每次成功的概率为 $p$，失败的概率为 $1-p$
在 $n$ 次实验中恰好有 $k$ 次成功的概率为
$k$ 个成功乘以 $n-k$ 个失败

$$p^k (1-p)^{n-k}$$

而 $k$ 次成功可以出现在 $n$ 次实验中的任意位置，共有 $\binom{n}{k}$ 种情况
所以总概率为重复独立地同一实验 $n$ 次，其中某事件 $A$ 每次发生的概率为 $p$

$$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$

检查全概率和

$$\sum\limits_{k=0}^n P(X=k) = \sum\limits_{k=0}^n \binom{n}{k} p^k (1-p)^{n-k} = (p + 1 - p)^n = 1$$

$$\begin{aligned} E[X] &= \sum\limits_{k=0}^n k P(X=k) \\ &= \sum\limits_{k=0}^n k \binom{n}{k} p^k (1-p)^{n-k} \\ &= \sum\limits_{k=1}^n k \frac{n!}{k!(n-k)!} p^k (1-p)^{n-k} \\ &= \sum\limits_{k=1}^n \frac{n!}{(k-1)!(n-k)!} p^k (1-p)^{n-k} \\ &= np \sum\limits_{k=1}^n \frac{(n-1)!}{(k-1)!(n-k)!} p^{k-1} (1-p)^{(n-1)-(k-1)} \\ &= np \sum\limits_{j=0}^{n-1} \binom{n-1}{j} p^j (1-p)^{(n-1)-j} \\ &= np (p + 1 - p)^{n-1} \\ &= np \end{aligned}$$

$$\begin{aligned} V[X] &= E[(X - E[X])^2] \\ &= E[X^2] - (E[X])^2 \\ &= \sum\limits_{k=0}^n k^2 P(X=k) - (np)^2 \\ &= \sum\limits_{k=0}^n k^2 \binom{n}{k} p^k (1-p)^{n-k} - (np)^2 \\ &= \sum\limits_{k=1}^n k^2 \frac{n!}{k!(n-k)!} p^k (1-p)^{n-k} - (np)^2 \\ &= \sum\limits_{k=1}^n k \frac{n!}{(k-1)!(n-k)!} p^k (1-p)^{n-k} - (np)^2 \\ &= np \sum\limits_{k=1}^n k \frac{(n-1)!}{(k-1)!(n-k)!} p^{k-1} (1-p)^{(n-1)-(k-1)} - (np)^2 \\ &= np \sum\limits_{j=0}^{n-1} (j+1) \binom{n-1}{j} p^j (1-p)^{(n-1)-j} - (np)^2 \\ &= np \left( \sum\limits_{j=0}^{n-1} j \binom{n-1}{j} p^j (1-p)^{(n-1)-j} + \sum\limits_{j=0}^{n-1} \binom{n-1}{j} p^j (1-p)^{(n-1)-j} \right) - (np)^2 \\ &= np \left( (n-1)p + 1 \right) - (np)^2 \\ &= np(1-p) \end{aligned}$$

$$\begin{aligned} M_X(t) &= E[e^{tX}] \\ &= \sum\limits_{k=0}^n e^{tk} P(X=k) \\ &= \sum\limits_{k=0}^n e^{tk} \binom{n}{k} p^k (1-p)^{n-k} \\ &= \sum\limits_{k=0}^n \binom{n}{k} (pe^t)^k (1-p)^{n-k} \\ &= (pe^t + 1 - p)^n \end{aligned}$$

由于 Poisson 分布是基于二项分布的，所以首先有

$$P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$$

此处代入 $p = \frac{\lambda}{n} \quad$

$$P(X=k) = \binom{n}{k} \left(\frac{\lambda}{n}\right)^k \left(1-\frac{\lambda}{n}\right)^{n-k}$$

当 $n \to \infty$ 时

$$\frac{n!}{n^k(n-k)!} = (1 - \frac{1}{n})(1 - \frac{2}{n}) \cdots (1 - \frac{k-1}{n}) \to 1$$

$$(1 - \frac{\lambda}{n})^n \to e^{-\lambda}$$

所以当 $0 \leq k$ 被固定的时候

$$\lim\limits_{n \to \infty} \binom{n}{k} \left(\frac{\lambda}{n}\right)^k \left(1-\frac{\lambda}{n}\right)^{n-k} = \frac{\lambda^k e^{-\lambda}}{k!}$$

