算法乐园主页笔记刷题

点击开启白天模式：whitemode

概率论与数理统计

高数预备知识

\int_{- \infty}^{+ \infty} e^{- x^{2}} d x = \sqrt{π}

伽马函数

Γ (α) = \int_{0}^{+ \infty} x^{α - 1} e^{- x} d x = (α - 1) Γ (α - 1)

Γ (\frac{1}{2}) = \sqrt{π}

$(p>0,q>0)$

B (p, q) = \frac{Γ (p) Γ (q)}{Γ (p + q)} = \int_{0}^{1} x^{p - 1} (1 - x)^{q - 1} d x

分布函数

0-1分布

X \sim b (1, p)

P {X = k} = p^{k} (1 - p)^{1 - k}, k = 0, 1

E (X) = p

D (X) = p (1 - p)

二项分布

X \sim b (n, p)

P {X = k} = C_{n}^{k} p^{k} (1 - p)^{n - k}, k = 0, 1, . . ., n

E (X) = n p

D (X) = n p (1 - p)

泊松分布

X \sim π (λ)

P {X = k} = \frac{λ^{k} e^{- λ}}{k!}, k = 0, 1, 2, . . .

E (X) = λ

D (X) = λ

均匀分布

X \sim U (a, b)

\begin{matrix} f (x) = {\begin{cases} \begin{aligned} \frac{1}{b - a} & , b < x < a \\ 0 & , e l s e \end{aligned} \end{cases} \end{matrix}

E (X) = \frac{a + b}{2}

D (X) = \frac{(b - a)^{2}}{12}

指数分布

\begin{matrix} f (x) = {\begin{cases} \begin{aligned} \frac{1}{θ} e^{- \frac{x}{θ}} & , x > 0 \\ 0 & , e l s e \end{aligned} \end{cases} \end{matrix}

E (X) = θ

D (X) = θ^{2}

正态分布

X \sim N (μ, σ^{2})

f (x) = \frac{1}{\sqrt{2 π} σ} e^{- \frac{(x - μ)^{2}}{2 σ^{2}}}

E (X) = μ

D (X) = σ^{2}

抽样分布

卡方分布

$X_1,X_2,...,X_n$ $N(0,1)$ 的样本，则

χ^{2} = X_{1}^{2} + X_{2}^{2} + . . . + X_{n}^{2} \sim χ^{2} (n)

$\chi_1^2\sim\chi^2(n_1)$ $\chi_2^2\sim\chi^2(n_2)$ ，则

χ_{1}^{2} + χ_{2}^{2} \sim χ^{2} (n_{1} + n_{2})

$\chi^2\sim\chi^2(n_1)$ , 则

E (χ^{2}) = n

D (χ^{2}) = 2 n

上分位数

P {χ^{2} > χ_{α}^{2} (n)} = α

t分布

$X\sim N(0,1)$ $Y\sim\chi^2(n)$ ，则：

\frac{X}{\sqrt{\frac{Y}{n}}} \sim t (n)

上分位数

P {t > t_{α} (n)} = α

t_{1 - α} (n) = - t_{α} (n)

F分布

$U\sim \chi^2(n_1)$ $V\sim \chi^2(n_2)$ , 则：

\frac{\frac{U}{n_{1}}}{\frac{V}{n_{2}}} \sim F (n_{1}, n_{2})

\frac{1}{F} \sim F (n_{2}, n_{1})

上分位数

P {F > F_{α} (n_{1}, n_{2})} = α

F_{1 - α} (n_{1}, n_{2}) = \frac{1}{F_{α} (n_{2}, n_{1})}

正态分布的样本均值和样本方差的分布

$X_1,X_2,...,X_n$ $\mu$ $\sigma^2$ ，则

E (\overset{―}{X}) = μ

D (\overset{―}{X}) = \frac{σ^{2}}{n}

E (S^{2}) = σ^{2}

$X_1,X_2,...,X_n$ $N(\mu,\sigma ^2)$ 的样本，则

\overset{―}{X} \sim N (μ, \frac{σ^{2}}{n})

