[Probability Theory] Final Review Notes: Basic Concepts of Mathematical Statistics

1. Population and sample

Overall : the whole of the research objects or the whole of a certain (or some) quantitative indicators of the research objects , use XXX means (normal population:X ~ N ( μ , σ 2 ) \newcommand{\td}{\,\text{\large\textasciitilde}\,}X\td N(\mu,\sigma^2)X~N ( μ ,p2)
个体:总体的每个元素
有限总体:含有有限个个体的总体
无限总体:含有无限个个体的总体
总体分布:数量指标 X X X取不同值的比率(是客观存在的)

样本/子样:总体中取得的一部分个体
样本容量 n n n):样本中所含个体的个数
抽样:取得样本的过程
抽样法:抽样过程所采取的方法
随机抽样法:每一个个体是从总体中随机抽取的
随机样本:采用随机抽样法得到的样本
样本: n 维随机向量 ( X 1 , X 2 , ⋯   , X n ) ⟶ 观测 样本值:一组具体的实数 ( x 1 , x 2 , ⋯   , x n ) \text{样本:}n\text{维随机向量}(X_1,X_2,\cdots,X_n)\overset{\text{观测}}{\longrightarrow}\text{样本值:一组具体的实数}(x_1,x_2,\cdots,x_n) 样本:n维随机向量(X1,X2,,Xn)观测样本值:一组具体的实数(x1,x2,,xn)简单随机样本:各 X i X_i Xi X X X同分布相互独立(不做特殊声明,样本均指简单随机样本)
简单随机抽样:获得简单随机样本的方法
简单随机样本 ( X 1 , X 2 , ⋯   , X n ) (X_1,X_2,\cdots,X_n) (X1,X2,,Xn)的分布函数:设总体 X X X的分布函数为 F ( x ) F(x) F(x),则样本的分布函数为 F ( x 1 , x 2 , ⋯   , x n ) = P { X 1 ≤ x 1 , X 2 ≤ x 2 , ⋯   , X n ≤ x n } = ∏ i = 1 n P { X i ≤ x i } = ∏ i = 1 n F ( x i ) F(x_1,x_2,\cdots,x_n)=P\{X_1\le x_1,X_2\le x_2,\cdots,X_n\le x_n\}=\prod\limits_{i=1}^n P\{X_i\le x_i\}=\prod\limits_{i=1}^n F(x_i) F(x1,x2,,xn)=P{ X1x1,X2x2,,Xnxn}=i=1nP{ Xixi}=i=1nF(xi)

  • If overall XXX is a continuous random variable (the probability density isf ( x ) f(x)f ( x ) ), then samples( X 1 , X 2 , ⋯ , X n ) (X_1,X_2,\cdots,X_n)(X1,X2,,Xn) isf ( x 1 , x 2 , ⋯ , xn ) = ∏ i = 1 nf ( xi ) f(x_1,x_2,\cdots,x_n)=\prod\limits_{i=1}^nf( x_i)f(x1,x2,,xn)=i=1nf(xi)
  • If overall XXX is a discrete random variable (the distribution law isP { X = ai } = pi P\{X=a_i\}=p_iP{ X=ai}=pi), then the samples ( X 1 , X 2 , ⋯ , X n ) (X_1,X_2,\cdots,X_n)(X1,X2,,Xn)的分布律为 P { X 1 = x 1 , X 2 = x 2 , ⋯   , X n = x n } = ∏ i = 1 n P { X = x i } P\{X_1=x_1,X_2=x_2,\cdots,X_n=x_n\}=\prod\limits_{i=1}^n P\{X=x_i\} P{ X1=x1,X2=x2,,Xn=xn}=i=1nP{ X=xi}

2. Collation of sample data

1. Sample frequency distribution and frequency distribution

样本频数分布:样本值中不同数值在样本值中出现的频数(即次数)
样本频率分布:样本值中不同数值在样本值中出现的频率(即次数/样本容量)
设样本值中不同的数值记为 x 1 ∗ , x 2 ∗ , ⋯   , x l ∗ x_1^*,x_2^*,\cdots,x_l^* x1,x2,,xl(递增),相应的频数为 m 1 , m 2 , ⋯   , m l m_1,m_2,\cdots,m_l m1,m2,,ml ∑ i = 1 l m i = n \sum\limits_{i=1}^l m_i=n i=1lmi=n),则样本频数分布表:

指标 X X X x 1 ∗ x_1^* x1 x 2 ∗ x_2^* x2 ⋯ \cdots x l ∗ x_l^* xl
频数 m i m_i mi m 1 m_1 m1 m 2 m_2 m2 ⋯ \cdots m l m_l ml

样本频率分布表:

指标 X X X x 1 ∗ x_1^* x1 x 2 ∗ x_2^* x2 ⋯ \cdots x l ∗ x_l^* xl
频率 m i n \frac{m_i}{n} nmi m 1 n \frac{m_1}{n} nm1 m 2 n \frac{m_2}{n} nm2 ⋯ \cdots m l n \frac{m_l}{n} nml

如果总体 X X X is a discrete random variable, then the event{ X = xi ∗ } \{X=x_i^*\}{ X=xi} frequencymin \frac{m_i}{n}nmishould be close to the probability pi p_i of its occurrencepi.
If overall XXX is a continuous random variable, then the event{ X = xi ∗ } \{X=x_i^*\}{ X=xi} The probability of occurrence is0 00 , it is meaningless to examine the frequency distribution of the sample at this time, and it is necessary to examine the frequency histogram of the sample.

