【Probability and Statistics】 Final review

introduction

Normative: $P\left(\Omega \right)=1$,$P(\varnothing )=0$
$P(A)=\frac{Geometric}{Geometric\; measure\; ofG}$
For pairwise mutually exclusive random events,
$$\begin{gathered} P(\sum _{i=1}^{\infty }{A}_i)=\sum _{i=1}^{\infty }P(A_i) \\ P(A\cup B)=P\left(A\right)+P(B)-P(AB) \end{gathered}$$

Conditional Probability

$P(B|A)=\frac{P(AB)}{P(A)}$
Total rate formula:
$$P\left(A\right)=\sum _{i=1}^{n}P(B_i)P(A|B_i)=P(B_1\left)P\right(A|B_1)+P\left(B_2\right)P(A|B_2)+...+P(B_n\left)P\right(A|B_n)$$
of which,$B_1,B_2,...,B_i,...,B_n$is the
division Bayesian formula:
$$P(B_i|A)=\frac{P\left(B_i\right)P(A|B_i)}{{\sum }_{j=1}^{n}P(B_j)P(A|B_j)}$$
$P(AB)=P\left(A\right)P\left(B\right)$ ,$AB$ function
$$\{\begin{array}{l}P(AB)=P(A)P(B)\\ P\left(BC\right)=P(B)P(C)\\ P\left(AC\right)=P(A)P(C)\end{array}$$
Then $ABC$ pairwise independent
$P(ABC)=P\left(A\right)P\left(B\right)P\left(C\right)$ , then they are mutually independent
$P_n(k)=C_n^k{p}^k(1-p{)}^{n-k}$ , 则$A$ innnExactly kk$occurs\; in\; n$ experiments$k$ times

Probability distributions

distribution name	mark	official	expect	variance
01 distribution	$B(1,p)$	$P(X=1)=p,P(X=0)=1-p$	$p$	$p(1-p)$
binomial distribution	$B(n,p)$	$P(X=k)=C_n^k{p}^k{q}^{n-k}$	$np$	$np(1-p)$
Poisson distribution	$P\left(\lambda \right)$	$P(X=k)=\frac{l^k}{k!}{e}^{-\lambda }(k=0,...,+\infty )$	$l$	$l$
hypergeometric distribution	$H(N,m,M)$	$P(X=k)\frac{C_M^k{C}_{N-M}^{n-k}}{C_N^n}(k=0,1,...,\min (M,n)$	$\frac{nM}{N}$	$\frac{nM}{N}(1-\frac{M}{N})(\frac{N-n}{N-1})$

distribution name	mark	Probability Density	cumulative probability distribution	expect	variance
Evenly distributed	$U(a,b)$	$f(x)=\{\begin{array}{ll}\frac{1}{b-a}& a\le x\le b\\ 0& \end{array}$	$F(x)=\{\begin{array}{ll}0& x<a\\ \frac{x-a}{b-a}& a\le x<b\\ 1& x\ge b\end{array}$	$\frac{a+b}{2}$	$\frac{(b-a{)}^2}{12}$
index distribution	$Exp\left(\lambda \right)$	$f(x)=\{\begin{array}{ll}\lambda e^{-\lambda x}& x>0\\ 0& x\le 0\end{array}$	$F(x)=\{\begin{array}{ll}0& x<0\\ 1-e^{\lambda x}& x\ge 0\end{array}$	$\frac{1}{l}$	$\frac{1}{l^2}$
normal distribution	$N(\mu ,p^2)$	$f(x)=\frac{1}{\sqrt{2p.m}p}{e}^{-\frac{(x-\mu {)}^2}{2p^2}}$		$m$	$p^2$
standard normal distribution	$N(0,1)$	$\varphi \left(x\right)=\frac{1}{\sqrt{2p.m}}{e}^{-\frac{x^2}{2}}$	$\Phi \left(x\right)=\frac{1}{\sqrt{2p.m}}{\int }_{-\infty }^x{e}^{-\frac{t^2}{2}}dt$	0	1

Definition: $P(a<X\le b)=\varphi (\frac{b-}{p})-\varphi \left(\frac{a-}{p}\right)$

two-dimensional probability distribution

$Y=g(X)$,求$f_X(y)$:
$X=h(Y)$,$f_X(y)=f_x[h(y)]|h^{\prime }(y)|$
若$P(X=x_i,Y=y_i)=P_{ij}$,则$P(X=x_i)={\sum }_{j=1}^{\infty }{P}_{ij}$

