概率良定性
这是显然的，因为每个取值的概率均为 $\dfrac{1}{n}$，所以概率和为 $1$

期望值

$$\begin{aligned} E[X] &= \sum\limits_{i=1}^n x_i P(X = x_i) \\ &= \sum\limits_{i=1}^n x_i \dfrac{1}{n} \\ &= \dfrac{1}{n} \sum\limits_{i=1}^n x_i \\ &= \dfrac{1}{n} \cdot \dfrac{n(n+1)}{2} \\ &= \dfrac{n+1}{2} \end{aligned}$$

方差

$$\begin{aligned} V[X] &= E[(X - E[X])^2] \\ &= E[X^2] - (E[X])^2 \\ &= \sum\limits_{i=1}^n x_i^2 P(X = x_i) - \left(\dfrac{n+1}{2}\right)^2 \\ &= \sum\limits_{i=1}^n x_i^2 \dfrac{1}{n} - \dfrac{(n+1)^2}{4} \\ &= \dfrac{1}{n} \cdot \dfrac{n(n+1)(2n+1)}{6} - \dfrac{(n+1)^2}{4} \\ &= \dfrac{(n+1)(2n+1)}{6} - \dfrac{(n+1)^2}{4} \\ &= \dfrac{(n+1)(4n+2 - 3n - 3)}{12} \\ &= \dfrac{(n+1)(n-1)}{12} \\ &= \dfrac{n^2 - 1}{12} \end{aligned}$$

矩母函数

$$\begin{aligned} M_X(t) &= E[e^{tX}] = \sum\limits_{i=1}^n e^{t x_i} P(X = x_i) \\ &= \sum\limits_{i=1}^n e^{t x_i} \cdot \dfrac{1}{n} \\ &= \dfrac{1}{n} \sum\limits_{i=1}^n e^{t x_i} \\ &= \dfrac{1}{n} e^t \cdot \dfrac{1 - e^{nt}}{1 - e^t} \\ &= \dfrac{e^t (1 - e^{nt})}{n(1 - e^t)} \end{aligned}$$

$\square$

概率良定性
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

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

期望值

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

方差

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

矩母函数

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

$\square$

概率良定性
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

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

期望值

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

方差

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

矩母函数

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

$\square$

概率良定性
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

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

期望值

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

方差

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

矩母函数

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

$\square$

$$\begin{aligned} P(X \gt m + n \mid X \gt m) &= \frac{P(X \gt m + n \cap X \gt m)}{P(X \gt m)} \\ &= \frac{P(X \gt m + n)}{P(X \gt m)} \\ &= \frac{\sum\limits_{k=m+n+1}^\infty (1-p)^{k-1} p}{\sum\limits_{k=m+1}^\infty (1-p)^{k-1} p} \\ &= \frac{(1-p)^{m+n} \sum\limits_{k=1}^\infty (1-p)^{k-1} p}{(1-p)^m \sum\limits_{k=1}^\infty (1-p)^{k-1} p} \\ &= (1-p)^n \\ &= P(X \gt n) \end{aligned}$$

$\square$

概率良定性
均匀分布的概率密度函数处处相等，所以如果设某一点的概率为 $c$ 的话，根据全概率和为 1

$$\int_a^b c \, dx = c(b-a) = 1 \implies c = \frac{1}{b-a}$$

所以概率密度函数为

$$f(x) = \begin{cases} \frac{1}{b-a}, & a \leq x \leq b \\ 0, & \text{otherwise} \end{cases}$$

期望值

$$\begin{aligned} E[X] &= \int_{-\infty}^{+\infty} x f(x) \, dx \\ &= \int_a^b x \cdot \frac{1}{b-a} \, dx \\ &= \frac{1}{b-a} \cdot \frac{b^2 - a^2}{2} \\ &= \frac{a+b}{2} \end{aligned}$$