2. Frequency histogram

Let the overall XXX is a continuous random variable with probability densityf ( x ) f(x)f(x) ( x 1 , x 2 , ⋯   , x n ) (x_1,x_2,\cdots,x_n) (x1,x2,,xn) is from the overallXXA sample value of X. The method for making a frequency histogram is:

  1. Organize data : put sample values ​​x 1 , x 2 , ⋯ , xn x_1,x_2,\cdots,x_nx1,x2,,xnSort from small to large x ( 1 ) ≤ x ( 2 ) ≤ ⋯ ≤ x ( n ) x_{(1)}\le x_{(2)}\le\cdots\le x_{(n)}x(1)x(2)x(n)

  2. Grouping : in the interval [ a , b ] containing all observations [a,b][a,b]中插入一些分点 a = t 0 < t 1 < ⋯ < t l − 1 < t l = b a=t_0<t_1<\cdots<t_{l-1}<t_l=b a=t0<t1<<tl1<tl=b grip[ a , b ] [a,b][a,b ] divided intolll个小区间: t 0 ↑ a t 1 t 2 ⋯ t l − 1 t l ↑ b \underset{\underset{a}{\uparrow}}{t_0}\qquad t_1\qquad t_2\qquad\cdots\qquad t_{l-1}\qquad\underset{\underset{b}{\uparrow}}{t_l} at0t1t2tl1btlSome concepts:

    • Group distance : the length between cells di = ti − ti − 1 d_i=t_i-t_{i-1}di=titi1
    • group median : the midpoint of the interval
    • Number of groups : the number of cells between lll

    Generally, equal division is adopted (the distance between each group is equal), at this time di = b − al d_i=\frac{ba}{l}di=lba. Group distance llThe choice of l :

    • n > 100 n>100 n>100lll take10 1010 to20 2020
    • n ≈ 50 n\approx 50 n50lll take5 55 or6 66

    Pay attention to the principle of division: make sure that there are sample observations falling into each interval.

  3. Column grouping frequency distribution table : by mi m_imiIndicates that the observed value falls into ( ti − 1 , ti ] (t_{i-1},t_i](ti1,ti] (that is, the frequencyof this interval or this group),fi = min f_i=\frac{m_i}{n}fi=nmiFor the frequency of this group , remember yi = fidi = mindi y_i=\frac{f_i}{d_i}\textcolor{#aaaaaa}{=\frac{m_i}{nd_i}}yi=difi=ndimi, to list the grouped data into a table:

group group median Frequency mi m_imi frequency fi f_ifi y i y_i yi
[ 27 , 30 ] [27,30] [27,30] 28.5 28.5 28.5 8 8 8 0.105 0.105 0.105 0.035 0.035 0.035
( 30 , 33 ] (30,33] (30,33] 31.5 31.5 31.5 10 10 10 0.132 0.132 0.132 0.044 0.044 0.044
⋯ \cdots ⋯ \cdots ⋯ \cdots ⋯ \cdots ⋯ \cdots
  1. Make a frequency histogram : at x O y xOyOn the x O y coordinate plane, usexxEach interval on the x- axis ( ti − 1 , ti ] (t_{i-1},t_i](ti1,ti]为底,以yi = fidi y_i=\frac{f_i}{d_i}yi=difiDraw a row of vertical rectangles for the height to get the frequency histogram. Note that the height of the rectangle is yi = fidi y_i=\frac{f_i}{d_i}yi=difiInstead of frequency fi f_ifi, is to be divided by the group distance, the purpose is to make the sum of the areas of all rectangles be 1 11 . At this time the overallXXX falls into the interval( ti − 1 , ti ) (t_{i-1},t_i)(ti1,ti) probabilitypi ≈ fi p_i\approx f_ipifi
  2. Make a probability density curve : smoothly connect the midpoints on the sides of each rectangle in the frequency histogram to obtain a curve, when nnn andllWhen l is sufficiently large, this curve approximatesXXProbability density curve of X y = f ( x ) y=f(x)y=f(x)

3. Empirical distribution function

Set sample values ​​( x 1 , x 2 , ⋯ , xn ) (x_1,x_2,\cdots,x_n)(x1,x2,,xn) , its empirical distribution function is F n ( x ) = 1 n ∑ i = 1 n [ xi ≤ x ] F_n(x)=\frac{1}{n}\sum\limits_{i=1}^n\ left[x_i\le x\right]Fn(x)=n1i=1n[xix]其中 [ x i ≤ x ] \left[x_i\le x\right] [xix ] means whenxi ≤ x x_i\le xxiTake 1 when x is 11 x i > x x_i>x xi>0when x0 . To sum up,F n ( x ) F_n(x)Fn( x ) isnnLess than or equal to xx among n sample valuesx ofxi x_ixiThe number of divided by the sample size nnn . In other words, it is less than or equal toxxThe ratio of the number of sample values ​​of x to the total number of samples.

The empirical distribution function has the following properties:
(1) Monotonically increasing;
(2) Right continuous;
(3) F n ( − ∞ ) = 0 F_n(-\infty)=0Fn()=0F n ( + ∞ ) = 1 F_n(+\infty)=1Fn(+)=1

If the sample values ​​are given in a frequency distribution table, the empirical distribution function F n ( x ) F_n(x)Fn(x)可具体表达为 F n ( x ) = { 0 , x < x i ∗ m 1 + m 2 + ⋯ + m i n , x i ∗ ≤ x < x i + 1 ∗ ,   ( i = 1 , 2 , ⋯   , l − 1 ) 1 , x ≥ x l ∗ F_n(x)=\begin{cases} 0,&x<x_i^*\\ \frac{m_1+m_2+\cdots+m_i}{n},&x_i^*\le x<x_{i+1}^*,\,(i=1,2,\cdots,\textcolor{dodgerblue}{l-1})\\ 1,&x\ge x_l^* \end{cases} Fn(x)= 0,nm1+m2++mi,1,x<xixix<xi+1,(i=1,2,,l1)xxlObviously F n ( x ) F_n(x)Fn( x ) is a step function, at eachxi ∗ x_i^*xiThere is a jump here.

The empirical distribution function is not only related to the sample size, but also related to the obtained sample value ( x 1 , x 2 , ⋯ , xn ) (x_1,x_2,\cdots,x_n)(x1,x2,,xn) related.

