.$ Hence, by denoting $a = \int Y d\mu,$ we have $\mathbf{V}\mathrm{ar}(Y) = \|Y - a\|_{\mathscr{L}^2(\mu)}^2=\int(Y-a)^2 d\mu = \int Y^2 d\mu - a^2.$ So yes.