Common formulas of probability theory

Chapter one

The addition formula

$$P(A\cup B)=P(A)+P(B)-P(AB)$$

And so on （ Classical pattern ）

$$P(A)=\frac{k}{n}=\frac{AinpackagecontainOfthingsPieces\; ofCount}{SinThe\; baseBenthingsPieces\; ofOftotalCount}$$

Hypergeometric distribution

$$P=\frac{C_D^k{C}_{N_D}^{n-k}}{C_N^n}$$

event A The occurrence of a conditional event B Probability of occurrence

$$P(B|A)=\frac{P(AB)}{P(A)}$$

set up $B_1,B_2\dots$ Not compatible with each other , Yes

$$P(\bigcup _{i=1}^{\infty }{B}_i|A)=\sum _{i=1}^{\infty }P(B_i|A)$$

Such as

$$P(B_1\cup B_2|A)=P(B_1|A)P(A)+P(B_2|A)P(A)-P(B_1{B}_2)|P(A)$$

Multiplication principle

$$P(AB)=P(B|A)P(A)$$

set up $B_1,B_2,\cdots ,B_n$ Is the sample space $S$ A division of , Then there are

All probability formula

$$P(A)=P(A|B_1)P(B_1)+P(A|B_2)P(B_2)+\cdots +P(A|B_n)P(B_n)$$

Bayes' formula

$$P(B_i|A)=\frac{P(A|B_i)P(B_i)}{\sum _{j=1}^{n}P(A|B_j)P(B_j)}$$

Special , When $n=2$ when ,

$$P(A)=P(A|B)P(B)+P(A|\overline{B})P(\overline{B})$$

$$P(B|A)=\frac{P(AB)}{P(A)}=\frac{P(A|B)P(B)}{P(A|B)P(B)+P(A|\overline{B})P(\overline{B})}$$

$A、B$ When they are independent of each other , Yes

$$P(AB)=P(B|A)P(A)=P(A)P(B)$$

from $n$ There is no put back extraction in different elements $m$ Arrange the elements into An orderly column when , obtain

$$A_n^m=\frac{n!}{(n-m)!}$$

A different arrangement

from $n$ There is no put back extraction in different elements $m$ Elements Form a group regardless of order when , obtain

$$C_n^m=\frac{n!}{m!(n-m)!}$$

Two different combinations

if $A\subset B$（ Notice the big decrease ）, be
$$P(B-A)=P(B)-P(A)$$

Independent and Compatible with It could happen at the same time

13/11/2021 09:56

Chapter two

Discrete random variables ： All possible values are finite or infinite

The binomial distribution （n Heavy Bernoulli experiment ）：$X\sim b(n,p)$

$$P\{X=k\}=C_n^k{p}^k{q}^{n-k},k=0,1,2,\cdots ,n$$

among $q=1-p$

When $n=2$ when , The binomial distribution becomes 0-1 Distribution ：

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

Poisson distribution ：$X\sim \pi (\lambda )$
$$P\{X=k\}=\frac{{\lambda }^k{e}^{-\lambda }}{k!}$$

When n Big enough ,p Enough hours , The binomial distribution can be estimated by Poisson distribution , There is a

$$C_n^k{p}^k(1-p{)}^{n-k}\approx \frac{{\lambda }^k{e}^{-\lambda }}{k!}$$

among $\lambda =np$

Distribution function

$$F(x)=P\{X\le x\},-\infty <x<+\infty$$

For any real number $x_1<x_2$, Yes

$$P\{x_1<X\le x_2\}=P\{X\le x_2\}-P\{X\le x_1\}=F(x_2)-F(x_1)$$

This property can be used to prove that the distribution function is an undiminished function

Continuous random variable : Distribution function $F(x)$ Satisfy

$$F(x)={\int }_{-\infty }^{x}f(t)dt$$

call $f(x)$ by $X$ The probability density function of , abbreviation Probability density

Yes :

$$P\{x_1<X\le x_2\}=F(x_2)-F(x_1)={\int }_{x_1}^{x_2}f(x)dx$$

$$F^{\prime }(x)=f(x)$$

For continuous random variables , The probability of a single point is 0, That is to say

$$P\{X=a\}=0$$

Uniform distribution $X\sim U(a,b)$:

$$f(x)=\{\begin{array}{lr}\frac{1}{b-a},& a<x<b\\ 0,& ItsHe\end{array}$$

$$F(x)=\{\begin{array}{lr}0,& x<a\\ \frac{x-a}{b-a},& a\le x<b\\ 1,& x\ge b\end{array}$$

