最小值期望

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

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英文解析

Denote by $M$ the minimum of $X_{1},X_{2},\ldots ,X_{n}$ . The expectation of a positive random variable with differentiable cumulative distribution function can be computed using the formula

$$\mathbb{E}[M] = \int_{0}^{\infty}\mathbb{P}(M > t)d t.$$

Since $\mathbb{P}(M > 1) = 0$ , it follows from (2.261) that

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

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

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

From (2.262) and (2.263), we conclude that

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