统计代写| Connections between Poisson and Binomial stat代写
统计代考
4.8 Connections between Poisson and Binomial
The Poisson and Binomial distributions are closely connected, and their relationship is exactly parallel to the relationship between the Binomial and Hypergeometric distributions that we examined in the previous chapter: we can get from the Poisson to the Binomial by conditioning, and we can get from the Binomial to the Poisson by taking a limit.
Our results will rely on the fact that the sum of independent Poissons is Poisson, just as the sum of independent Binomials is Binomial. We’ll prove this result using the law of total probability for now; in Chapter 6 we’ll learn a faster method that uses a tool called the moment generating function. Chapter 13 gives further insight into these results.
Theorem $4.8 .1$ (Sum of independent Poissons). If $X \sim \operatorname{Pois}\left(\lambda_{1}\right), Y \sim \operatorname{Pois}\left(\lambda_{2}\right)$, and $X$ is independent of $Y$, then $X+Y \sim \operatorname{Pois}\left(\lambda_{1}+\lambda_{2}\right)$
Proof. To get the PMF of $X+Y$, condition on $X$ and use the law of total probability:
$$
\begin{aligned}
P(X+Y=k) &=\sum_{j=0}^{k} P(X+Y=k \mid X=j) P(X=j) \
&=\sum_{j=0}^{k} P(Y=k-j) P(X=j) \
&=\sum_{j=0}^{k} \frac{e^{-\lambda_{2}} \lambda_{2}^{k-j}}{(k-j) !} \frac{e^{-\lambda_{1}}}{j_{1}^{j}} \
&=\frac{e^{-\left(\lambda_{1}+\lambda_{2}\right)}}{k !} \sum_{j=0}^{k}\left(\begin{array}{c}
k \
j
\end{array}\right) \lambda_{1}^{j} \lambda_{2}^{k-j} \
&=\frac{e^{-\left(\lambda_{1}+\lambda_{2}\right)}\left(\lambda_{1}+\lambda_{2}\right)^{k}}{k !}
\end{aligned}
$$
182
The last step used the binomial theorem. Since we’ve arrived at the Pois $\left(\lambda_{1}+\lambda_{2}\right)$ PMF, we have $X+Y \sim \operatorname{Pois}\left(\lambda_{1}+\lambda_{2}\right)$.
The story of the Poisson distribution provides intuition for this result. If there are two different types of events occurring at rates $\lambda_{1}$ and $\lambda_{2}$, independently, then the overall event rate is $\lambda_{1}+\lambda_{2}$.
Theorem 4.8.2 (Poisson given a sum of Poissons). If $X \sim$ Pois $\left(\lambda_{1}\right), Y \sim$ Pois $\left(\lambda_{2}\right)$, and $X$ is independent of $Y$, then the conditional distribution of $X$ given $X+Y=n$ is $\operatorname{Bin}\left(n, \lambda_{1} /\left(\lambda_{1}+\lambda_{2}\right)\right)$.
Proof. Exactly as in the corresponding proof for the Binomial and Hypergeometric, we use Bayes’ rule to compute the conditional PMF $P(X=k \mid X+Y=n)$ :
$P(X=k \mid X+Y=n)=\frac{P(X+Y=n \mid X=k) P(X=k)}{P(X+Y=n)}$ $=\frac{P(Y=n-k) P(X=k)}{P(X+Y=n)} .$