An index distribution ：

$$f(x)=\{\begin{array}{lr}\frac{1}{\theta }{e}^{-\frac{x}{\theta }},& x>0\\ 0,& ItsHe\end{array}$$

$$F(x)=\{\begin{array}{lr}1-e^{-\frac{x}{\theta }},& x>0\\ 0,& ItsHe\end{array}$$

Exponential distribution has no memory , That is to say ：

$$P\{X>s+t|X>s\}=P\{x>t\}$$

Normal distribution ：$X\sim N(\mu ,{\sigma }^2)$

$$f(x)=\frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{{(x-\mu )}^2}{2{\sigma }^2}},-\infty <x<+\infty$$

$$F(x)=\frac{1}{\sqrt{2\pi }\sigma }{\int }_{-\infty }^x{e}^{-\frac{{(t-\mu )}^2}{2{\sigma }^2}}dt$$

Special , When $\mu =0,\sigma =1$ It is called random variable $X$ obey Standard normal distribution , Its probability density and distribution function are respectively used $\phi (x),\Phi (x)$ Express , Yes

$$\phi (x)=\frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{x^2}{2}},-\infty <x<+\infty$$

$$\Phi (x)=\frac{1}{\sqrt{2\pi }\sigma }{\int }_{-\infty }^x{e}^{-\frac{t^2}{2}}dt$$

obviously , Yes

$$\Phi (-x)=1-\Phi (x)$$

If there are random variables $X\sim N(\mu ,{\sigma }^2)$, Then there are random variables Z, bring

$$Z=\frac{X-\mu }{\sigma }\sim N(0,1)$$

When solving the distribution of the function of random variables , A more general method is to first deform by inequality $F_X(x)$ or $f_X(x)$ solve $F_Y(y)$, Right again $F_Y(y)$ Take the derivative and get $f_Y(y)$

if $X$ With probability density $f_X(x),-\infty <x<+\infty$, And set up $X$ To $Y$ Transformation of $Y=g(X)$ You can lead everywhere , Monotonous everywhere , be $Y$ Probability density of

$$f_Y(y)=\{\begin{array}{lr}{f}_X[h(y)]|h^{\prime }(y)|,& \alpha <y<\beta \\ 0,& ItsHe\end{array}$$

among $h(y)$ yes $g(x)$ The inverse function of

If there are random variables $X\sim N(\mu ,{\sigma }^2)$, Then its linear function $Y=aX+b$ It also follows a normal distribution

13/11/2021 16:14

The third chapter

（ For the conclusion of symmetry , Just give one , Another similar conclusion can be drawn ）

Joint distribution function : set up $(X,Y)$ It's a two-dimensional random variable , For any real number $x$,$y$, Dual function
$$F(x,y)=P\{(X\le x)\cap (Y\le y)\}=P\{X\le x,Y\le y\}$$
by $(X,Y)$ The joint distribution function of

Joint distribution can be understood as $(X,Y)$ The probability surrounded by the point to the infinite distance at the bottom left .

Yes
$$0\le F(x,y)\le 1$$
$$\forall solidsetOfy,F(-\infty ,y)=0$$
$$\forall solidsetOfx,F(x,-\infty )=0$$
$$F(-\infty ,-\infty )=0,F(+\infty ,+\infty )=1$$

about $\forall (x_1,y_1),(x_2,y_2),x_1<x_2,y_1<y_2$, Yes
$$F(x_2,y_2)-F(x_2,y_1)+F(x_1,y_1)-F(x_1,y_2)\ge 0$$

14/11/2021 16:02

Two dimensional discrete random variables ： Two-dimensional random variable $(X,Y)$ All possible values are finite pairs or countable infinite pairs

Joint distribution law of two-dimensional random variables ：$P\{X=x_i,Y=y_i\}=p_{ij},i,j=1,2,\cdots$

Or use a table to show

[ Failed to transfer the external chain picture , The origin station may have anti-theft chain mechanism , It is suggested to save the pictures and upload them directly (img-nKddJcCx-1637058108176)(:/c4a60e283b3243a3a2f472358ccdfb14)]

Two dimensional continuous random variables ： If there is a nonnegative integrable function , send
$$F(x,y)={\int }_{-\infty }^y{\int }_{-\infty }^{x}f(u,v)dudv$$
be $(X,Y)$ It is a two-dimensional continuous random variable ,$f(x,y)$ Is the probability density