3. Statistics

1. The concept of statistics

Statistics : Let ( X 1 , X 2 , ⋯ , X n ) (X_1,X_2,\cdots,X_n)(X1,X2,,Xn) is from the overallXXA sample of X , T = g ( X 1 , X 2 , ⋯ , X n ) T=g(X_1,X_2,\cdots,X_n)T=g(X1,X2,,Xn) ( X 1 , X 2 , ⋯   , X n ) (X_1,X_2,\cdots,X_n) (X1,X2,,Xn) of a real-valued function, andggg does not contain any unknown parameters, it is calledTTT is a sample( X 1 , X 2 , ⋯ , X n ) (X_1,X_2,\cdots,X_n)(X1,X2,,Xn) is a statistic.
Observed value of statistics: if( x 1 , x 2 , ⋯ , xn ) (x_1,x_2,\cdots,x_n)(x1,x2,,xn) is the sample( X 1 , X 2 , ⋯ , X n ) (X_1,X_2,\cdots,X_n)(X1,X2,,Xn) , thent = g ( x 1 , x 2 , ⋯ , xn ) t=g(x_1,x_2,\cdots,x_n)t=g(x1,x2,,xn) is called the statisticTTAn observed value of T.

2. Several commonly used statistics

( X 1 , X 2 , ⋯   , X n ) (X_1,X_2,\cdots,X_n) (X1,X2,,Xn) is from the overallXXA sample of X , ( x 1 , x 2 , ⋯ , xn ) (x_1,x_2,\cdots,x_n)(x1,x2,,xn) is the observed value of this sample.

1) Sample mean

Sample mean : X ‾ = 1 n ∑ i = 1 n X i \overline{X}=\frac{1}{n}\sum\limits_{i=1}^n X_iX=n1i=1nXi(Its observed value is recorded as x ‾ \overline{x}x

E ( X ) = μ E(X)=\mu E ( X )=μ D ( X ) = σ 2 D(X)=\sigma^2 D(X)=p2 exists, then

2) Sample variance and sample standard deviation

样本方差 S 2 = 1 n − 1 ∑ i = 1 n ( X i − X ‾ ) 2 = 1 n − 1 ( ∑ i = 1 n X i 2 − n X ‾ 2 ) S^2=\frac{1}{\textcolor{red}{n-1}}\sum\limits_{i=1}^n(X_i-\overline{X})^2=\frac{1}{\textcolor{red}{n-1}}\left(\sum\limits_{i=1}^n X_i^2-n\overline{X}^2\right) S2=n11i=1n(XiX)2=n11(i=1nXi2nX2 )(the observed value is recorded ass 2 s^2s2 )
Sample standard deviation:S = S 2 = 1 n − 1 ∑ i = 1 n ( X i − X ‾ ) 2 S=\sqrt{S^2}=\sqrt{\frac{1}{\textcolor{ red}{n-1}}\sum\limits_{i=1}^n(X_i-\overline{X})^2}S=S2 =n11i=1n(XiX)2 (The observed value is denoted as sss )
They are quantities that reflect the degree of dispersion of the sample values.

E ( X ) = μ E(X)=\mu E ( X )=μ D ( X ) = σ 2 D(X)=\sigma^2 D(X)=p2 exists, then

  • E ( S 2 ) = σ 2 E(S^2)=\sigma^2 E(S2)=p2
  • ( p ) lim ⁡ n → ∞ S 2 = σ 2 (p)\lim\limits_{n\to\infty}S^2=\sigma^2 (p)nlimS2=p2

3) Sample moments

Sample kkk阶原点矩 A k = 1 n ∑ i = 1 n X i k A_k=\frac{1}{n}\sum\limits_{i=1}^n X_i^k Ak=n1i=1nXik(The observed value is recorded as ak a_kak)
sample kkK -order central moment:B k = 1 n ∑ i = 1 n ( X i − X ‾ ) k B_k=\frac{1}{n}\sum\limits_{i=1}^n(X_i-\overline{ X})^kBk=n1i=1n(XiX)k (its observed value is denoted asbk b_kbk

Obviously, A 1 = X ‾ A_1=\overline{X}A1=X B 1 = 0 B_1=0 B1=0 B 2 = n − 1 n S 2 B_2=\textcolor{red}{\frac{n-1}{n}}S^2 B2=nn1S2

Let the overall XXX 'skkK -order origin momentα k = E ( X k ) \alpha_k=E(X^k)ak=E ( Xk )exists, then

  • E ( X k ) = α k E(X^k)=\alpha_kE ( Xk)=ak
  • ( p ) lim ⁡ n → ∞ A k = α k (p)\lim\limits_{n\to\infty}A_k=\alpha_k (p)nlimAk=ak

4) Order statistics

( X 1 , X 2 , ⋯   , X n ) (X_1,X_2,\cdots,X_n) (X1,X2,,Xn) is from the overallXXA sample of X , ( x 1 , x 2 , ⋯ , xn ) (x_1,x_2,\cdots,x_n)(x1,x2,,xn) is an observation for this sample. The observationsx 1 , x 2 , ⋯ , xn x_1,x_2,\cdots,x_nx1,x2,,xnArranged from small to large as x ( 1 ) ≤ x ( 2 ) ≤ ⋯ ≤ x ( n ) x_{(1)}\le x_{(2)}\le\dots\le x_{(n)}x(1)x(2)x(n)

Define the statistic X ( k ) X_{(k)}X(k)The value is x ( k ) x_{(k)}x(k) k = 1 , 2 , ⋯   , n k=1,2,\cdots,n k=1,2,,n ), thus obtainingnnn statisticsX ( 1 ) , X ( 2 ) , ⋯ , X ( n ) X_{(1)},X_{(2)},\cdots,X_{(n)}X(1),X(2),,X(n), and they satisfy X ( 1 ) ≤ X ( 2 ) ≤ ⋯ ≤ X ( n ) X_{(1)}\le X_{(2)}\le\dots\le X_{(n)}X(1)X(2)X(n),称 X ( 1 ) , X ( 2 ) , ⋯   , X ( n ) X_{(1)},X_{(2)},\cdots,X_{(n)} X(1),X(2),,X(n)is the order statistic or order statistic for this sample .