\frac{(n - 1) S^{2}}{σ^{2}} \sim χ^{2} (n - 1)

$\overline X$ $S^2$ 相互独立

\frac{\overset{―}{X} - μ}{\frac{S}{\sqrt{n}}} \sim t (n - 1)

$X_1,X_2,...,X_{n_1}$ $N(\mu_1,\sigma_1 ^2)$ $Y_1,Y_2,...,Y_{n_2}$ $N(\mu_2,\sigma_2 ^2)$ 的样本, 两个样本相互独立，则

两个样本的平均值：

\overset{―}{X} = \frac{1}{n_{1}} \sum_{i = 1}^{n_{1}} X_{i}

\overset{―}{Y} = \frac{1}{n_{2}} \sum_{i = 1}^{n_{2}} Y_{i}

两个样本的样本方差

S_{1}^{2} = \frac{1}{n_{1} - 1} \sum_{i = 1}^{n_{1}} (X_{i} - \overset{―}{X})^{2}

S_{2}^{2} = \frac{1}{n_{2} - 1} \sum_{i = 1}^{n_{2}} (Y_{i} - \overset{―}{Y})^{2}

满足以下性质

\frac{\frac{S_{1}^{2}}{S_{2}^{2}}}{\frac{σ_{1}^{2}}{σ_{2}^{2}}} \sim F (n_{1} - 1, n_{2} - 1)

$\sigma_1^2=\sigma_2^2=\sigma^2$ 时，

\frac{(\overset{―}{X} - \overset{―}{Y}) - (μ_{1} - μ_{2})}{S_{W} \sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}}} \sim t (n_{1} + n_{2} - 2)

其中

S_{W}^{2} = \frac{(n_{1} - 1) S_{1}^{2} + (n_{2} - 1) S_{2}^{2}}{n_{1} + n_{2} - 2}

正态总体均值、方差的置信区间与单侧置信限

$1-\alpha$ ）

单个正态总体

$\mu$ $\sigma^2$

根据

\frac{\overset{―}{X} - μ}{σ} \sqrt{n} \sim N (0, 1)

置信区间

(\overset{―}{X} - \frac{σ}{\sqrt{n}} z_{\frac{α}{2}}, \overset{―}{X} + \frac{σ}{\sqrt{n}} z_{\frac{α}{2}})

置信上界

\overset{―}{μ} = \overset{―}{X} + \frac{σ}{\sqrt{n}} z_{α}

置信下界

\underset{―}{μ} = \overset{―}{X} - \frac{σ}{\sqrt{n}} z_{α}

$\mu$ $\sigma^2$

根据

\frac{\overset{―}{X} - μ}{S} \sqrt{n} \sim t (n - 1)

置信区间

(\overset{―}{X} - \frac{S}{\sqrt{n}} t_{\frac{α}{2}} (n - 1), \overset{―}{X} + \frac{S}{\sqrt{n}} t_{\frac{α}{2}} (n - 1))

置信上界

\overset{―}{μ} = \overset{―}{X} + \frac{S}{\sqrt{n}} t_{α} (n - 1)

置信下界

\underset{―}{μ} = \overset{―}{X} - \frac{S}{\sqrt{n}} t_{α} (n - 1)

$\sigma^2$ $\mu$

根据

\frac{(n - 1) S^{2}}{σ^{2}} \sim χ^{2} (n - 1)

置信区间

(\frac{(n - 1) S^{2}}{χ_{\frac{α}{2}}^{2} (n - 1)}, \frac{(n - 1) S^{2}}{χ_{1 - \frac{α}{2}}^{2} (n - 1)})

置信上界

\overset{―}{σ^{2}} = \frac{(n - 1) S^{2}}{χ_{1 - α}^{2} (n - 1)}

置信下界

\underset{―}{σ^{2}} = \frac{(n - 1) S^{2}}{χ_{α}^{2} (n - 1)}

两个正态总体

$\mu_1-\mu_2$ $\sigma_1^2$ $\sigma_2^2$

根据

\frac{(\overset{―}{X} - \overset{―}{Y}) - (μ_{1} - μ_{2})}{\sqrt{\frac{σ_{1}^{2}}{n_{1}} + \frac{σ_{2}^{2}}{n_{2}}}} \sim N (0, 1)