方差

$$\begin{aligned} V[X] &= E[X^2] - (E[X])^2 \\ &= \int_{-\infty}^{+\infty} x^2 f(x) \, dx - \left(\frac{a+b}{2}\right)^2 \\ &= \int_a^b x^2 \cdot \frac{1}{b-a} \, dx - \frac{(a+b)^2}{4} \\ &= \frac{1}{b-a} \cdot \frac{b^3 - a^3}{3} - \frac{(a+b)^2}{4} \\ &= \frac{b^2 + ab + a^2}{3} - \frac{(a+b)^2}{4} \\ &= \frac{b^2 + ab + a^2}{3} - \frac{a^2 + 2ab + b^2}{4} \\ &= \frac{4(b^2 + ab + a^2) - 3(a^2 + 2ab + b^2)}{12} \\ &= \frac{b^2 - 2ab + a^2}{12} = \frac{(b-a)^2}{12} \end{aligned}$$

矩母函数

$$\begin{aligned} M_X(t) &= E[e^{tX}] = \int_{-\infty}^{+\infty} e^{tx} f(x) \, dx \\ &= \int_a^b e^{tx} \cdot \frac{1}{b-a} \, dx \\ &= \frac{1}{b-a} \cdot \frac{e^{tb} - e^{ta}}{t},\quad t \neq 0 \\ &= \frac{e^{tb} - e^{ta}}{t(b-a)} \end{aligned}$$

$\square$

概率良定性
设 $\lambda>0$。在 $x < 0$ 时，事件 $A$ 不可能发生，所以概率密度函数为 $0$。
在 $x \geq 0$ 时，定义指数分布的尾分布为 $P(X > x)=e^{-\lambda x}$。因此

$$P(X \leq x) = 1 - P(X > x) = 1 - e^{-\lambda x}$$

对其求导即得概率密度函数

$$f(x) = \begin{cases} \lambda e^{-\lambda x}, & x \geq 0 \\ 0, & \text{otherwise} \end{cases}$$

检查全概率和

$$\int_{-\infty}^{+\infty} f(x) dx = \int_0^{+\infty} \lambda e^{-\lambda x} dx = 1$$

期望值

$$\begin{aligned} E[X] &= \int_{-\infty}^{+\infty} x f(x) \, dx \\ &= \int_0^{+\infty} x \cdot \lambda e^{-\lambda x} \, dx \\ &= \left[-x e^{-\lambda x}\right]_0^{+\infty} + \int_0^{+\infty} e^{-\lambda x} \, dx \\ &= 0 + \left[-\frac{1}{\lambda} e^{-\lambda x}\right]_0^{+\infty} \\ &= \frac{1}{\lambda} \end{aligned}$$

方差

$$\begin{aligned} V[X] &= E[X^2] - (E[X])^2 \\ &= \int_{-\infty}^{+\infty} x^2 f(x) \, dx - \left(\frac{1}{\lambda}\right)^2 \\ &= \int_0^{+\infty} x^2 \cdot \lambda e^{-\lambda x} \, dx - \frac{1}{\lambda^2} \\ &= \left[-x^2 e^{-\lambda x}\right]_0^{+\infty} + \int_0^{+\infty} 2x e^{-\lambda x} \, dx - \frac{1}{\lambda^2} \\ &= 0 + 2\left[-\frac{x}{\lambda} e^{-\lambda x}\right]_0^{+\infty} + \int_0^{+\infty} \frac{2}{\lambda} e^{-\lambda x} \, dx - \frac{1}{\lambda^2} \\ &= 0 + 0 + \left[-\frac{2}{\lambda^2} e^{-\lambda x}\right]_0^{+\infty} - \frac{1}{\lambda^2} \\ &= \frac{2}{\lambda^2} - \frac{1}{\lambda^2} = \frac{1}{\lambda^2} \end{aligned}$$

矩母函数

$$\begin{aligned} M_X(t) &= E[e^{tX}] = \int_{-\infty}^{+\infty} e^{tx} f(x) \, dx \\ &= \int_0^{+\infty} e^{tx} \cdot \lambda e^{-\lambda x} \, dx \\ &= \lambda \int_0^{+\infty} e^{-(\lambda - t)x} \, dx \\ &= \lambda \left[-\frac{1}{\lambda - t} e^{-(\lambda - t)x}\right]_0^{+\infty},\quad t < \lambda \\ &= \frac{\lambda}{\lambda - t} \end{aligned}$$

$\square$