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# 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}
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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

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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。

$$\开始{对齐} P(X+Y=k) &=\sum_{j=0}^{k} P(X+Y=k \mid X=j) P(X=j) \ &=\sum_{j=0}^{k} P(Y=k-j) P(X=j) \ &=\sum_{j=0}^{k} \frac{e^{-\lambda_{2}} \lambda_{2}^{kj}}{(kj) !} \frac{e^{-\ lambda_{1}}}{j_{1}^{j}} \ &=\frac{e^{-\left(\lambda_{1}+\lambda_{2}\right)}}{k !} \sum_{j=0}^{k}\left(\begin{array }{C} k \ j \end{数组}\right) \lambda_{1}^{j} \lambda_{2}^{k-j} \ &=\frac{e^{-\left(\lambda_{1}+\lambda_{2}\right)}\left(\lambda_{1}+\lambda_{2}\right)^{k}}{克！} \end{对齐}$$
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$P(X=k \mid X+Y=n)=\frac{P(X+Y=n \mid X=k) P(X=k)}{P(X+Y=n)}$ $=\frac{P(Y=nk) P(X=k)}{P(X+Y=n)} .$
$\operatorname{Pois}\left(\lambda_{1}+\lambda_{2}\right)$ 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}