置信区间

(\overset{―}{X} - \overset{―}{Y} - z_{\frac{α}{2}} \sqrt{\frac{σ_{1}^{2}}{n_{1}} + \frac{σ_{2}^{2}}{n_{2}}}, \overset{―}{X} - \overset{―}{Y} + z_{\frac{α}{2}} \sqrt{\frac{σ_{1}^{2}}{n_{1}} + \frac{σ_{2}^{2}}{n_{2}}})

置信上界

\overset{―}{μ_{1} - μ_{2}} = \overset{―}{X} - \overset{―}{Y} + z_{α} \sqrt{\frac{σ_{1}^{2}}{n_{1}} + \frac{σ_{2}^{2}}{n_{2}}}

置信下界

\underset{―}{μ_{1} - μ_{2}} = \overset{―}{X} - \overset{―}{Y} - z_{α} \sqrt{\frac{σ_{1}^{2}}{n_{1}} + \frac{σ_{2}^{2}}{n_{2}}}

$\mu_1-\mu_2$ $\sigma_1^2=\sigma_2^2=\sigma^2$

根据

\frac{(\overset{―}{X} - \overset{―}{Y}) - (μ_{1} - μ_{2})}{S_{W} \sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}}} \sim t (n_{1} + n_{2} - 2)

其中：

S_{W}^{2} = \frac{(n_{1} - 1) S_{1}^{2} + (n_{2} - 1) S_{2}^{2}}{n_{1} + n_{2} - 2}

置信区间

(\overset{―}{X} - \overset{―}{Y} - S_{W} \sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}} t_{\frac{α}{2}} (n_{1} + n_{2} - 2), \overset{―}{X} - \overset{―}{Y} + S_{W} \sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}} t_{\frac{α}{2}} (n_{1} + n_{2} - 2))

置信上界

\overset{―}{μ_{1} - μ_{2}} = \overset{―}{X} - \overset{―}{Y} + S_{W} \sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}} t_{α} (n_{1} + n_{2} - 2)

置信下界

\underset{―}{μ_{1} - μ_{2}} = \overset{―}{X} - \overset{―}{Y} - S_{W} \sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}} t_{α} (n_{1} + n_{2} - 2)

$\frac{\sigma_1^2}{\sigma_2^2}$ $\mu_1$ $\mu_2$

根据

\frac{\frac{S_{1}^{2}}{S_{2}^{2}}}{\frac{σ_{1}^{2}}{σ_{2}^{2}}} \sim F (n_{1} - 1, n_{2} - 1)

置信区间

(\frac{\frac{S_{1}^{2}}{S_{2}^{2}}}{F_{\frac{α}{2}} (n_{1} - 1, n_{2} - 1)}, \frac{\frac{S_{1}^{2}}{S_{2}^{2}}}{F_{1 - \frac{α}{2}} (n_{1} - 1, n_{2} - 1)})

置信上界

\overset{―}{(\frac{σ_{1}^{2}}{σ_{2}^{2}})} = \frac{\frac{S_{1}^{2}}{S_{2}^{2}}}{F_{1 - α} (n_{1} - 1, n_{2} - 1)}

置信下界

\underset{―}{(\frac{σ_{1}^{2}}{σ_{2}^{2}})} = \frac{\frac{S_{1}^{2}}{S_{2}^{2}}}{F_{α} (n_{1} - 1, n_{2} - 1)}

正态总体均值、方差的检验法

$\alpha$ )

$H_0$ $H_1$ 为备择假设

单个正态总体均值的检验

$X\sim N(\mu,\sigma^2)$

如果方差已知

$\sigma^2$ $\mu$

利用检验统计量

Z = \frac{\overset{―}{X} - μ_{0}}{σ} \sqrt{n} \sim N (0, 1)