Minimum order statistic : X ( 1 ) = min ⁡ { X ( 1 ) , X ( 2 ) , ⋯ , X ( n ) } X_{(1)}=\min\{X_{(1)},X_{ (2)},\cdots,X_{(n)}\}X(1)=min{ X(1),X(2),,X(n)}
Maximum order statistics:X ( n ) = max ⁡ { X ( 1 ) , X ( 2 ) , ⋯ , X ( n ) } X_{(n)}=\max\{X_{(1)},X_ {(2)},\cdots,X_{(n)}\}X(n)=max{ X(1),X(2),,X(n)}

5) Extremely poor samples

Sample range : R = X ( n ) − X ( 1 ) R=X_{(n)}-X_{(1)}R=X(n)X(1)(The observed value is recorded as r = x ( n ) − x ( 1 ) r=x_{(n)}-x_{(1)}r=x(n)x(1)

6) Sample ppp -quantile

sample ppp quantile: for0 < p < 1 0<p<10<p<1 , statistic M p = { X ( ⌈ np ⌉ ) , np is not an integer 1 2 ( X ( np ) + X ( np + 1 ) ) , np is an integer M_p=\begin{cases} X_{(\lceil np \rceil)},&np\text{is not an integer}\\ \frac{1}{2}\left(X_{(np)}+X_{(np+1)}\right),&np\text{is an integer } \end{cases}Mp={ X(⌈np⌉),21(X(np)+X(np+1)),n p is not an integern p is an integerwhere ⌈ np ⌉ \lceil np\rceiln p stands fornp npn p is rounded up, it is also equivalent tonp + 1 np+1np+1 is rounded down.
Sample median:p = 1 2 p=\frac{1}{2}p=21The sample median when ( nnWhen n is odd, it is equal to X ( ⌈ n 2 ⌉ ) X_{\left(\left\lceil\frac{n}{2}\right\rceil\right)}X(2n) n n n为偶数时等于 1 2 ( X ( n 2 ) + X ( n 2 + 1 ) ) \frac{1}{2}\left(X_{\left(\frac{n}{2}\right)}+X_{\left(\frac{n}{2}+1\right)}\right) 21(X(2n)+X(2n+1))

四、抽样分布

抽样分布:统计量的概率分布

1. Γ \Gamma Γ分布

X X X服从参数为 α , λ \alpha,\lambda α,λ Γ \Gamma Γ分布: X   ~   Γ ( α , λ ) \newcommand{\td}{\,\text{\large\textasciitilde}\,}X\td\Gamma(\alpha,\lambda) X~Γ(α,λ),其中 α > 0 , λ > 0 \alpha>0,\lambda>0 α>0,λ>0

性质:

  1. X   ~   Γ ( α , λ )    ⟹    E ( X ) = α λ \newcommand{\td}{\,\text{\large\textasciitilde}\,}X\td\Gamma(\alpha,\lambda)\implies E(X)=\frac{\alpha}{\lambda} X~Γ(α,λ)E(X)=λα D ( X ) = α λ 2 D(X)=\frac{\alpha}{\lambda^2} D(X)=λ2α
  2. 设随机变量 X 1 , X 2 , ⋯   , X m X_1,X_2,\cdots,X_m X1,X2,,Xm相互独立,且 X i   ~   Γ ( α i , λ ) \newcommand{\td}{\,\text{\large\textasciitilde}\,}X_i\td\Gamma(\alpha_i,\lambda) Xi~C ( ai,λ),则 ∑ i = 1 m X i   ~   Γ ( ∑ i = 1 m α i , λ ) \newcommand{\td}{\,\text{\large\textasciitilde}\,}\sum\limits_{i=1}^m X_i\td\Gamma\left(\sum\limits_{i=1}^m\alpha_i,\lambda\right) i=1mXi~C(i=1mai,l )

2. χ 2 \chi^2h2 distribution

In Γ \GammaTake α = n 2 \alpha=\frac{n}{2}in the Γ distributiona=2nλ = 1 2 \lambda=\frac{1}{2}l=21 Γ \Gamma The Γ distribution isnnχ 2 \chi^2of nh2 distribution.
ZZZ obeysnnχ 2 \chi^2of nh2Structure :Z ~ χ 2 ( n ) \newcommand{\td}{\,\text{\large\texttitle}\,}Z\td\chi^2(n)Z~h2(n)

nature:

  1. Z ~ χ 2 ( n ) ⟹ E ( Z ) = n implies E(Z) . =nZ~h2(n)E(Z)=nD ( Z ) = 2n D(Z)=2nD(Z)=2 n
  2. If the random variable Z 1 , Z 2 , ⋯ , Z m Z_1,Z_2,\cdots,Z_mZ1,Z2,,Zmare independent of each other, and Z i ~ χ 2 ( ni ) \newcommand{\td}{\,\text{\large\textasciitilde}\,}Z_i\td\chi^2(n_i)Zi~h2(ni),则 ∑ i = 1 m Z i   ~   χ 2 ( ∑ i = 1 m n i ) \sum\limits_{i=1}^\newcommand{\td}{\,\text{\large\textasciitilde}\,}m Z_i\td\chi^2\left(\sum\limits_{i=1}^m n_i\right) i=1mZi~χ2(i=1mni)
  3. 设随机变量 X 1 , X 2 , ⋯   , X n X_1,X_2,\cdots,X_n X1,X2,,Xn相互独立,且都服从标准正态分布 N ( 0 , 1 ) N(0,1) N(0,1),则随机变量 χ 2 = ∑ i = 1 n X i 2 \chi^2=\sum\limits_{i=1}^n X_i^2 χ2=i=1nXi2服从自由度为 n n n χ 2 \chi^2 χ2分布,即 χ 2   ~   χ 2 ( n ) \newcommand{\td}{\,\text{\large\textasciitilde}\,}\chi^2\td\chi^2(n) χ2~χ2(n)特别地,若 X   ~   N ( 0 , 1 ) \newcommand{\td}{\,\text{\large\textasciitilde}\,}X\td N(0,1) X~N(0,1),则 X 2   ~   χ 2 ( 1 ) \newcommand{\td}{\,\text{\large\textasciitilde}\,}X^2\td\chi^2(1) X2~χ2(1)

