Suppose $E[X|Y=y] = a + by$ and $V[X|Y=y] = C + dy^2$ where $Y$ is normally distributed with mean μ and variance $σ^2$ . What is $V[X]$?
2025-01-13 01:39:53.1736732393
Expectations and Moments Variance
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You already have that \begin{align*} E(X \mid Y) &= a + bY\\ E(X^2 \mid Y) &= V(X|Y) + E(X\mid Y)^2 \\ &= c+ d Y^2 + (a+bY)^2. \end{align*} By tower law, you have $E(X)$ and $E(X^2)$, and subsequently $V(X)$.