Yes ：
$${\int }_{-\infty }^{+\infty }{\int }_{-\infty }^{+\infty }f(x,y)dxdy=F(+\infty ,+\infty )=1$$

$$\frac{{\partial }^{2}F(x,y)}{\partial x\partial y}=f(x,y)$$
Yes $xOy$ The area on the plane $G$, Yes

$$P\{(X,Y)\in G\}=\underset{G}{\iint }f(x,y)dxdy$$

Edge distribution function : Make a variable tend to infinity in the distribution function
$$F_X(x)=F(x,\infty )$$
$$F_Y(y)=F(\infty ,y)$$

Marginal distribution law of discrete random variables ：
$$p_{i\cdot }=\sum _{j=1}^{\infty }{p}_{ij}=P\{X=x_i\}$$

Edge distribution function of continuous random variables ：
$$F_X(x)=F(x,\infty )={\int }_{-\infty }^x[{\int }_{-\infty }^{\infty }f(x,y)dy]dx$$

Marginal probability density of continuous random variables ：
$$f_X(x)={\int }_{-\infty }^{\infty }f(x,y)dy$$

stay $Y=y_i$( fixed ) Random variables under conditions $X$ Conditional distribution law of :
$$P\{X=x_i|Y=y_j\}=\frac{P\{X=x_i,Y=y_i\}}{P\{Y=y_i\}}=\frac{p_{ij}}{p_{\cdot j}}$$

Conditional probability density of continuous random variables :

$$f_{X|Y}(x|y)=\frac{f(x,y)}{f_Y(y)}$$

Conditional distribution function of continuous random variables :

$$F_{X|Y}(x|y)=P\{X⩽x|Y=y\}={\int }_{-\infty }^x\frac{f(x,y)}{f_Y(y)}dy$$

Uniform distribution of two-dimensional random variables ：

$$f(x,y)=\{\begin{array}{lr}\frac{1}{A},& (x,y)\in G\\ 0,& ItsHe\end{array}$$
among $G$ Is a bounded region on a plane , Area is $A$

Independent random variables
$$F(x,y)=F_X(x)F_Y(y)$$

For continuous random variables , also
$$f(x,y)=f_X(x)f_Y(y)$$

For two random variables ：

$Z=X+Y$ when
$$f_{X+Y}(z)={\int }_{-\infty }^{\infty }f(z-y,y)dy$$
if $X$,$Y$ Are independent of each other
$$f_{X+Y}(z)={\int }_{-\infty }^{\infty }{f}_X(z-y)f_Y(y)dy$$

$Z=\frac{Y}{X}$ or $Z=XY$ when
$$f_{\frac{Y}{X}}(z)={\int }_{-\infty }^{\infty }|x|f(x,xz)dx$$
$$f_{XY}(z)={\int }_{-\infty }^{\infty }\frac{1}{|x|}f(x,\frac{z}{x})dx$$

When they are independent of each other

$$f_{\frac{Y}{X}}(z)={\int }_{-\infty }^{\infty }|x|f_X(x)f_Y(xz)dx$$
$$f_{XY}(z)={\int }_{-\infty }^{\infty }\frac{1}{|x|}{f}_X(x)f_Y(\frac{z}{x})dx$$

$M=max\{X,Y\}$ and $N=min\{X,Y\}$（ important ）

（ When they are independent of each other ）
$$F_{max}(z)=F_X(z)F_Y(z)$$
$$F_{min}(z)=1-[1-F_X(z)][1-F_Y(z)]$$

Extension

Yes $M=max\{X_1,X_2,\cdots ,X_n\}$ and $N=min\{X_1,X_2,\cdots ,X_n\}$
$$F_{max}(z)=F_{X_1}(z)F_{X_2}(z)\cdots F_{X_n}(z)$$
$$F_{min}(z)=1-[1-F_{X_1}(z)][1-F_{X_2}(z)]\cdots [1-F_{X_n}(z)]$$

Homodistribution

$$F_{max}(z)=[F(z){]}^n$$
$$F_{min}(z)=1-[1-F(z){]}^n$$

15/11/2021 21:47

Chapter four

Mathematical expectation : The average value of the characterization variable

（ discrete ）

$$E(X)=\sum _{k=1}^{\infty }{x}_k{p}_k$$

When the series of the distribution law $\sum _{k=1}^{\infty }{x}_k{p}_k$ It is true when it is absolutely convergent

（ Continuous type ）

$$E(X)={\int }_{-\infty }^{\infty }xf(x)dx$$

When the probability density is integral ${\int }_{-\infty }^{\infty }xf(x)dx$ It is true when it is absolutely convergent

set up $Y=g(X)$,$g$ It's a continuous function

（ discrete ）

$$E(Y)=E[g(X)]=\sum _{k=1}^{\infty }g(x_k)p_k$$

( Continuous type )

$$E(Y)=E[g(X)]={\int }_{-\infty }^{\infty }g(x)f(x)dx$$

For random variables $Z=g(X,Y)$,$Z$ It's one-dimensional , Then you can apply the above formula , set up $(X,Y)$ The probability density of is $f(x,y)$, Yes ：

$$E(Z)=E[g(X,Y)]={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }g(x,y)f(x,y)dxdy$$