3. t t t分布

T T T服从自由度为 n n n t t t分布: T   ~   t ( n ) \newcommand{\td}{\,\text{\large\textasciitilde}\,}T\td t(n) T~t(n)
t t t分布又称为学生氏分布。
t t t分布的概率密度关于 x = 0 x=0 x=0 symmetry (Γ \GammaΓ distribution,χ 2 \chi^2h2 distribution,FFThe probability density of the F distribution is only atx > 0 x>0x>0 is positive), andlim ⁡ n → ∞ t ( x ; n ) = 1 2 π e − x 2 2 \lim\limits_{n\to\infty} t(x;n)=\frac{1} {\sqrt{2\pi}}e^{-\frac{x^2}{2}}nlimt(x;n)=2 p.m 1e2x2, then n → ∞ n\to\inftyn degrees of freedom arennn'tt __The t distribution converges to the standard normal distributionN ( 0 , 1 ) N(0,1)N(0,1)

性质:若 X   ~   N ( 0 , 1 ) \newcommand{\td}{\,\text{\large\textasciitilde}\,}X\td N(0,1) X~N(0,1) Y   ~   χ 2 ( n ) \newcommand{\td}{\,\text{\large\textasciitilde}\,}Y\td\chi^2(n) Y~χ2(n),且 X X X Y Y Y相互独立,则 T = X Y / n   ~   t ( n ) \newcommand{\td}{\,\text{\large\textasciitilde}\,} T=\frac{X}{\sqrt{Y/n}}\td t(n) T=Y/n X~t(n)

4. F F F分布

F F F服从自由度为 ( n 1 , n 2 ) (n_1,n_2) (n1,n2) F F F分布: F   ~   F ( n 1 , n 2 ) \newcommand{\td}{\,\text{\large\textasciitilde}\,}F\td F(n_1,n_2) F~F(n1,n2)

性质:

  1. X   ~   χ 2 ( n 1 ) \newcommand{\td}{\,\text{\large\textasciitilde}\,}X\td\chi^2(n_1) X~χ2(n1) Y   ~   χ 2 ( n 2 ) \newcommand{\td}{\,\text{\large\textasciitilde}\,}Y\td\chi^2(n_2) Y~χ2(n2),且 X X X Y Y Y相互独立,则 F = X / n 1 Y / n 2   ~   F ( n 1 , n 2 ) \newcommand{\td}{\,\text{\large\textasciitilde}\,} F=\frac{X/n_1}{Y/n_2}\td F(n_1,n_2) F=Y/n2X/n1~F(n1,n2)
  2. F   ~   F ( n 1 , n 2 )    ⟹    1 F   ~   F ( n 2 , n 1 ) \newcommand{\td}{\,\text{\large\textasciitilde}\,}F\td F(n_1,n_2)\implies\frac{1}{F}\td F(n_2,n_1) F~F(n1,n2)F1~F(n2,n1) (just put XXin property 1XYYIt can be proved by exchanging Y )

A detailed definition of the above distribution

  1. Γ \Gamma Γ distribution: if the random variableXXX infinitesimal function f ( x ; α , λ ) = { λ α Γ ( α ) x α − 1 e − λ x , x > 0 0 , x ≤ 0 f(x;\alpha,\lambda)=\begin {cases} \frac{\lambda^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\lambda x},&x>0\\ 0,&x\le 0 \end {cases}f(x;a ,l )={ C ( a )laxa 1 eλx,0,x>0x0where α > 0 \alpha>0a>0λ > 0 \lambda>0l>0 is a constant, it is calledXXX obeys parameters asα , λ \alpha,\lambdaa ,λΓ \GammaΓ type, defineX ~ Γ ( α , λ ) \newcommand{\td}{\,\text{\large\textscript}\,}X\td \Gamma(\alpha,\lambda)X~C ( a ,l ) .
  2. χ 2 \chi^2h2 distribution: if the random variableZZZ has probability density χ 2 ( x ; n ) = { 1 2 n 2 Γ ( n 2 ) xn 2 − 1 e − x 2 , x > 0 0 , x ≤ 0 \chi^2(x;n)=\ begin{cases} \frac{1}{2^{\frac{n}{2}}\Gamma\left(\frac{n}{2}\right)}x^{\frac{n}{2} -1}e^{-\frac{x}{2}},&x>0\\ 0,&x\le 0 \end{cases}h2(x;n)={ 22nC (2n)1x2n1 e2x,0,x>0x0ZZ _Z obeysnnχ 2 \chi^2of nh2 distribution, recorded asZ ~ χ 2 ( n ) \newcommand{\td}{\,\text{\large\textasciitilde}\,}Z\td\chi^2(n)Z~h2(n)
  3. t t t distribution: if the random variableTTT具有概率密度 t ( x ; n ) = Γ ( n + 1 2 ) n π Γ ( n 2 ) ( 1 + x 2 n ) − n + 1 2 t(x;n)=\frac{\Gamma\left(\frac{n+1}{2}\right)}{\sqrt{n\pi}\Gamma\left(\frac{n}{2}\right)}{\left(1+\frac{x^2}{n}\right)}^{-\frac{n+1}{2}} t(x;n)= C(2n)C(2n+1)(1+nx2)2n+1TT _T obeysnnn'tt __t distribution, recorded asT ~ t ( n ) \newcommand{\td}{\,\text{\large\textasciitilde}\,}T\td t(n)T~t(n)
  4. F F F distribution: if the random variableFFF具有概率密度 f ( x ; n 1 , n 2 ) = { Γ ( n 1 + n 2 2 ) Γ ( n 1 2 ) Γ ( n 2 2 ) ( n 1 n 2 ) ( n 1 n 2 x ) n 1 2 − 1 ( 1 + n 1 n 2 x ) − n 1 + n 2 2 f(x;n_1,n_2)=\begin{cases}\frac{\Gamma\left(\frac{n_1+n_2}{2}\right)}{\Gamma\left(\frac{n_1}{2}\right)\Gamma\left(\frac{n_2}{2}\right)}\left(\frac{n_1}{n_2}\right){\left(\frac{n_1}{n_2}x\right)}^{\frac{n_1}{2}-1}{\left(1+\frac{n_1}{n_2}x\right)}^{-\frac{n_1+n_2}{2}}\end{cases} f(x;n1,n2)={ C (2n1) C (2n2)C (2n1+n2)(n2n1)(n2n1x)2n11(1+n2n1x)2n1+n2FF _F obeys degrees of freedom( n 1 , n 2 ) (n_1,n_2)(n1,n2) ofFFF distribution, recorded asF ~ F ( n 1 , n 2 ) \newcommand{\td}{\,\text{\large\textasciitilde}\,}F\td F(n_1,n_2)F~F(n1,n2)