$H_0: \mu\le \mu_0$ $H_1: \mu> \mu_0$ , 拒绝域为

z = \frac{\overset{―}{x} - μ_{0}}{σ} \sqrt{n} \geq z_{α}

$H_0: \mu\ge \mu_0$ $H_1: \mu< \mu_0$ , 拒绝域为

z = \frac{\overset{―}{x} - μ_{0}}{σ} \sqrt{n} \leq - z_{α}

$H_0: \mu= \mu_0$ $H_1: \mu\ne \mu_0$ , 拒绝域为

| z | = | \frac{\overset{―}{x} - μ_{0}}{σ} \sqrt{n} | \geq z_{\frac{α}{2}}

如果方差未知

$\sigma^2$ $\mu$

利用检验统计量

t = \frac{\overset{―}{X} - μ_{0}}{S} \sqrt{n} \sim t (n - 1)

$H_0: \mu\le \mu_0$ $H_1: \mu> \mu_0$ , 拒绝域为

t = \frac{\overset{―}{x} - μ_{0}}{S} \sqrt{n} \geq t_{α} (n - 1)

$H_0: \mu\ge \mu_0$ $H_1: \mu< \mu_0$ , 拒绝域为

t = \frac{\overset{―}{x} - μ_{0}}{S} \sqrt{n} \leq - t_{α} (n - 1)

$H_0: \mu= \mu_0$ $H_1: \mu\ne \mu_0$ , 拒绝域为

| t | = | \frac{\overset{―}{x} - μ_{0}}{S} \sqrt{n} | \geq t_{\frac{α}{2}} (n - 1)

两个正态总体均值差的检验

如果两个总体方差已知

$\sigma_1^2$ $\sigma_2^2$ $\mu_1-\mu_2$

利用检验统计量

Z = \frac{\overset{―}{X} - \overset{―}{Y} - δ}{\sqrt{\frac{σ_{1}^{2}}{n_{1}} + \frac{σ_{1}^{2}}{n_{1}}}} \sim N (0, 1)

$H_0: \mu_1-\mu_2\le \delta$ $H_1: \mu_1-\mu_2> \delta$ , 拒绝域为

z = \frac{\overset{―}{x} - \overset{―}{y} - δ}{\sqrt{\frac{σ_{1}^{2}}{n_{1}} + \frac{σ_{1}^{2}}{n_{1}}}} \geq z_{α}

$H_0: \mu_1-\mu_2\ge \delta$ $H_1: \mu_1-\mu_2< \delta$ , 拒绝域为

z = \frac{\overset{―}{x} - \overset{―}{y} - δ}{\sqrt{\frac{σ_{1}^{2}}{n_{1}} + \frac{σ_{1}^{2}}{n_{1}}}} \leq - z_{α}

$H_0: \mu_1-\mu_2= \delta$ $H_1: \mu_1-\mu_2\ne \delta$ , 拒绝域为

| z | = | \frac{\overset{―}{x} - \overset{―}{y} - δ}{\sqrt{\frac{σ_{1}^{2}}{n_{1}} + \frac{σ_{1}^{2}}{n_{1}}}} | \geq z_{\frac{α}{2}}

如果两个总体方差未知

$\sigma_1^2$ $\sigma_2^2$ $\mu_1-\mu_2$

利用检验统计量

t = \frac{\overset{―}{X} - \overset{―}{Y} - δ}{S_{W} \sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}}} \sim t (n_{1} + n_{2} - 2)

其中

S_{W}^{2} = \frac{(n_{1} - 1) S_{1}^{2} + (n_{2} - 1) S_{2}^{2}}{n_{1} + n_{2} - 2}

$H_0: \mu_1-\mu_2\le \delta$ $H_1: \mu_1-\mu_2> \delta$ , 拒绝域为

t = \frac{\overset{―}{x} - \overset{―}{y} - δ}{S_{W} \sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}}} \geq t_{α} (n_{1} + n_{2} - 2)

$H_0: \mu_1-\mu_2\ge \delta$ $H_1: \mu_1-\mu_2< \delta$ , 拒绝域为

t = \frac{\overset{―}{x} - \overset{―}{y} - δ}{S_{W} \sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}}} \leq - t_{α} (n_{1} + n_{2} - 2)

