[Probability Theory] Final Review Notes: Parameter Estimation

1. Point estimate

1. The concept of estimators

Point estimation : Let the population XXThe distribution function of$X$$F(x;i_1,i_2,\cdots ,i_l)$ ,includeθ$i_1,i_2,\cdots ,i_l$is the unknown parameter to be estimated, $(X_1,X_2,\cdots ,X_n)$ is from the overall$X$的样本，$(x_1,x_2,\cdots ,x_n)$是相应的样本值，点估计问题就是要构造$l$个适当的统计量${\hat{\theta }}_i(X_1,X_2,\cdots ,X_n)(i=1,2,\cdots ,l)$，分别用观测值${\hat{\theta }}_i(x_1,x_2,\cdots ,x_n)$作为未知参数${\theta }_i$的估计值。
估计量：估计用的统计量${\hat{\theta }}_i(X_1,X_2,\cdots ,X_n)$
estimated value:the observed value${\hat{i}}_i(x_1,x_2,\cdots ,x_n)$ are collectively referred to as estimators and estimates
as estimates without confusion, and are abbreviated asθ ^ i \hat{\theta}_i${\hat{i}}_i$。

The estimator is a function of the sample and is a random variable, and the estimated values ​​obtained by different sample values ​​are often different.

2. Finding method of estimator

method of moments

Let the overall XXex llof$x$$L$ -order origin moment$a_k=E\left(X^k\right)(k=1,2,\cdots ,l)$ Specific,dimensionθ$i_1,i_2,\cdots ,i_l$For example, $a_k=a_k(i_1,i_2,\cdots ,i_l)$ . Replace the overall origin moment with the sample origin moment ($a_k\to A_k$), the unknown parameters are replaced by their estimators ( $i_i\to {\hat{i}}_i$)Then $$\{\begin{array}{l}{a}_1\left({\hat{i}}_1,{\hat{i}}_2,\cdots ,{\hat{i}}_l\right)=A_1\\ a_2\left({\hat{i}}_1,{\hat{i}}_2,\cdots ,{\hat{i}}_l\right)=A_2\\ \cdots \\ a_l\left({\hat{i}}_1,{\hat{i}}_2,\cdots ,{\hat{i}}_l\right)=A_l\end{array}$$Solving this system of equations can get ${\hat{i}}_1,{\hat{i}}_2,\cdots ,{\hat{i}}_l$（是$A_1,A_2,\cdots ,A_k$function), and take them as $i_1,i_2,\cdots ,i_l$estimate of . $A_1$Generally written $\overline{X}$。

The theoretical basis of the moment estimation method is the law of large numbers, when $When\; n$ is sufficiently large, the sample moment$A_k$falls in the overall moment α k \alpha_k with a high probability$a_k$The vicinity of , so $A_k$As $a_k$The moment estimator of .

例 $X\text{~}U(0,\theta )$ ,ifθ\thetaThe moment estimator of$\theta \; .$
Solution: We know that$$a_1=E\left(X\right)=\frac{}{2}$$Put $a_1$Change to $A_1$（$\overline{X}$），$\theta$ is replaced by$\hat{i}$得$$\overline{X}=\frac{\hat{i}}{2}$$Therefore θ \thetaThe moment estimator of$\theta$$\hat{i}=2\overline{X}$。

The moment estimation method does not need to know the distribution of the population. The advantage is that it is simple and direct, but the disadvantage is that it only uses the local characteristics of the population and does not make full use of the information of the population.

Maximum Likelihood Estimation

Idea: If there is a certain distribution, so that under this distribution $(x_1,x_2,\cdots ,x_n)$ has the largest probability, then$(x_1,x_2,\cdots ,x_n)$ from this distribution.

