https://sichen-wang.github.io/tags/expected-value/Expected ValueHugoBlox Kit (https://hugoblox.com)en-usTue, 27 Aug 2024 19:42:49 +0000https://sichen-wang.github.io/media/icon_hu_9dfcb674aa885f38.pnghttps://sichen-wang.github.io/tags/expected-value/https://sichen-wang.github.io/blog/%E5%A4%9A%E6%AC%A1%E5%8F%96-0-1-%E4%B9%8B%E9%97%B4%E7%9A%84%E9%9A%8F%E6%9C%BA%E5%AE%9E%E6%95%B0-%E6%9C%9F%E6%9C%9B%E5%A4%9A%E5%B0%91%E6%AC%A1%E5%90%8E%E5%85%B6%E5%92%8C%E6%81%B0%E4%B8%8D%E5%B0%8F%E4%BA%8E-1/Tue, 27 Aug 2024 19:42:49 +0000https://sichen-wang.github.io/blog/%E5%A4%9A%E6%AC%A1%E5%8F%96-0-1-%E4%B9%8B%E9%97%B4%E7%9A%84%E9%9A%8F%E6%9C%BA%E5%AE%9E%E6%95%B0-%E6%9C%9F%E6%9C%9B%E5%A4%9A%E5%B0%91%E6%AC%A1%E5%90%8E%E5%85%B6%E5%92%8C%E6%81%B0%E4%B8%8D%E5%B0%8F%E4%BA%8E-1/<p>设事件 \(\omega_{n}\) 表示随机取 \(n\) 个 \((0, 1)\) 中的实数, 其和小于 \(1\).</p> \[ \begin{aligned} E &= \sum_{n = 1}^{\infty} n P(\omega_{n - 1} \overline{\omega_n}) \\ &= \sum_{n = 1}^{\infty} n \left[ P(\omega_{n - 1}) - P(\omega_{n - 1}\omega_n) \right] \\ &= \sum_{n = 1}^{\infty} n \left[ P(\omega_{n - 1}) - P(\omega_n) \right] \\ &= 1 + \sum_{n = 1}^{\infty} P(\omega_n) \end{aligned} \]<p>考虑根据定义求 \(\omega_{n}\):</p> \[ \omega_n = \int_{0}^1 \mathrm{d}x_1 \int_{0}^{1 - x_1} \mathrm{d}x_2 \int_{0}^{1 - x_1 -x_2} \mathrm{d}x_3 \cdots \int_{0}^{1 - \sum_{i \lt n} x_i} \mathrm{d}x_n \]<p>将 \(1\) 用形式变量 \(x\) 代替, 即得</p> \[ \omega_n(x) = \int_{0}^x \mathrm{d}x_1 \int_{0}^{x - x_1} \mathrm{d}x_2 \int_{0}^{x - x_1 -x_2} \mathrm{d}x_3 \cdots \int_{0}^{x - \sum_{i \lt n} x_i} \mathrm{d}x_n \]<p>考虑寻找 \(\omega_{n + 1}\) 和 \(\omega_{n}\) 之间的关系, 有</p> \[ \omega_{n + 1}(x) = \int_{0}^{x} \omega_n(x - x_{n + 1}) \mathrm{d}x_{n + 1} \]<p>注意到在积分内部, \(\mathrm{d}(x - x_{n + 1})\) 与 \(\mathrm{d}x_{n + 1}\) 实际上是等价的, 于是得到递推式</p> \[ \omega_{n + 1}(x) = \int_{0}^{x} \omega_n(t) \mathrm{d}t \]<p>根据定义 \(\omega_{1}(x) = x\), 于是归纳得</p> \[ \omega_{n}(x) = \int_{0}^{x} \frac{t^{n - 1}}{(n - 1)!} \mathrm{d}t = \frac{x^n}{n!} \]<p>将 \(x\) 换回 \(1\), 回代得</p> \[ E = 1 + \sum_{n = 1}^{\infty} \frac{1}{n!} = \mathrm{e} \]