最大值期望

设 $M=\\max(X_1,\\ldots,X_n)$。则\n\n$$\n\\mathbb{P}(M\\le t)=t^n,\\quad 0\\le t\\le 1.\n$$\n\n因此\n\n$$\n\\boxed{\\mathbb{E}[M]=\\int_0^1(1-t^n)dt=\\frac{n}{n+1}}.\n$$

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英文解析

Denote by $M$ the maximum of $X_{1}, X_{2}, \ldots , X_{n}$ . Since $M$ is a positive random variable with differentiable cumulative distribution function and $\mathbb{P}(M > 1) = 0$ ,

$$\mathbb{E}[M] = \int_{0}^{1}\mathbb{P}(M > t)dt$$

Since $X_{1}, X_{2}, \ldots , X_{n}$ , are independent,

$$\begin{array}{r l} & {\mathbb{P}(M > t) = 1 - \mathbb{P}(M\leq t)}\\ & {\qquad = 1 - \mathbb{P}\left(X_{1}\leq t,X_{2}\leq t,\ldots ,X_{n}\leq t\right)}\\ & {\qquad = 1 -}\\ & {\qquad \mathbb{P}\left(X_{1}\leq t\right)\mathbb{P}\left(X_{2}\leq t\right)\dots \mathbb{P}\left(X_{n}\leq t\right)}\\ & {\qquad = 1 - t^{n}} \end{array}$$

From (2.265) and (2.266), we conclude that

$$\begin{array}{l}{\mathbb{E}[M] = \int_{0}^{1}\left(1 - t^{n}\right)d t}\\ {= \left(1 - \frac{1}{n + 1} t^{n + 1}\right)\bigg|_{t = 0}^{t = 1}}\\ {= \frac{n}{n + 1}.} \end{array}$$