Likelihood function : if the overall $X$ is a discrete or continuous random variable, and its distribution law isP { X = x } = p ( x ; θ 1 , θ 2 , ⋯ , θ l ) P\{X=x\}=p(x;\ theta_1,\theta_2,\cdots,\theta_l)P{ X=x}=p(x;i1​,i2​,⋯,il​) , or its probability density is$f(x;i_1,i_2,\cdots ,i_l)$ ,includeθ$i_1,i_2,\cdots ,i_l$is an unknown parameter, in the parameter space Θ \ThetaValue in$\Theta$$x$ in the random variableXXTake a value within the range of possible values ​​for$X.$Let$(X_1,X_2,\cdots ,X_n)$ is from the overall$A\; sample\; of\; X$ , then$(X_1,X_2,\cdots ,X_n)$ Feedback L ( x 1 , x 2 , ⋯ , xn ; θ 1 , θ 2 , ⋯ , θ l ) = P { X 1 = x 1 , X 2 = x 2 , ⋯ , X n = xn } = ∏ i = 1 np ( xi ; θ 1 , θ 2 , ⋯ , θ l ) \begin{aligned} L(x_1,x_2,\cdots,x_n;\theta_1,\theta_2,\cdots,\theta_l)&= P\{X_1=x_1,X_2=x_2,\cdots,X_n=x_n\}\\&=\prod\limits_{i=1}^np(x_i;\theta_1,\theta_2,\cdots,\theta_l)\ end{aligned}L(x1​,x2​,⋯,xn​;i1​,i2​,⋯,il​)​=P{ X1​=x1​,X2​=x2​,⋯,Xn​=xn​}=i=1∏n​p(xi​;i1​,i2​,⋯,il​)​Let $$L(x_1,x_2,\cdots ,x_n;i_1,i_2,\cdots ,i_l)=\prod _{i=1}^{n}f(x_i;i_1,i_2,\cdots ,i_l)$$ when fixed$(x_1,x_2,\cdots ,x_n)$ , put$Levels$ θ 1 , θ 2 , ⋯ , θ l \theta_1,\theta_2,\cdots,\theta_l$i_1,i_2,\cdots ,i_l$is defined in $\Theta$ , it is called parameters$i_1,i_2,\cdots ,i_l$The likelihood function of , and abbreviated as $L(i_1,i_2,\cdots ,i_l)$ . That is:the likelihood function is the distribution law/probability density of the sample, and then regarded as a function of the parameters.
Log likelihood function: logarithm of the likelihood function$\ln L(i_1,i_2,\cdots ,i_l)$ is called the log-likelihood function.

Maximum likelihood estimation method : get sample values $(x_1,x_2,\cdots ,x_n)$后,inθ${\hat{i}}_1,{\hat{i}}_2,\cdots ,{\hat{i}}_n$Let $$L({\hat{i}}_1,{\hat{i}}_2,\cdots ,{\hat{i}}_n)=\underset{(i_1,i_2,\cdots ,i_l)\in \Theta }{\max }L(i_1,i_2,\cdots ,i_l)$$ obtained in this way${\hat{i}}_1,{\hat{i}}_2,\cdots ,{\hat{i}}_n$with sample values $(x_1,x_2,\cdots ,x_n)$ with respect to${\hat{i}}_i={\hat{i}}_i(x_1,x_2,\cdots ,x_n)$ , and called parameter$i_i(i=1,2,\cdots ,l)$ ,and the corresponding statisticsθ${\hat{i}}_i={\hat{i}}_i(X_1,X_2,\cdots ,X_n)(i=1,2,\cdots ,l)$ is called the parameter$i_i$The maximum likelihood estimator of .