quantile

Let random variable XXThe distribution function of X is F ( x ) = P { X ≤ x } F(x)=P\{X\le x\}F(x)=P{ Xx } .
lower sideppp quantile: for0 < p < 1 0<p<10<p<1 , ifxp x_pxp使 P { X ≤ x p } = F ( x p ) = p P\{X\le x_p\}=F(x_p)=p P{ Xxp}=F(xp)=p , then it is calledxp x_pxpis the distribution F ( x ) F(x)F ( x ) (or random variableXXX ) on the lower sideppp -quantile.
upper sideα \alphaα quantile: for0 < α < 1 0<\alpha<10<a<1 , ifx α x_\alphaxa使P { X > x α } = 1 − F ( x α ) = α P\{X>x_\alpha\}=1-F(x_\alpha)=\alphaP{ X>xa}=1F(xa)=α , then sayx α x_\alphaxais the distribution F ( x ) F(x)F ( x ) (or random variableXXX ) upper sideα \alphaalpha quantile.

upper side α \alphaα quantile = lower side1 − α 1-\alpha1α quantile;
lower sideppp quantile = upper side1 − p 1-p1p -quantile.

Overall, the upper side α \alphaThe α quantile is such thatXXThe probability that X is greater than it isα \alphaThe number of α .

Standard normal distribution N ( 0 , 1 ) N(0,1)N(0,1 ) on the upper sideα \alphaα quantile: foru α u_\alphaua运动,1 − Φ ( u α ) = α 1-\Phi(u_\alpha)=\alpha1Φ ( ua)=α ;u 1 − α = − u α u_{1-\alpha}=-u_\alphau1 - a=ua
t ( n ) t(n) The upper side of the t ( n ) distributionα \alphaAlpha quantile: uset α ( n ) t_\alpha(n)ta( n ) means;t 1 − α = − t α t_{1-\alpha}=-t_\alphat1 - a=ta
χ 2 ( n ) \chi^2(n)h2 (n)the upper side of the distributionα \alphaAlpha quantile: useχ α 2 ( n ) \chi^2_\alpha(n)ha2( n ) means
F ( n 1 , n 2 ) F(n_1,n_2)F(n1,n2) distribution on the upper sideα \alphaAlpha quantile: useF α ( n 1 , n 2 ) F_\alpha(n_1,n_2)Fa(n1,n2) display;F α ( n 1 , n 2 ) = 1 F 1 − α ( n 2 , n 1 ) F_\alpha(n_1,n_2)=\frac{1}{F_{1-\alpha}(n_2, n_1)}Fa(n1,n2)=F1 - a(n2,n1)1

If the probability density function of the distribution is about x = 0 x=0x=0 is symmetrical, then its upper side1 − α 1-\alpha1The α quantile is equal to the upper sideα \alphaThe opposite of the alpha quantile. Taking the standard normal distribution as an example, we know thatΦ ( u α ) = 1 − α \Phi(u_\alpha)=1-\alphaΦ ( ua)=1αΦ ( u 1 − α ) = α \Phi(u_{1-\alpha})=\alphaΦ ( u1 - a)=α,则Φ ( u α ) + Φ ( u 1 − α ) = 1 \Phi(u_\alpha)+\Phi(u_{1-\alpha})=1Φ ( ua)+Φ ( u1 - a)=1 LetΦ ( u α ) = ∫ − ∞ u α ϕ ( x ) ⁣ dx = ∫ − u α + ∞ φ ( x ) ⁣ dx \newcommand{\dif}{\origin{}\!\mathrm{d }}\Phi(u_\easy)=\int_{-\infty}^{u_\infty}\varphi(x)\dif x=\int_{-u_\easy}^{+\infty}\varphi(x). )\dif xΦ ( ua)=uaφ ( x )dx=ua+φ ( x )d x ,Φ ( u 1 − α ) = ∫ − ∞ u 1 − α φ ( x ) ⁣ dx \alpha})=\int_{-\infty}^{u_{1-\alpha}}\varphi(x)\dif xΦ ( u1 - a)=u1 - aφ ( x )d x , the sum of the two is1 11 , indicating that the lower limit of the former integral is equal to the upper limit of the latter integral, sou 1 − α = − u α u_{1-\alpha}=-u_\alphau1 - a=ua. Similarly t 1 − α = − t α t_{1-\alpha}=-t_\alphat1 - a=ta