$\operatorname{Pois}\left(\lambda_{1}+\lambda_{2}\right)$ by the previous theorem. This gives $$ \begin{aligned} P(X=k \mid X+Y=n) &=\frac{\left(\frac{e^{-\lambda_{2}} \lambda_{2}^{n-k}}{(n-k) !}\right)\left(\frac{e^{-\lambda_{1}} \lambda_{1}^{k}}{k !}\right)}{\frac{e^{-\left(\lambda_{1}+\lambda_{2}\right)}\left(\lambda_{1}+\lambda_{2}\right)^{n}}{n !}} \ &=\left(\begin{array}{l}n \ k\end{array}\right) \frac{\lambda_{1}^{k} \lambda_{2}^{n-k}}{\left(\lambda_{1}+\lambda_{2}\right)^{n}} \end{aligned} $$
统计代考
4
4.8 泊松和二项式之间的联系
Poisson 分布和 Binomial 分布紧密相连，它们的关系与我们在前一章研究的 Binomial 和 Hypergeometric 分布之间的关系完全平行：我们可以通过条件化从 Poisson 到 Binomial，我们可以从通过取极限对泊松进行二项式。
我们的结果将依赖于独立泊松之和是泊松这一事实，就像独立二项式之和是二项式一样。我们现在将使用全概率定律来证明这个结果；在第 6 章中，我们将学习一种使用称为矩生成函数的工具的更快方法。第 13 章对这些结果进行了进一步的了解。
定理 $4.8 .1$（独立泊松之和）。如果$X \sim \operatorname{Pois}\left(\lambda_{1}\right)，Y \sim \operatorname{Pois}\left(\lambda_{2}\right)$，并且$X$独立于$Y$，然后 $X+Y \sim \operatorname{Pois}\left(\lambda_{1}+\lambda_{2}\right)$
证明。要获得 $X+Y$ 的 PMF，以 $X$ 为条件并使用总概率定律：
$$
\开始{对齐}
P(X+Y=k) &=\sum_{j=0}^{k} P(X+Y=k \mid X=j) P(X=j) \
&=\sum_{j=0}^{k} P(Y=k-j) P(X=j) \
&=\sum_{j=0}^{k} \frac{e^{-\lambda_{2}} \lambda_{2}^{kj}}{(kj) !} \frac{e^{-\ lambda_{1}}}{j_{1}^{j}} \
&=\frac{e^{-\left(\lambda_{1}+\lambda_{2}\right)}}{k !} \sum_{j=0}^{k}\left(\begin{array }{C}
k \
j
\end{数组}\right) \lambda_{1}^{j} \lambda_{2}^{k-j} \
&=\frac{e^{-\left(\lambda_{1}+\lambda_{2}\right)}\left(\lambda_{1}+\lambda_{2}\right)^{k}}{克！}
\end{对齐}
$$
182
最后一步使用了二项式定理。由于我们已经到达 Pois $\left(\lambda_{1}+\lambda_{2}\right)$ PMF，我们有 $X+Y \sim \operatorname{Pois}\left(\lambda_{1} +\lambda_{2}\right)$。
泊松分布的故事为这个结果提供了直觉。如果有两种不同类型的事件以 $\lambda_{1}$ 和 $\lambda_{2}$ 的速率独立发生，则总体事件速率为 $\lambda_{1}+\lambda_{2}$。
定理 4.8.2（泊松给定泊松的总和）。如果$X \sim$ Pois $\left(\lambda_{1}\right)，Y \sim$ Pois $\left(\lambda_{2}\right)$，并且$X$独立于$Y$，那么给定 $X+Y=n$ 的 $X$ 的条件分布是 $\operatorname{Bin}\left(n, \lambda_{1} /\left(\lambda_{1}+\lambda_{2}\right )\对)$。
证明。正如在二项式和超几何的相应证明中一样，我们使用贝叶斯规则来计算条件 PMF $P(X=k \mid X+Y=n)$ ：
$P(X=k \mid X+Y=n)=\frac{P(X+Y=n \mid X=k) P(X=k)}{P(X+Y=n)}$ $ =\frac{P(Y=nk) P(X=k)}{P(X+Y=n)} .$
$\operatorname{Pois}\left(\lambda_{1}+\lambda_{2}\right)$ 由前面的定理。这给出了 $$ \begin{aligned} P(X=k \mid X+Y=n) &=\frac{\left(\frac{e^{-\lambda_{2}} \lambda_{2}^{ nk}}{(nk) !}\right)\left(\frac{e^{-\lambda_{1}} \lambda_{1}^{k}}{k !}\right)}{\frac{ e^{-\left(\lambda_{1}+\lambda_{2}\right)}\left(\lambda_{1}+\lambda_{2}\right)^{n}}{n !}} \ \ &=\left(\begin{array}{l}n \ k\end{array}\right) \frac{\lambda_{1}^{k} \lambda_{2}^{nk}}{\左(\lambda_{1}+\lambda_{2}\right)^{n}} \end{对齐} $$
。
R语言代写
统计代写|SAMPLE SPACES AND PEBBLE WORLD stat 代写 请认准UprivateTA™. UprivateTA™为您的留学生涯保驾护航。