Marginal probability density:
$$\begin{gathered} f_X(x)={\int }_{-\infty }^{\infty }f(x,y)dy \\ {f}_Y(y)={\int }_{-\infty }^{\infty }f(x,y)dx \end{gathered}$$
若$P(X\le x,Y\le y)=P(X\le x)P(Y\le y),F(x,y)=F_X(x)F_Y\left(y\right)$ , then$X$和$Y$ are independent of each other

条件分布：
$$\begin{gathered} P(X=x_i|Y=y_j)=\frac{P(X=x_i,Y=y_j)}{P(Y)=y_j)}=\frac{P_{ij}}{P_{\cdot j}} \\ P(Y)=y_j|X=x_i)=\frac{P(X=x_i,Y=y_j)}{P(X=x_i)}=\frac{P_{ij}}{P_{i\cdot }} \\ f(x,y)=f_X(x)f_{Y|X}(y|x),f_{X|Y}(x|y)=\frac{f(x,y)}{f_Y(y)} \end{gathered}$$
$Z=g(X,Y)$ probability density

$$Z=G(X,Y),F_Z(z)=P(Z\le z)=P(g(X,Y)\le z)={\iint }_{D}f(x,y)dxdy$$
其中$D=\{(x,y)|g(x,y)\le z\}$
即
$$F_Z(z)={\iint }_{g(x,y)\le z}f(x,y)dxdy,f_Z\left(z\right)=\frac{d}{dz}{F}_Z\left(z\right)$$
For example,$Z=X+Y$时
$$f_Z\left(z\right)={\int }_{-\infty }^{\infty }f(x,z-x)dx={\int }_{-\infty }^{\infty }{f}_X(x)f_Y(z-x)dx$$
$${\int }_{-\infty }^{\infty }{e}^{-t^2}dt=\sqrt{Pi}$$
$$F_{\mathrm{MAX}}(z)=F_1(z)F_2(z),F_{\mathrm{MIN}}(z)=1-(1-F_1(z))(1-F_2(z))$$

expectation and variance

$$E\left(X\right)=\sum _{k=1}^{\infty }{x}_k{p}_k$$
$$E\left(X\right)={\int }_{-\infty }^{\infty }xf(x)dx$$
$$E[g(x)]={\int }_{-\infty }^{\infty }g(x)f(x)dx$$
$$D(x)=And\{[X-E\left(X\right){]}^2\}$$
$$D\left(x\right)=\sum _{k=1}^{\infty }[x_k-E(X){]}^2{p}_k$$
$$D\left(x\right)={\int }_{-\infty }^{\infty }[x-E(X){]}^{2}f(x)dx$$
$$D\left(x\right)=E(X^2)-\left[E\right(X){]}^2$$
$$D(CX)=C^{2}D(X)$$
$$D(X\pm Y)=D(X)\pm D(Y)$$

Theorem of large numbers

Infinitesimal function:
$$P(|x-a|\ge )\_\le \frac{D(x)}{{ϵ}^2}$$
The central limit theorem refers to $Number\; of\; occurrences\; of\; n$ -fold Bernoulli experiments:
$$P(X\le x)=\varphi (\frac{x-np}{\sqrt{npq}})$$

Sample distribution and statistics

$x_1,x_2,...,x_n$Identical distribution independent, $E=m,D=p^2$ ,则
$$P(\frac{{\sum }_{i=1}^n{x}_i-n}{\sqrt{n}p}\le x)=\varphi \left(x\right)$$
overall$X$ has a probability density$f(x)$,则$f(x_1,x_2,...,x_n)={\prod }_{i=1}^{n}f\left(x_i\right)$
sample mean:
$$\bar{X}=\frac{1}{n}\sum _{i=1}^n{X}_n$$
Sample variance:
$$S^2=\frac{1}{n-1}\sum _{i=1}^n(x_i-\bar{x}{)}^2$$
中心矩
$${\hat{C}}_k=\frac{1}{n}\sum _{i=1}^n(x_i-\bar{x}{)}^k$$
原点矩
$${\hat{M}}_k=\frac{1}{n}\sum _{i=1}^n{x}_i^k$$
$x_1,x_2,...,x_n$Independent, subject to $h^2\left(1\right)$,${\sum }_{i=1}^n{x}_i\sim h^2(n)$
$x_1,x_2,...,x_n$Independent, subject to $N(0,1)$ ,${\sum }_{i=1}^n{x}_i^2\sim h^2\left(n\right)$
$x_1,x_2,...,x_n$独立，$X_i\sim N(m_i,p_i^2)$,则${\sum }_{i=1}^n(\frac{x_i-m_i}{p_i}{)}^2\sim h^2(n)$

