趣题: 包含(1, 0)的圆弧长度的期望和方差

趣题: 包含(1, 0)的圆弧 #

题目 #

在单位圆 $x^2 +y^2 =1$ 上按均匀分布随机取 $n$ 个点($n \geq 2$), 这 $n$ 个点可以把单位圆分成 $n$ 段圆弧, 求包含点 $(1.0)$ 的圆弧的长度的数学期望和方差.

Solution 1 #

不妨设 $n$ 个点的弧度分别为 $0\leq \xi_1, \xi_2, …\xi_n \leq 2\pi$ , 记 $\xi_1^{*}=min{\xi_1, …, \xi_n}$ , $\xi_n^{*}=max{\xi_1, …, \xi_n}$ , 那么包含 $(1, 0)$ 的圆弧的长度为 $\xi_1^{*} + 2\pi - \xi_n^{*}$ . $$P(\xi_1^{*} > x) = \prod_{1\leq i\leq n}P(\xi_i > x) = (1 - \frac{x}{2\pi})^n$$ $$P(\xi_n^{*} < x) = \prod_{1\leq i\leq n}P(\xi_i < x) = (\frac{x}{2\pi})^n$$ 计算 $E[\xi_1^{*}]$ : $$ E[\xi_1^{*}] = \int_{0}^{2\pi}xf(x)dx = \int_{0}^{2\pi}xdF(x) = \int_{0}^{2\pi}\int_{0}^{x}dydF(x) = \int_{0}^{2\pi}\int_{y}^{2\pi}df(x)dy = \int_{0}^{2\pi}P(\xi_1^{*} > y)dy = \int_{0}^{2\pi}(1 - \frac{y}{2\pi})^ndy = \frac{2\pi}{n+1} $$ 类似地, $$E[\xi_n^{*}] = 2\pi - \frac{2\pi}{n+1}$$ 故 $$E[\xi_1^{*} + 2\pi - \xi_n^{*}] = \frac{4\pi}{n + 1}$$

对于方差, 有 $Var[X] = E[X^2] - E^2[X]$ . 考虑计算 $E[(\xi_1^{*} + 2\pi - \xi_n^{*})^2] = 4\pi^2+4\pi E[xi_1^{*}] - 4\pi E[\xi_n^{*}] + E[{\xi_1^{*}}^2] + E[{\xi_n^{*}}^2] - 2E[\xi_1^{*} \xi_n^{*}]$ . 重点计算 $E[\xi_1^{*} \xi_n^{*}]$ . 考虑 $$F(x, y) = P(\xi_1^{*} \leq x, \xi_n^{*} \leq y) \= P(\xi_n^{*}\leq y) - P(x< \xi_1^{*}, \xi_n^{*}\leq y) \= P(\xi_n^{*}\leq y) - \prod_{1\leq i\leq n}P(x<\xi_i\leq y) \= (\frac{y}{2\pi})^n - (\frac{y - x}{2\pi})^n$$ 有 $$ f(x, y) = \frac{\partial^2 F(x, y)}{\partial x \partial y} =\frac{n(n - 1)}{(2\pi)^n}(y - x)^{n-2}$$ 计算 $$E[\xi_1^{*} \xi_n^{*}] = \int_{0}^{2\pi}\int_{0}^{2\pi}xyf(x, y)dxdy =\frac{4\pi^2}{n+2} $$ 对于 $E[{\xi_1^{*}}^2]$ , 类似 $E[\xi_1^{*}]$ , 计算得 $E[{\xi_1^{*}}^2] = \frac{8\pi^2}{(n + 1)(n+ 2)}$ . 同理 $E[{\xi_n^{*}}^2] = 4\pi^2 - \frac{8\pi^2}{n+ 2}$ . 故 $$Var(x) = \frac{8\pi^2(n -1)}{(n+1)^2(n+2)}$$