Then the expectation of a single two-dimensional random variable can be regarded as $Z=X$, Thus there are ：
$$E(X)={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }xf(x,y)dxdy$$

For discrete random variables , There are similar conclusions ：
$$E(Z)=E[g(X,Y)]=\sum _{j=1}^{\infty }\sum _{i=1}^{\infty }g(x_i,y_i)p_{ij}$$

( Are independent of each other )
$$E(XY)=E(X)E(Y)$$

variance ： Represent the deviation degree of the variable from the mean , In essence, it is the average $E(X)$ Function of

$$D(X)=Var(X)=E\{[X-E(X){]}^2\}$$

( discrete )

$$D(X)=\sum _{k=1}^{\infty }[x_k-E(X){]}^2{p}_k$$

( Continuous type )

$$D(X)={\int }_{-\infty }^{\infty }[x-E(X){]}^{2}f(x)dx$$

Yes ：

$$D(X)=E(X^2)-[E(X){]}^2$$

The nature of variance ：

• $D(C)=0$
• $D(CX)=C^{2}D(X)$,$D(X+C)=D(X)$
• $D(X+Y)=D(X)+D(Y)+2E\{[X-E(X)][Y-E(Y)]\}$
  , Special , When $X$,$Y$ When they are independent of each other ,$D(X+Y)=D(X)+D(Y)$
• $D(X)=0\iff P\{X=E(X)\}=1$

Chebyshev inequality ： Set the random variable $X$,$E(X)=\mu$,$D(X)={\sigma }^2$, On the other hand $\forall \epsilon >0$
$$P\{|X-\mu |\ge \epsilon \}\le \frac{{\sigma }^2}{{\epsilon }^2}$$

Focus on the six distributions （ Memorize ）

Distribution	Distribution law / Probability density	$E(X)$	$D(X)$
(0-1) Distribution	$P\{X=k\}=p^k(1-p{)}^{1-k}$ $k=0,1$	$p$	$p(1-p)$
The binomial distribution $X\sim b(n,p)$	$P\{X=k\}=C_n^k{p}^k{q}^{n-k}$ $k=0,1,2,\cdots ,n$	$np$	$np(1-p)$
Poisson distribution $X\sim \pi (\lambda )$	$P\{X=k\}=\frac{{\lambda }^k{e}^{-\lambda }}{k!}$	$\lambda$	$\lambda$
Uniform distribution $X\sim U(a,b)$	$f(x)=\{\begin{array}{lr}\frac{1}{b-a},& a<x<b\\ 0,& ItsHe\end{array}$	$\frac{a+b}{2}$	$\frac{(b-a{)}^2}{12}$
An index distribution	$f(x)=\{\begin{array}{lr}\frac{1}{\theta }{e}^{-\frac{x}{\theta }},& x>0\\ 0,& ItsHe\end{array}$	$\theta$	${\theta }^2$
Normal distribution $X\sim N(\mu ,{\sigma }^2)$	$f(x)=\frac{1}{\sqrt{2\pi }\sigma }{e}^{-\frac{{(x-\mu )}^2}{2{\sigma }^2}}$ $-\infty <x<+\infty$	$\mu$	${\sigma }^2$

covariance ：
$$Cov(X,Y)=E\{[X-E(X)][Y-E(Y)]\}$$

The correlation coefficient :
$${\rho }_{XY}=\frac{Cov(X,Y)}{\sqrt{D(X)}\sqrt{D(Y)}}$$

Yes ：
$$Cov(X,X)=D(X)$$
$$Cov(X,Y)=E(XY)-E(X)E(Y)$$
$$Cov(aX,bY)=abCov(X,Y)$$
$$Cov(X_1+X_2,Y)=Cov(X_1,Y)+Cov(X_2,Y)$$
$$|{\rho }_{XY}|\le 1$$
$$|{\rho }_{XY}|=1\iff \exists a,b\in \mathbb{R},P\{Y=a+bX\}=1$$

When ${\rho }_{XY}=0$ when , call $X$ and $Y$ No （ linear ） relevant

Irrelevant and independent are different , Are independent of each other $\Rightarrow$ Unrelated , Unrelated $\nRightarrow$ Are independent of each other

But for two-dimensional normal distribution , Are independent of each other $\iff$ Unrelated