$H_0: \mu_1-\mu_2= \delta$ $H_1: \mu_1-\mu_2\ne \delta$ , 拒绝域为

| t | = | \frac{\overset{―}{x} - \overset{―}{y} - δ}{S_{W} \sqrt{\frac{1}{n_{1}} + \frac{1}{n_{2}}}} | \geq t_{\frac{α}{2}} (n_{1} + n_{2} - 2)

正态总体方差的检验

单个总体的情况

$\sigma^2$

利用检验统计量

χ^{2} = \frac{(n - 1) S^{2}}{σ_{0}^{2}} \sim χ^{2} (n - 1)

$H_0: \sigma^2\le \sigma_0^2$ $H_1: \sigma^2 >\sigma_0^2$ , 拒绝域为

χ^{2} = \frac{(n - 1) S^{2}}{σ_{0}^{2}} \geq χ_{α}^{2} (n - 1)

$H_0: \sigma^2\ge \sigma_0^2$ $H_1: \sigma^2 <\sigma_0^2$ , 拒绝域为

χ^{2} = \frac{(n - 1) S^{2}}{σ_{0}^{2}} \leq χ_{1 - α}^{2} (n - 1)

$H_0: \sigma^2= \sigma_0^2$ $H_1: \sigma^2 \ne \sigma_0^2$ , 拒绝域为

χ^{2} = \frac{(n - 1) S^{2}}{σ_{0}^{2}} \geq χ_{\frac{α}{2}}^{2} (n - 1)

或

χ^{2} = \frac{(n - 1) S^{2}}{σ_{0}^{2}} \leq χ_{1 - \frac{α}{2}}^{2} (n - 1)

两个总体的情况

$\frac{\sigma_1^2}{\sigma_2^2}$

利用检验统计量

F = \frac{\frac{S_{1}^{2}}{S_{2}^{2}}}{\frac{σ_{1}^{2}}{σ_{2}^{2}}} \sim F (n_{1} - 1, n_{2} - 1)

$H_0: \sigma_1^2\le \sigma_2^2$ $H_1: \sigma_1^2> \sigma_2^2$ , 拒绝域为

F = \frac{S_{1}^{2}}{S_{2}^{2}} \geq F_{α} (n_{1} - 1, n_{2} - 1)

$H_0: \sigma_1^2\ge \sigma_2^2$ $H_1: \sigma_1^2 < \sigma_2^2$ , 拒绝域为

F = \frac{S_{1}^{2}}{S_{2}^{2}} \leq F_{1 - α} (n_{1} - 1, n_{2} - 1)

$H_0: \sigma_1^2 = \sigma_2^2$ $H_1: \sigma_1^2 \ne \sigma_2^2$ , 拒绝域为

F = \frac{S_{1}^{2}}{S_{2}^{2}} \leq F_{1 - \frac{α}{2}} (n_{1} - 1, n_{2} - 1)

或

F = \frac{S_{1}^{2}}{S_{2}^{2}} \geq F_{\frac{α}{2}} (n_{1} - 1, n_{2} - 1)

基于成对数据的检验

$D_i=X_i-Y_i$ $i=1,2,...,n$

$D_i\sim N(\mu_D,\sigma^2_D)$

利用检验统计量

t = \frac{\overset{―}{D} - 0}{S_{D}} \sqrt{n} \sim t (n - 1)

$H_0: \mu_D\le 0$ $H_1: \mu_D>0$ , 拒绝域为

t = \frac{\overset{―}{D} - 0}{S_{D}} \sqrt{n} \geq t_{α} (n - 1)

$H_0: \mu_D\ge 0$ $H_1: \mu_D<0$ , 拒绝域为

t = \frac{\overset{―}{D} - 0}{S_{D}} \sqrt{n} \leq - t_{α} (n - 1)

$H_0: \mu_D= 0$ $H_1: \mu_D\ne 0$ , 拒绝域为

| t | = | \frac{\overset{―}{D} - 0}{S_{D}} \sqrt{n} | \geq t_{\frac{α}{2}} (n - 1)