检查全概率和

$$\sum\limits_{k=0}^\infty P(X=k) = \sum\limits_{k=0}^\infty \frac{\lambda^k e^{-\lambda}}{k!} = e^{-\lambda} \sum\limits_{k=0}^\infty \frac{\lambda^k}{k!} = e^{-\lambda} e^{\lambda} = 1$$

$$\begin{aligned} E[X] &= \sum\limits_{k=0}^\infty k P(X=k) \\ &= \sum\limits_{k=0}^\infty k \frac{\lambda^k e^{-\lambda}}{k!} \\ &= \sum\limits_{k=1}^\infty k \frac{\lambda^k e^{-\lambda}}{k!} \\ &= \sum\limits_{k=1}^\infty \frac{\lambda^k e^{-\lambda}}{(k-1)!} \\ &= \lambda e^{-\lambda} \sum\limits_{k=1}^\infty \frac{\lambda^{k-1}}{(k-1)!} \\ &= \lambda e^{-\lambda} \sum\limits_{j=0}^\infty \frac{\lambda^j}{j!} \\ &= \lambda e^{-\lambda} e^{\lambda} \\ &= \lambda \end{aligned}$$

$$\begin{aligned} V[X] &= E[(X - E[X])^2] \\ &= E[X^2] - (E[X])^2 \\ &= \sum\limits_{k=0}^\infty k^2 P(X=k) - \lambda^2 \\ &= \sum\limits_{k=0}^\infty k^2 \frac {\lambda^k e^{-\lambda}}{k!} - \lambda^2 \\ &= \sum\limits_{k=1}^\infty k^2 \frac{\lambda^k e^{-\lambda}}{k!} - \lambda^2 \\ &= \sum\limits_{k=1}^\infty k \frac{\lambda^k e^{-\lambda}}{(k-1)!} - \lambda^2 \\ &= \lambda e^{-\lambda} \sum\limits_{k=1}^\infty k \frac{\lambda^{k-1}}{(k-1)!} - \lambda^2 \\ &= \lambda e^{-\lambda} \sum\limits_{j=0}^\infty (j+1) \frac{\lambda^j}{j!} - \lambda^2 \\ &= \lambda e^{-\lambda} \left( \sum\limits_{j=0}^\infty j \frac{\lambda^j}{j!} + \sum\limits_{j=0}^\infty \frac{\lambda^j}{j!} \right) - \lambda^2 \\ &= \lambda e^{-\lambda} (\lambda e^{\lambda} + e^{\lambda}) - \lambda^2 \\ &= \lambda (\lambda + 1) - \lambda^2 \\ &= \lambda \end{aligned} \quad$$

$$\begin{aligned} M_X(t) &= E[e^{tX}] \\ &= \sum\limits_{k=0}^\infty e^{tk} P(X=k) \\ &= \sum\limits_{k=0}^\infty e^{tk} \frac{\lambda^k e^{-\lambda}}{k!} \\ &= e^{-\lambda} \sum\limits_{k=0}^\infty \frac{(\lambda e^t)^k}{k!} \\ &= e^{-\lambda} e^{\lambda e^t} \\ &= e^{\lambda (e^t - 1)} \end{aligned}$$

每次失败的概率为 $1-p$，在前 $k-1$ 次实验中均失败，第 $k$ 次实验成功，所以直接用乘积得到概率

$$(1-p)^{k-1} p$$

检查全概率和

$$\sum\limits_{k=1}^\infty P(X=k) = \sum\limits_{k=1}^\infty (1-p)^{k-1} p = p \sum\limits_{k=0}^\infty (1-p)^k = p \cdot \frac{1}{1-(1-p)} = 1$$

$$\begin{aligned} E[X] &= \sum\limits_{k=1}^\infty k P(X=k) \\ &= \sum\limits_{k=1}^\infty k (1-p)^{k-1} p \\ &= p \sum\limits_{k=1}^\infty k (1-p)^{k-1} \\ &= p \cdot \frac{1}{(1-(1-p))^2} \\ &= \frac{1}{p} \end{aligned}$$

$$\begin{aligned} V[X] &= E[X^2] - (E[X])^2 \\ &= p \sum\limits_{k=1}^\infty k^2 (1-p)^{k-1} - \left(\frac{1}{p}\right)^2 \\ &= p \cdot \frac{1+(1-p)}{(1-(1-p))^3} - \frac{1}{p^2} \\ &= \frac{1-p}{p^2} \end{aligned}$$

$$\begin{aligned} M_X(t) &= E[e^{tX}] \\ &= \sum\limits_{k=1}^\infty e^{tk} P(X=k) \\ &= \sum\limits_{k=1}^\infty e^{tk} (1-p)^{k-1} p \\ &= p \cdot \frac{e^{t}}{1-(1-p)e^{t}},\quad t < -\ln(1-p) \end{aligned}$$