1.6 Computing Variance by Conditioning
\(Var(X/Y)\) is a random variable. It’s a function of \(Y\).
When \(Y = y,\hspace {0.3cm} Var(X/Y)\) assumes value \(\, Var(X/Y = y)\). \begin {align*} Var(X/Y) & = E(X^2/Y = y) - \left ( E(X/ Y = y)\right )^2\\ Var(X) & = E\left (Var(X/Y)\right ) + Var\left (E(X/Y)\right ) \end {align*}
Proof. \(\begin {aligned}[t] E\left (Var(X/Y)\right ) & = E\left (E\left (X^2/Y\right )\right ) - E\left (E(X/Y)\right )^2\\\\ & = E\left (X^2\right ) - E\left (E(X/Y)\right )^2\hspace {0.5cm}\cdots \cdots \cdots \hspace {0.5cm} (1) \end {aligned}\). \begin {align*} Var\left (E(X/Y)\right ) & = E\left (\left (E(X/Y)\right )^2\right ) - \left (E\left (E(X/Y)\right )\right )^2\\\\ & = E\left (\left (E\left (X/Y\right )\right )^2\right ) - \left (E(X)\right )^2. \end {align*}
\[E\left (\left (E(X/Y)\right )^2\right ) = Var\left (E(X/Y)\right ) + \left (E(X)\right )^2\hspace {0.5cm}\cdots \cdots \cdots \hspace {0.5cm}(2)\] Substituting in (1) we get \[E\left (Var(X/Y)\right ) = E(X^2) - Var\left (E(X/Y)\right ) - \left (E(X)\right )^2.\] \[E\left (Var(X/Y)\right ) + Var\left (E(X/Y)\right ) = E(X^2) - \left (E(X)\right )^2 = Var(X).\] □
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