$X\sim N(0,1),Y\sim h^2\left(n\right)$,则
$$\frac{X}{\sqrt{and/n}}\sim t(n)$$

$X\sim h^2(m),Y\sim h^2(n)$,则：
$\frac{X/m}{and/n}\sim F(m,n)$
$\frac{1}{F}\sim F(n,m)$
$F_{1-}(n_1,n_2)=\frac{1}{F_{}(n_2,n_1)}$

$\bar{x}\sim N(\mu ,p^2/n),(n-1)s^2/p^2\sim h^2(n-1)$
$\bar{x}$ and$s^2$ independent
$U=\sqrt{n}\frac{\bar{x}-}{p}\sim N(0,1),T=\sqrt{n}\frac{\bar{x}-}{s}\sim t(n-1)$

Parameter Estimation

Form: $E(\hat{i})=\theta variance$
:$D({\hat{i}}_1)<D\left({\hat{i}}_2\right)$
definition:${\lim }_{n\to \infty }p(|\hat{i}-\theta |<)\_=1$
$$\hat{E}(X^k)=\frac{1}{n}{X}_i^k$$

known conditions	Estimation formula	$1-alpha$ confidence interval
$p^2$ Known, estimated$m$	$U=\frac{\bar{x}-}{s/\sqrt{n}}\sim N(0,1)$	( $\bar{x}-\frac{}{\sqrt{n}}{u}_{\frac{}{2}},\bar{x}+\frac{}{\sqrt{n}}{u}_{\frac{}{2}})$
$p^2$ unknown, estimate$m$	$T=\frac{\bar{x}-}{s/\sqrt{n}}\sim t(n-1)$	$(\bar{x}-\frac{s}{\sqrt{n}}{t}_{\frac{}{2}},\bar{x}+\frac{s}{\sqrt{n}}{t}_{\frac{}{2}})$
$\mu$ is known, estimate$p^2$		$(\frac{{\sum }_{i=1}^n(x_i-\mu {)}^2}{{\chi }_{\frac{a}{2}}^2(n)},\frac{{\sum }_{i=1}^n(x_i-\mu {)}^2}{{\chi }_{1-\frac{a}{2}}^2(n)})$
$\mu$未知，求${\sigma }^2$		$(\frac{(n-1)s^2}{{\chi }_{\frac{a}{2}}^2(n-1)},\frac{(n-1)s^2}{{\chi }_{1-\frac{a}{2}}^2(n-1)})$

其中，$S^2$为样本方差，分母$n-1$,$Q=\frac{{\sum }_{i=1}^n(x_i-\bar{x}{)}^2}{{\sigma }^2}=\frac{(n-1)s^2}{{\sigma }^2}\sim {\chi }^2(n-1)$

假设检验

$$\begin{array}{ccc}\mathrm{已知条件}& \mathrm{值}& \mathrm{拒绝域}\\ \mathrm{已知}{\sigma }^2，假设检验\mu & U=\frac{\bar{x}-{\mu }_0}{\sigma /\sqrt{n}}\sim N(0,1)& |U|=\frac{\bar{x}-{\mu }_0}{\sigma /\sqrt{n}}\ge u_{\frac{\alpha }{2}}\\ \mathrm{未知}{\sigma }^2,\mathrm{检验}\mu & T=\frac{\bar{x}-{\mu }_0}{s/\sqrt{n}}\sim t(n-1)& |T|=|\frac{\bar{x}-{\mu }_0}{s/\sqrt{n}}|\ge t_{\frac{\alpha }{2}}(n-1)\\ {\chi }^2\mathrm{检验}& {\chi }^2=\frac{(n-1)S^2}{{\sigma }_o^2}\sim {\chi }^2(n-1)& \frac{(n-1)s^2}{{\sigma }_o^2}\le {\chi }_{1-\frac{\alpha }{2}}^2(n-1)\mathrm{或}\frac{(n-1)s^2}{{\sigma }_o^2}\ge {\chi }_{\frac{a}{2}}^2(n-1)\end{array}$$

先检验$\mu$再检验$\sigma$
流程：提出原假设，确定检验水平，构造检验统计量，确定拒绝域。