給定 $X_1,...,X_n$ 為一組 i.i.d. 非負隨機變數,現在令 $M:=\max\{X_1,...,X_n\}$我們想問 \[ E[M] =? \] 首先我們回憶 \[E[M] = \sum\limits_{k \geqslant 0} P (M > k) = \sum\limits_{k \geqslant 0} {\left( {1 - P(M \leqslant k)} \right)} \]上述第一等式利用 非負隨機變數的期望值的性質(亦即若 $X$ 為非負隨機變數,則 $E[X] = \int_0^\infty P(X>x)dx$),在此不做贅述。現在注意到 \begin{align*} P(M \leqslant k) &= P\left( {\max \left\{ {{X_1},...,{X_n}} \right\} \leqslant k} \right) \hfill \\ &= P\left( {\bigcap\limits_{i = 1}^n {{X_i} \leqslant k} } \right) \hfill \\ & = P{\left( {{X_1} \leqslant k} \right)^n} \hfill \\ \end{align*} 上述最後一條等式成立 因為 i.i.d 性質。故我們得到 \begin{align*} E[M] &= \sum\limits_{k \geqslant 0} {\left( {1 - P(M \leqslant k)} \right)} \hfill \\ &= \sum\limits_{k \geqslant 0} {\left( {1 - P{{\left( {{X_1} \leqslant k} \right)}^n}} \right)} \hfill \\ \end{align*}
If you can’t solve a problem, then there is an easier problem you can solve: find it. -George Polya