About F α ( n 1 , n 2 ) = 1 F 1 − α ( n 2 , n 1 ) F_\alpha(n_1,n_2)=\frac{1}{F_{1-\alpha}(n_2,n_1) }Fa(n1,n2)=F1 - a(n2,n1)1, the proof is as follows: Let X ~ F ( n 1 , n 2 ) \newcommand{\td}{\,\text{\large\textasciitilde}\,}X\td F(n_1,n_2)X~F(n1,n2),则P { X > F α ( n 1 , n 2 ) } = α P\{X>F_\alpha(n_1,n_2)\}=\alphaP{ X>Fa(n1,n2)}=α P { 1 X < 1 F α ( n 1 , n 2 ) } = α P\left\{\frac{1}{X}<\frac{1}{F_\alpha(n_1,n_2)}\right\}=\alpha P{ X1<Fa(n1,n2)1}=α ,而1 X ~ F ( n 2 , n 1 ) \newcommand{\td}{\,\text{\large\texttitle}\,}\frac{1}{X}\td F(n_2,n_1)X1~F(n2,n1),故 P { 1 X < F 1 − α ( n 2 , n 1 ) } = 1 − ( 1 − α ) = α P\left\{\frac{1}{X}<F_{1-\alpha}(n_2,n_1)\right\}=1-(1-\alpha)=\alpha P{ X1<F1 - a(n2,n1)}=1(1a )=α,so1F α ( n 1 , n 2 ) = F 1 − α ( n 2 , n 1 ) \frac{1}{F_\alpha(n_1,n_2)}=F_{1-\alpha}(n_2 ,n_1)Fa(n1,n2)1=F1 - a(n2,n1)

Sampling distribution of a normal population

( X 1 , X 2 , ⋯   , X n ) (X_1,X_2,\cdots,X_n) (X1,X2,,Xn) is from the normal populationN ( μ , σ 2 ) N(\mu,\sigma^2)N ( μ ,p2 ),X ‾ \overline{X}Xis the sample mean, S 2 S^2S2 is the sample variance, then:

  1. X ‾ ~ N ( μ , σ 2 n ) \newcommand{\td}{\,\text{\large\textasciitilde}\,}\overline{X}\td N\left(\mu,\frac{\sigma ^2}{n}\right)X~N( m ,np2)
  2. ( n − 1 ) S 2 σ 2 = ∑ i = 1 n ( X i − X ‾ ) 2 σ 2   ~   χ 2 ( n − 1 ) \newcommand{\td}{\,\text{\large\textasciitilde}\,}\frac{(n-1)S^2}{\sigma^2}=\frac{\sum\limits_{i=1}^n\left(X_i-\overline{X}\right)^2}{\sigma^2}\td\chi^2(n-1) p2(n1)S2=p2i=1n(XiX)2~h2(n1)
  3. X ‾ \overline{X} Xwith S 2 S^2S2 independent of each other
  4. T = n ( X ‾ − μ ) S   ~   t ( n − 1 ) \newcommand{\td}{\,\text{\large\textasciitilde}\,}T=\frac{\sqrt{n}\left(\overline{X}-\mu\right)}{S}\td t(n-1) T=Sn (Xm ).~t(n1)

( X 1 , X 2 , ⋯   , X n 1 ) (X_1,X_2,\cdots,X_{n_1}) (X1,X2,,Xn1)( Y 1 , Y 2 , ⋯ , Y n 2 ) (Y_1,Y_2,\cdots,Y_{n_2})(Y1,Y2,,Yn2) are fromN ( μ 1 , σ 2 ) N(\mu_1,\sigma^2)N ( m1,p2 )N ( μ 2 , σ 2 ) N(\mu_2,\sigma^2)N ( m2,p2 )samples (note that the variances are equal), and the two samples are independent of each other,X ‾ = 1 n 1 ∑ i = 1 n 1 X i \overline{X}=\frac{1}{n_1}\sum\limits_ {i=1}^{n_1}X_iX=n11i=1n1Xi Y ‾ = 1 n 2 ∑ i = 1 n 2 Y i \overline{Y}=\frac{1}{n_2}\sum\limits_{i=1}^{n_2}Y_i Y=n21i=1n2Yi S 1 n 1 2 = 1 n 1 − 1 ∑ i = 1 n 1 ( X i − X ‾ ) 2 S_{1n_1}^2=\frac{1}{n_1-1}\sum\limits_{i=1}^{n_1}{\left(X_i-\overline{X}\right)}^2 S1n12=n111i=1n1(XiX)2 S 2 n 2 2 = 1 n 2 − 1 ∑ i = 1 n 2 ( Y i − Y ‾ ) 2 S_{2n_2}^2=\frac{1}{n_2-1}\sum\limits_{i=1}^{n_2}{\left(Y_i-\overline{Y}\right)}^2 S2 n22=n211i=1n2(YiY)2 , then:

  1. T = ( X ‾ − Y ‾ ) − ( μ 1 − μ 2 ) S W 1 n 1 + 1 n 2   ~   t ( n 1 + n 2 − 2 ) ( S W = ( n 1 − 1 ) S 1 n 1 2 + ( n 2 − 1 ) S 2 n 2 2 n 1 + n 2 − 2 ) \newcommand{\td}{\,\text{\large\textasciitilde}\,}T=\frac{\left(\overline{X}-\overline{Y}\right)-(\mu_1-\mu_2)}{S_W\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\td t(n_1+n_2-2)\quad\left(S_W=\sqrt{\frac{(n_1-1)S_{1n_1}^2+(n_2-1)S_{2n_2}^2}{n_1+n_2-2}}\right) T=SWn11+n21 (XY)( m1m2)~t(n1+n22) SW=n1+n22(n11)S1n12+(n21)S2 n22
  2. F = σ 2 2 σ 1 2 S 1 n 1 2 S 2 n 2 2   ~   F ( n 1 − 1 , n 2 − 1 ) \newcommand{\td}{\,\text{\large\textasciitilde}\,}F=\frac{\sigma_2^2}{\sigma_1^2}\frac{S_{1n_1}^2}{S_{2n_2}^2}\td F(n_1-1,n_2-1) F=p12p22S2 n22S1n12~F(n11,n21)

explain:

  1. E ( X ‾ ) = μ E(\overline{X})=\muE(X)=μD ( X ‾ ) = σ 2 n D(\overline{X})=\frac{\sigma^2}{n}D(X)=np2easy.
  2. It needs complex linear algebra knowledge to prove, omitted.
  3. omitted.
  4. We know that X ~ N ( 0 , 1 ) \newcommand{\td}{\,\text{\large\textasciitilde}\,}X\td N(0,1)X~N(0,1 ) ,Y ~ χ 2 ( n ) \newcommand{\td}{\,\text{\large\texttitle}\,}Y\td\chi^2(n)Y~h2 (n),且X , YX,YX,Y can be independently derivedT = XY / n ~ t ( n ) \newcommand{\td}{\,\text{\large\textasciitilde}\,}T=\frac{X}{\sqrt{Y/n}} \td t(n)T=and / n X~t ( n ) . Now we know thatn ( X ‾ − μ ) σ ~ N ( 0 , 1 ) \newcommand{\td}{\,\text{\large\textasciitilde}\,}\frac{\sqrt{n}\left(\ overline{X}-\mu\right)}{\sigma}\td N(0,1)pn (Xm ).~N(0,1 ) ,( n − 1 ) S 2 σ 2 ~ χ 2 ( n − 1 ) \newcommand{\td}{\,\text{\large\texttitle}\,}\frac{(n-1)S^ 2}{\sigma^2}\td\chi^2(n-1)p2(n1)S2~h2(n1 ) , and both are independent, andn ( X ‾ − μ ) / σ ( n − 1 ) S 2 / σ 2 / ( n − 1 ) = n ( X ‾ − μ ) S \frac{\sqrt{ n}\left(\overline{X}-\mu\right)/\sigma}{\sqrt{(n-1)S^2/\sigma^2/(n-1)}}=\frac{\ sqrt{n}\left(\overline{X}-\mu\right)}{S}(n1)S2 /p2/(n1) n (Xm ) / p=Sn (Xm ).,故 n ( X ‾ − μ ) S   ~   t ( n − 1 ) \newcommand{\td}{\,\text{\large\textasciitilde}\,}\frac{\sqrt{n}\left(\overline{X}-\mu\right)}{S}\td t(n-1) Sn (Xm ).~t(n1)
  5. What you need to know is X ‾ − Y ‾ ~ N ( μ 1 − μ 2 , 1 n 1 + 1 n 2 σ 2 ) \newcommand{\td}{\,\text{\large\textasciitilde}\,}\overline {X}-\overline{Y}\td N\left(\mu_1-\mu_2,\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}\sigma^2\right)XY~N( m1m2,n11+n21 p2 ), so thatU = ( X ‾ − Y ‾ ) − ( μ 1 − μ 2 ) σ 1 n 1 + 1 n 2 ~ N ( 0 , 1 ) \newcommand{\td}{\,\text{\large \textasciitilde}\,}U=\frac{\left(\overline{X}-\overline{Y}\right)-(\mu_1-\mu_2)}{\sigma\sqrt{\frac{1}{n_1 }+\frac{1}{n_2}}}\td N(0,1)U=pn11+n21 (XY)( m1m2)~N(0,1 ) N又V = ( n 1 − 1 ) S 1 n 1 2 σ 2 + ( n 2 − 1 ) S 2 n 2 2 σ 2 ~ χ 2 ( n 1 + n 2 − 2 ) \newcommand{\td }{\,\text{\large\texttitle}\,}V=\frac{(n_1-1)S_{1n_1}^2}{\sigma^2}+\frac{(n_2-1)S_{2n_2 }^2}{\sigma^2}\td\chi^2(n_1+n_2-2)V=p2(n11)S1n12+p2(n21)S2 n22~h2(n1+n22),故 T = U V / ( n 1 + n 2 − 2 )   ~   t ( n 1 + n 2 − 2 ) \newcommand{\td}{\,\text{\large\textasciitilde}\,}T=\frac{U}{\sqrt{V/(n_1+n_2-2)}}\td t(n_1+n_2-2) T=V/(n1+n22) U~t(n1+n22)
  6. We know ( n 1 − 1 ) S 1 n 1 2 σ 1 2 ~ χ 2 ( n 1 − 1 ) \newcommand{\td}{\,\text{\large\textasciitilde}\,}\frac{(n_1 -1)S_{1n_1}^2}{\sigma_1^2}\td\chi^2(n_1-1)p12(n11)S1n12~h2(n11 ) ,( n 2 − 1 ) S 2 n 2 2 σ 2 2 ~ χ 2 ( n 2 − 1 ) \newcommand{\td}{\,\text{\large\textset}\,}\frac{( n_2-1)S_{2n_2}^2}{\sigma_2^2}\td\chi^2(n_2-1)p22(n21)S2 n22~h2(n21 ) , and the two are independent of each other, so according toFFF function functionF = ( n 1 − 1 ) S 1 n 1 2 σ 1 2 / ( n 1 − 1 ) ( n 2 − 1 ) S 2 n 2 2 σ 2 2 / ( n 2 − 1 ) ~ F ( n 1 − 1 , n 2 − 1 ) \newcommand{\td}{\,\text{\large\texttitle}\,}F=\frac{\left.\frac{(n_1-1)S_{ 1n_1}^2}{\sigma_1^2}\right/(n_1-1)}{\left.\frac{(n_2-1)S_{2n_2}^2}{\sigma_2^2}\right/(n_2 -1)}\td F(n_1-1,n_2-1)F=p22(n21)S2 n22/(n21)p12(n11)S1n12/(n11)~F(n11,n21)

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Origin blog.csdn.net/qaqwqaqwq/article/details/128454175