Since $\ln x$ is$x$的单调增函数，所以$L$取最大的时候$\ln L$也取最大，我们也可以考察$\ln L$的最大值。

在很多时候，$L$和$\ln L$关于参数${\theta }_1,{\theta }_2,\cdots ,{\theta }_l$的偏导数存在，此时${\hat{\theta }}_1,{\hat{\theta }}_2,\cdots ,{\hat{\theta }}_n$可从似然方程$$\{\begin{array}{l}\frac{\partial L(i_1,i_2,\cdots ,i_l)}{\partial {\theta }_1}=0\\ \frac{\partial L(i_1,i_2,\cdots ,i_l)}{\partial {\theta }_2}=0\\ \cdots \\ \frac{\partial L(i_1,i_2,\cdots ,i_l)}{\partial {\theta }_l}=0\end{array}$$Constraints { ∂ ln ⁡ L ( θ 1 , θ 2 , ⋯ , θ l ) ∂ θ 1 = 0 ∂ ln ⁡ L ( θ 1 , θ 2 , ⋯ , θ l ) ∂ θ 2 = ⋯ $$\{\begin{array}{l}\frac{\partial \ln L(i_1,i_2,\cdots ,i_l)}{\partial {\theta }_1}=0\\ \frac{\partial \ln L(i_1,i_2,\cdots ,i_l)}{\partial {\theta }_2}=0\\ \cdots \\ \frac{\partial \ln L(i_1,i_2,\cdots ,i_l)}{\partial {\theta }_l}=0\end{array}$$solved in.

Definition X ~ U ( 0 , θ ) \newcommand{\td}{\,\text{\large\texttitle}\,}X\td U(0,\theta )$X\text{~}U(0,\theta )$ ,ifθ\thetaThe maximum likelihood estimator of$\theta \; .$
Solution:XXThe probability density of$X$$$f(x;i)=\{\begin{array}{ll}\frac{1}{i},& 0\le x\le i\\ 0,& \end{array}$$Then sample $(X_1,X_2,\cdots ,X_n)$ is$$f(x_1,x_2,\cdots ,x_n;i)=\prod _{i=1}^{n}f(x_i;i)=\{\begin{array}{ll}\frac{1}{i^n},& 0\le x_1,x_2,\cdots ,x_n\le i\\ 0,& \end{array}$$Think of it as $Function\; of\; \theta$ ($x_1,x_2,\cdots ,x_n$is known), then $\theta$的似然函数为 L ( θ ) = { 1 θ n , θ ≥ max ⁡ { x 1 , x 2 , ⋯   , x n } 0 , 其他 L(\theta)=\begin{cases} \frac{1}{\theta^n},&\theta\ge\max\{x_1,x_2,\cdots,x_n\}\\ 0,&\text{其他} \end{cases} L(θ)={ θn1​,0,​θ≥max{ x1​,x2​,⋯,xn​}其他​这个函数我们不用求导就能求出最大值。首先，它在 θ ≥ max ⁡ { x 1 , x 2 , ⋯   , x n } \theta\ge\max\{x_1,x_2,\cdots,x_n\} θ≥max{ x1​,x2​,⋯,xn​}时才是正数；其次，在$\theta$满足这个条件的情况下，因为${\theta }^n$在分母，所以我们希望$\theta$ should be as small as possible. So whenθ = max ⁡ { x 1 , x 2 , ⋯ , xn } \theta=\max\{x_1,x_2,\cdots,x_n\}i=max{ x1​,x2​,⋯,xn​}时$L\left(\theta \right)$ takes the maximum value. θ \thetaThe maximum likelihood estimator of$\theta$$\hat{i}=X_{(n)}$. This is different from the estimator obtained by the method of moments estimation.

2. Estimated selection criteria

1. Unbiasedness

Unbiased estimator : Let $(X_1,X_2,\cdots ,X_n)$ is from the overallXXA sample of$X$$\theta$ is contained inXXUnknown parameters in the distribution of$X$$The\; value\; range\; of\; \theta$ is$\Theta$，$\hat{i}=\hat{i}(X_1,X_2,\cdots ,X_n)$是θ \thetaAn estimator of$\theta \; .$If$\forall \theta \in \Theta$ ,$E\left(\hat{i}\right)=\theta$ , then$\hat{i}$ flow\thetaAn unbiased estimator of$\theta \; .$
Biased estimator: A biased estimator, where the bias (referred to as bias) is equal to$E\left(\hat{i}\right)-\theta$。
渐进无偏估计量：若$E\left(\hat{\theta }\right)-\theta \ne 0$，但当样本容量$n\to \infty$时，有$\underset{n\to \infty }{\lim }\left[E\left(\hat{\theta }\right)-\theta \right]=0$，则称$\hat{\theta }$是$\theta$的渐近无偏估计量。

