56 lines
2.5 KiB
TeX
56 lines
2.5 KiB
TeX
\chapter{组合概率}
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\section{事件与概率}
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略。
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\section{独立重复试验}
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略。
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\section{大数定律}
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抛一枚硬币$n$次，正面朝上的次数记为$X$，有$X \sim B \left(n, \dfrac{1}{2}\right)$。对$k \in [0, n]$，有$P(X = k) = \dfrac{\binom{n}{k}}{2^n}$。
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\begin{theorem}[Bernolli大数定律]\label{Bernolli's law of large numbers}
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$\forall \varDelta > 0$，当$n \to \infty$有 \[P\left(\left|\dfrac{X}{n} - \dfrac{1}{2}\right| < \varepsilon\right) \to 1 \eqper\]
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\end{theorem}
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\begin{theorem}\label{corollary of Bernolli's law of large numbers}
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若$X \sim B \left(2n, \dfrac{1}{2}\right)$，则$\forall t \in [0, n]$，有
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\[P(\vert x - n \vert > t) \leq e^{-\frac{t^2}{n+t}} \eqper\]
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\end{theorem}
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我们先证明定理\ref{corollary of Bernolli's law of large numbers}。
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\begin{proof}
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注意到
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\begin{equation*}
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\begin{aligned}
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P(\vert X - n \vert > t) & = P(X < n - t \text{或} X > n + t)\\
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& = P(X = 0) + P(X = 1) + \cdots + P(X = n - t - 1)\\
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& \ \ \ + P(X = n + t + 1) + \cdots + P(X = 2n)\\
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& = 2[P(X = 0) + \cdots + P(X = n - t - 1)]\\
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& = \frac{2\left[\binom{2n}{0} + \cdots + \binom{2n}{n-t-1}\right]}{2^{2n}}\\
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\end{aligned}
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\end{equation*}
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应用引理\ref{lemma for Bernolli's law of large numbers}
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\begin{equation*}
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\begin{aligned}
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P(\vert X - n \vert > t) & < \binom{2n}{n-t} \bigg/ \binom{2n}{n}\\
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& \leq e^{-\frac{t^2}{n+t}}\eqper
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\end{aligned}
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\end{equation*}
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\end{proof}
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我们可利用定理\ref{corollary of Bernolli's law of large numbers}证明定理\ref{Bernolli's law of large numbers}。
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\begin{proof}
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我们需要证明
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\[n \to \infty \text{时} P\left(\left|\frac{X}{2n} - \frac{1}{2} \right| < \varepsilon \right) \to 1\]
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只需证
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\[n \to \infty \text{时} P\left(\left|\frac{X}{2n} - \frac{1}{2} \right| > \varepsilon \right) \to 0\]
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注意到
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\begin{equation*}
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P\left(\left|\frac{X}{2n} - \frac{1}{2} \right| > \varepsilon \right) = P\left(\left|X - n\right| > 2n\varepsilon \right)
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\end{equation*}
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应用\ref{corollary of Bernolli's law of large numbers}，这里的$2n\varepsilon$即为定理中的$t$，从而
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\begin{equation*}
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P\left(\left|X - n\right| > 2n\varepsilon \right) \leq e^{- \frac{4n^2\varepsilon^2}{n+2n\varepsilon}}
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\end{equation*}
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当$n \to \infty$时，$e^{- \frac{4n^2\varepsilon^2}{n+2n\varepsilon}} \to 0$，因此$P\left(\left|X - n\right| > 2n\varepsilon \right) \to 0$。
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\end{proof} |