Proof. Taking expectations in \(\Var (X \mid Y) = E(X^2 \mid Y) - \bigl (E(X \mid Y)\bigr )^2\) and using Result 4.2.1, \[ E\bigl (\Var (X \mid Y)\bigr ) = E(X^2) - E\bigl [\bigl (E(X \mid Y)\bigr )^2\bigr ] . \qquad (1) \] Also, since \(E\bigl (E(X \mid Y)\bigr ) = E(X)\), \[ \Var \bigl (E(X \mid Y)\bigr ) = E\bigl [\bigl (E(X \mid Y)\bigr )^2\bigr ] - \bigl (E(X)\bigr )^2 . \qquad (2) \] Adding (1) and (2), the middle terms cancel: \[ E\bigl (\Var (X \mid Y)\bigr ) + \Var \bigl (E(X \mid Y)\bigr ) = E(X^2) - \bigl (E(X)\bigr )^2 = \Var (X) . \] □

Solution. Let \(N(t)\) be the number checked in by time \(t\), so the quantity of interest is \(N(X)\). Given \(X = t\), \(N(X) = N(t)\) is Poisson with mean \(\mu t\), because \(X\) is independent of the arrivals, so \[ E\bigl (N(X) \mid X = t\bigr ) = \mu t , \qquad \Var \bigl (N(X) \mid X = t\bigr ) = \mu t . \] That is, \(E(N(X) \mid X) = \mu X\) and \(\Var (N(X) \mid X) = \mu X\). By Result 4.2.1, \[ E\bigl (N(X)\bigr ) = E(\mu X) = \mu E(X) = \frac {\mu T}{2} , \] and by the conditional variance formula, with \(\Var (X) = T^2/12\), \begin {align*} \Var \bigl (N(X)\bigr ) &= E\bigl (\Var (N(X) \mid X)\bigr ) + \Var \bigl (E(N(X) \mid X)\bigr ) \\ &= E(\mu X) + \Var (\mu X) \\ &= \frac {\mu T}{2} + \mu ^2\frac {T^2}{12} = \frac {\mu T}{12}(6 + \mu T) . \end {align*}