Solution. By independence, \[ f_{X,Y}(x, y) = f_X(x)f_Y(y) = \frac 1L\cdot \frac 1L = \frac {1}{L^2} , \qquad 0 < x < L , \quad 0 < y < L , \] and we want \(E\bigl (g(X, Y)\bigr )\) with \(g(x, y) = |x - y|\): \[ E\bigl (|X - Y|\bigr ) = \int _0^L\int _0^L|x - y|\,\frac {1}{L^2}\,dx\,dy = \frac {1}{L^2}\int _0^L\int _0^L|x - y|\,dx\,dy . \] Since \(|x - y| = y - x\) for \(x < y\) and \(x - y\) for \(x > y\), the inner integral is \begin {align*} \int _0^L|x - y|\,dx &= \int _0^y(y - x)\,dx + \int _y^L(x - y)\,dx \\ &= \Bigl [xy - \frac {x^2}{2}\Bigr ]_0^y + \Bigl [\frac {x^2}{2} - xy\Bigr ]_y^L \\ &= \Bigl (y^2 - \frac {y^2}{2}\Bigr ) + \Bigl (\frac {L^2}{2} - yL - \frac {y^2}{2} + y^2\Bigr ) = y^2 - yL + \frac {L^2}{2} . \end {align*}

Therefore \begin {align*} E\bigl (|X - Y|\bigr ) &= \frac {1}{L^2}\int _0^L\Bigl (y^2 - yL + \frac {L^2}{2}\Bigr )dy = \frac {1}{L^2}\Bigl [\frac {y^3}{3} - \frac {y^2L}{2} + \frac {yL^2}{2}\Bigr ]_0^L \\ &= \frac {1}{L^2}\Bigl (\frac {L^3}{3} - \frac {L^3}{2} + \frac {L^3}{2}\Bigr ) = \frac L3 . \end {align*}

Proof. Let \[ X_j = \begin {cases} 1 , & \text {if } A_j \text { occurs} , \\ 0 , & \text {otherwise} , \end {cases} \qquad \text {and}\qquad X = \sum _{j=1}^kX_j , \] the number of the events that occur. Then \(E(X_j) = 0 \cdot P(A_j^c) + 1 \cdot P(A_j) = P(A_j)\), so \[ E(X) = \sum _{j=1}^kE(X_j) = \sum _{j=1}^kP(A_j) . \] Now let \(Y = 1\) if \(X \ge 1\) and \(Y = 0\) otherwise, the indicator that at least one event occurs. Then \begin {align*} E(Y) &= 0 \cdot P(X = 0) + 1 \cdot P(X \ge 1) \\ &= P(A_1 \cup A_2 \cup \dots \cup A_k) = P\Bigl (\bigcup _{j=1}^kA_j\Bigr ) . \end {align*}

Since \(Y \le X\) always, \(E(Y) \le E(X)\), which is the inequality. □

Proof. (i) \(aX + b - E(aX + b) = a\bigl (X - E(X)\bigr )\) and \(cY + d - E(cY + d) = c\bigl (Y - E(Y)\bigr )\), so \(\Cov (aX + b, cY + d) = E\bigl [ac(X - E(X))(Y - E(Y))\bigr ] = ac\Cov (X, Y)\).

(ii) Write \(U = \sum _{i=1}^nX_i\) and \(V = \sum _{j=1}^kY_j\), so that \(UV = \sum _i\sum _jX_iY_j\). Using \(\Cov (U, V) = E(UV) - E(U)E(V)\), \begin {align*} \Cov (U, V) &= E\Bigl (\sum _{i=1}^n\sum _{j=1}^kX_iY_j\Bigr ) - E(U)E(V) \\ &= \sum _{i=1}^n\sum _{j=1}^kE(X_iY_j) - \sum _{i=1}^n\sum _{j=1}^kE(X_i)E(Y_j) \\ &= \sum _{i=1}^n\sum _{j=1}^k\bigl \{E(X_iY_j) - E(X_i)E(Y_j)\bigr \} = \sum _{i=1}^n\sum _{j=1}^k\Cov (X_i, Y_j) . \end {align*}

(iii) By (ii) with \(Y_j = X_j\), \[ \Var \Bigl (\sum _{i=1}^nX_i\Bigr ) = \Cov \Bigl (\sum _{i=1}^nX_i, \sum _{j=1}^nX_j\Bigr ) = \sum _{i=1}^n\sum _{j=1}^n\Cov (X_i, X_j) . \] The terms with \(i = j\) are \(\Cov (X_i, X_i) = \Var (X_i)\), and the terms with \(i \ne j\) come in equal pairs, \(\Cov (X_i, X_j) = \Cov (X_j, X_i)\), so \[ \Var \Bigl (\sum _{i=1}^nX_i\Bigr ) = \sum _{i=1}^n\Var (X_i) + 2\sum \sum _{i<j}\Cov (X_i, X_j) . \] □

Solution. By Example 1.3.2 (with \(m = n\), and the roles of \(X\) and \(Y\) exchanged), given \(X + Y = k\), \(Y\) has the hypergeometric distribution \[ P(Y = y \mid X + Y = k) = \frac {\binom ny\binom n{k-y}}{\binom {2n}k} , \] so \[ E(Y \mid X + Y = k) = \sum _yy\,\frac {\binom ny\binom n{k-y}}{\binom {2n}k} , \] the mean of a hypergeometric distribution: \(k\) draws from \(2n\) items of which \(n\) are of the kind counted. By the result below this mean is \(k \cdot n/(2n) = k/2\), as symmetry between \(X\) and \(Y\) demands.

Mean of the hypergeometric distribution

Suppose \(n\) balls are selected at random, without replacement, from an urn containing \(N\) balls of which \(m\) are white, and let \(X\) be the number of white balls selected, so that \[ P(X = x) = \frac {\binom mx\binom {N-m}{n-x}}{\binom Nn} , \qquad x = 0, 1, \dots , \min (n, m) . \] Number the white balls \(1, \dots , m\), and let \(X_i = 1\) if the \(i\)th white ball is selected and \(X_i = 0\) otherwise. Then \(X = X_1 + X_2 + \dots + X_m\), and \begin {align*} E(X_i) &= P(\text {the } i\text {th white ball is selected}) = \frac {\binom 11\binom {N-1}{n-1}}{\binom Nn} \\ &= \frac {(N - 1)!\,n!\,(N - n)!}{(n - 1)!\,(N - n)!\,N!} = \frac nN , \end {align*}

so \[ E(X) = \sum _{i=1}^mE(X_i) = \sum _{i=1}^m\frac nN = \frac {mn}N . \]