Proof. We give the proof for discrete \(Y\) (and discrete \(X\)); the continuous case replaces sums by integrals. \begin {align*} \sum _yE(X \mid Y = y)P(Y = y) &= \sum _y\sum _xx\,f_{X\mid Y}(x \mid y)f_Y(y) \\ &= \sum _y\sum _xx\,f_{X,Y}(x, y) = \sum _xx\sum _yf_{X,Y}(x, y) \\ &= \sum _xx\,f_X(x) = E(X) . \end {align*} □

Solution. Let \(X\) be the time until he reaches safety and \(Y\) the door he chooses first. If he chooses door \(2\) or \(3\), he is back where he started after \(5\) or \(7\) hours, so \(E(X \mid Y = 2) = 5 + E(X)\) and \(E(X \mid Y = 3) = 7 + E(X)\). By Result 4.2.1, \begin {align*} E(X) &= E\bigl (E(X \mid Y)\bigr ) \\ &= E(X \mid Y = 1)P(Y = 1) + E(X \mid Y = 2)P(Y = 2) \\ &\qquad + E(X \mid Y = 3)P(Y = 3) \\ &= \tfrac 13\bigl (3 + 5 + E(X) + 7 + E(X)\bigr ) , \end {align*}

so \(3E(X) = 15 + 2E(X)\), and \(E(X) = 15\) hours.

Solution. Let \(X\) be the number of black balls drawn, and write \(N(n, m) = E(X)\). Condition on the first draw: let \(Y = 1\) if it is white and \(Y = 0\) if it is black, so \[ P(Y = 1) = \frac {n}{n + m} , \qquad P(Y = 0) = \frac {m}{n + m} . \] If the first ball is white, no black balls are drawn: \(E(X \mid Y = 1) = 0\). If it is black, one black ball has been drawn and the urn now holds \(n\) white and \(m - 1\) black balls: \(E(X \mid Y = 0) = 1 + N(n, m - 1)\). Hence \begin {align*} N(n, m) &= 0 \cdot \frac {n}{n + m} + \bigl (1 + N(n, m - 1)\bigr )\frac {m}{n + m} \\ &= \bigl (1 + N(n, m - 1)\bigr )\frac {m}{n + m} . \end {align*}

With \(N(n, 0) = 0\), \begin {align*} N(n, 1) &= \bigl (1 + 0\bigr )\frac {1}{n + 1} = \frac {1}{n + 1} , \\ N(n, 2) &= \Bigl (1 + \frac {1}{n + 1}\Bigr )\frac {2}{n + 2} = \frac {n + 2}{n + 1}\cdot \frac {2}{n + 2} = \frac {2}{n + 1} , \\ N(n, 3) &= \Bigl (1 + \frac {2}{n + 1}\Bigr )\frac {3}{n + 3} = \frac {n + 3}{n + 1}\cdot \frac {3}{n + 3} = \frac {3}{n + 1} . \end {align*}

The pattern is \(N(n, m) = m/(n + 1)\), and it holds in general by induction: if \(N(n, m - 1) = (m - 1)/(n + 1)\), then \(N(n, m) = \frac {n + m}{n + 1}\cdot \frac {m}{n + m} = \frac {m}{n + 1}\). So \(E(X) = m/(n + 1)\).

Solution. Let \(Y = 1\) if the first trial is a success and \(Y = 0\) otherwise, so \(P(Y = 1) = p\) and \(P(Y = 0) = 1 - p\). If the first trial succeeds, \(X = 1\). If it fails, the process starts afresh after one wasted trial, so given \(Y = 0\), \(X\) has the distribution of \(1 + X\). Hence \begin {align*} E(X^2) &= E(X^2 \mid Y = 0)P(Y = 0) + E(X^2 \mid Y = 1)P(Y = 1) \\ &= E\bigl ((1 + X)^2\bigr )(1 - p) + 1^2 \cdot p \\ &= \bigl (1 + 2E(X) + E(X^2)\bigr )(1 - p) + p \\ &= \Bigl (1 + \frac 2p + E(X^2)\Bigr )(1 - p) + p . \end {align*}

Collecting the \(E(X^2)\) terms, \(p\,E(X^2) = (1 - p) + \frac {2(1 - p)}{p} + p = \frac {2 - p}{p}\), so \[ E(X^2) = \frac {2 - p}{p^2} , \] and \[ \Var (X) = E(X^2) - \bigl (E(X)\bigr )^2 = \frac {2 - p}{p^2} - \frac {1}{p^2} = \frac {1 - p}{p^2} . \]

Directly, by differentiating a series

Writing \(x^2 = x(x - 1) + x\), \begin {align*} E(X^2) &= \sum _{x=1}^\infty x^2p(1 - p)^{x-1} \\ &= p(1 - p)\sum _{x=1}^\infty x(x - 1)(1 - p)^{x-2} + \sum _{x=1}^\infty xp(1 - p)^{x-1} \\ &= p(1 - p)\sum _{x=1}^\infty \frac {d^2}{dp^2}(1 - p)^x - p\sum _{x=1}^\infty \frac {d}{dp}(1 - p)^x \\ &= p(1 - p)\frac {d^2}{dp^2}\sum _{x=1}^\infty (1 - p)^x - p\frac {d}{dp}\sum _{x=1}^\infty (1 - p)^x . \end {align*}

Since \(\sum _{x=1}^\infty (1 - p)^x = \frac {1}{1 - (1 - p)} - 1 = \frac 1p - 1\), \[ E(X^2) = p(1 - p)\cdot \frac {2}{p^3} - p\Bigl (-\frac {1}{p^2}\Bigr ) = \frac {2(1 - p)}{p^2} + \frac 1p = \frac {2 - p}{p^2} , \] as before.