多项分布与泊松分布

本文主要用途为测试一下MathJax插件hhh

多项分布

定义1（多项分布）

称 \((X_{1},\cdots,X_{r})\) 服从参数为 \((n;p_{1},\cdots,p_{r})\) 的多项分布，若 \[ \mathbb{P}(X_{1}=k_{1},\cdots,X_{r}=k_{r})=\frac{n!}{k_{1}!\cdots k_{r}!}p_{1}^{k_{1}}\cdots p_{r}^{k_{r}} \] 这里 \(k_{1},\cdots,k_{r}\) 为非负整数，且满足 \(k_{1}+\cdots+k_{r}=n,p_{1}+\cdots+p_{r}=1\) 实际上对于上式中的系数，我们有 \[\begin{align} \binom{n}{k_{1}}\binom{n-k_{1}}{k_{2}}\cdots \binom{n-(k_{1}+\cdots+k_{r-1})}{k_{r}}& =\frac{n!}{k_{1}!(n-k_{1})!} \frac{(n-k_{1})!}{k_{2}!(n-k_{2})}\cdots \frac{(n-k_{1}-\cdots-k_{r-1})!}{(k_{r})!0!}\\ & =\frac{n!}{k_{1}!\cdots k_{r}!} \end{align}\] 这表明 \((X_{1},\cdots,X_{r})\) 发生 \((k_{1},\cdots,k_{r})\) 次的事件共发生 \(\frac{n!}{k_{1}!\cdots k_{r}!}\) 次

命题1

设称 \((X_{1},\cdots,X_{r})\) 服从参数为 \((n;p_{1},\cdots,p_{r})\) 的多项分布，则 \[ X_{i}+X_{j}\sim B(n,p_{i}+p_{j}) \]

证明：

为方便表示，我们不妨假设 \(X_{i}+X_{j}=X_{1}+X_{2}\) ，因为 \[\begin{align} \mathbb{P}(X_{1}=k_{1},X_{2}=k_{2})& =\sum\limits_{k_{3}+\cdots+k_{r}=n-(k_{1}+k_{2})}\frac{n!}{k_{1}!\cdots k_{r}!}p_{1}^{k_{1}}\cdots p_{r}^{k_{r}}\\ & =\frac{p_{1}^{k_{1}}p_{2}^{k_{2}}}{k_{1}!k_{2}!}\sum\limits_{k_{3}+\cdots+k_{r}=n-(k_{1}+k_{2})}\frac{n!}{k_{3}!\cdots k_{r}!}p_{3}^{k_{3}}\cdots p_{r}^{k_{r}}\\ & =\frac{p_{1}^{k_{1}}p_{2}^{k_{2}}(1-p_{1}-p_{2})^{n-k_{1}-k_{2}}}{k_{1}!k_{2}!}\cdot \frac{n!}{(n-k_{1}-k_{2})!}\sum\limits_{k_{3}+\cdots+k_{r}=n-(k_{1}+k_{2})}\frac{(n-k_{1}-k_{2})!}{k_{3}!\cdots k_{r}!}\left (\frac{p_{3}}{1-p_{1}-p_{2}}\right )^{k_{3}}\cdots \left (\frac{p_{r}}{1-p_{1}-p_{2}}\right )^{k_{r}}\\ & =\frac{n!}{k_{1}!k_{2}!(n-k_{1}-k_{2})!}p_{1}^{k_{1}}p_{2}^{k_{2}}(1-p_{1}-p_{2})^{n-k_{1}-k_{2}} \end{align}\] 所以，对 \(\forall 0\leq l\leq n\)，我们有 \[\begin{align} \mathbb{P}(X_{1}+X_{2}=l)& =\sum\limits_{k=0}^{l}\mathbb{P}(X_{1}=k,X_{2}=l-k)\\ & =\sum\limits_{k=0}^{l}\mathbb{P}(X_{1}=k)\mathbb{P}(X_{2}=l-k)\\ & =\sum\limits_{k=0}^{l}\frac{n!}{k!(l-k)!(n-l)!}p_{1}^{k}p_{2}^{l-k}(1-p_{1}-p_{2})^{n-l}\\ & =\sum\limits_{k=0}^{l}\frac{n!}{l!(n-l)!}\cdot \frac{l!}{k!(l-k)!}p_{1}^{k}p_{2}^{l-k}(1-p_{1}-p_{2})^{n-l}\\ & =\binom{n}{l}(p_{1}+p_{2})^{l}(1-p_{1}-p_{2})^{n-l}\\ \end{align}\] 这就说明了\(X_{1}+X_{2}\sim B(n,p_{1}+p_{2})\)

推论1

设称 \((X_{1},\cdots,X_{r})\) 服从参数为 \((n;p_{1},\cdots,p_{r})\) 的多项分布，则对 \(\forall 1\leq i_{1},\cdots,i_{k}\leq r\)，均有 \[ X_{i_{1}}+\cdots+X_{i_{k}}\sim B(n,p_{i_{1}}+\cdots+p_{i_{k}}) \]

证明：

对 \(k\) 进行归纳即可

泊松分布

定义2

称非负整数值离散型随机变量 \(X\) 服从参数为 \(\lambda>0\) 的泊松分布，是指 \(X\) 的分布列为 \[ \mathbb{P}(X=k)=\frac{\lambda^{k}}{k!}e^{-\lambda},\quad \forall k\in \mathbb{N} \] 记为\(X\sim P(\lambda)\)

命题2

设\(X\sim P(\lambda_{1}),Y\sim P(\lambda_{2})\)，且\(X,Y\)相互独立，则 \[ X+Y\sim P(\lambda_{1}+\lambda_{2}) \]

证明：

对\(\forall n\in\mathbb{N}\) \[\begin{align} \mathbb{P}(X+Y=n)& =\sum\limits_{k=0}^{n}\mathbb{P}(X=k)\mathbb{P}(Y=n-k)\\ & =\sum\limits_{k=0}^{n}\frac{\lambda_{1}^{k}}{k!}e^{-\lambda_{1}}\cdot \frac{\lambda_{2}^{n-k}}{(n-k)!}e^{-\lambda_{2}}\\ & =e^{-(\lambda_{1}+\lambda_{2})}\cdot \frac{(\lambda_{1}+\lambda_{2})^{n}}{n!}\sum\limits_{k=0}^{n}\frac{n!}{k!(n-k)!}\left (\frac{\lambda_{1}}{\lambda_{1}+\lambda_{2}}\right )^{k}\left (\frac{\lambda_{2}}{\lambda_{1}+\lambda_{2}}\right )^{n-k}\\ & =\frac{(\lambda_{1}+\lambda_{2})^{n}}{n!}e^{-(\lambda_{1}+\lambda_{2})} \end{align}\] 这说明 \(X+Y\sim P(\lambda_{1}+\lambda_{2})\)