设$(X_1,X_2,\cdots ,X_n)$是来自总体$X$的样本，无论$What\; distribution\; does\; X$ obey?
(1) If$E\left(X\right)=\mu$ exists, then the sample mean$\overline{X}$Is E ( X ) E(X)Unbiased estimator of$E$$($$X$$)$
; (2) If$D(X)=p^2$ exists, then the sample variance$S^2$ is$p^{}$The unbiased estimator of${}^2$
; (3) If the population$k$ slightly$E\left(X^k\right)=a_k$exists, then $K$ -order sample origin moment$A_k=\frac{1}{k}\sum _{i=1}^n{X}_i^k$is $K$ -order overall origin moment$a_k$unbiased estimator of .

Example It can be proved that if the population $X\text{~}U(0,\theta )$ , parameter$i>0$ , then$2\overline{X}$和$\frac{n+1}{n}{X}_{(n)}$Both are θ \thetaAn unbiased estimator of$\theta \; .$

Although $S^2$ is$p^{}$An unbiased estimator of${}^2$$S$ is notσ \sigmaThe value of$\sigma$$\sqrt{\frac{n-1}{2}}\frac{C\left(\frac{n-1}{2}\right)}{C\left(\frac{n}{2}\right)}S$ isσ \sigmaAn unbiased estimator of$\sigma \; .$This shows that if$\hat{i}$ flow$The\; unbiased\; estimator\; of\; \theta$ , in general,$g\left(\hat{i}\right)$is notθ \thetaunbiased estimator of$\theta$$g$ is a linear function.

2. Effectiveness

The unbiased estimator is not necessarily unique, so we need to select the one with the most concentrated value, that is, the one with the smallest variance as the best estimator.

Validity : Let ${\hat{i}}_1$和${\hat{i}}_2$Both are θ \thetaThe unbiased estimator of$\theta$$D\left({\hat{i}}_1\right)\le D\left({\hat{i}}_2\right)$ , then${\hat{i}}_1$Page ${\hat{i}}_2$efficient.

Definition X ~ U ( 0 , θ ) \newcommand{\td}{\,\text{\large\texttitle}\,}X\td U(0,\theta )$X\text{~}U(0,\theta )$ ,则${\hat{i}}_2=\frac{n+1}{n}{X}_{(n)}$比${\hat{\theta }}_1=2\overline{X}$有效（$D\left({\hat{\theta }}_1\right)=\frac{{\theta }^2}{3n}>D\left({\hat{\theta }}_2\right)=\frac{{\theta }^2}{n(n+2)}$）。

最小方差无偏估计量：在所有估计量中方差最小的无偏估计量

3. 相合性

相合估计量/一致估计量：设$\hat{\theta }=\hat{\theta }(X_1,X_2,\cdots ,X_n)$是参数$\theta$的估计量，如果当$n\to \infty$时，$\hat{\theta }$ converges to$\theta$ ,for$$\forall \epsilon >0,\underset{n\to \infty }{\lim }P\left\{\left|\hat{i}-i\right|<e\right\}=1$$ is called$\hat{i}$ forθ \thetaConsistent estimator/consistent estimator of$\theta$$(p)\underset{n\to \infty }{\lim }\hat{i}=\theta$或$\hat{i}\stackrel{P}{⟶}i(n\to \infty )$ .
Mean square consistent estimator: if when$n\to \infty$时,$\hat{i}$ Mean square converges to$\theta$ ,in$$\underset{n\to \infty }{\lim }E\left[{\left(\hat{i}-i\right)}^2\right]=0$$ is called$\hat{i}$ forθ \thetaThe mean square consistent estimator of$\theta$$(\text{m. s. })\underset{n\to \infty }{\lim }\hat{i}=\theta$或$\hat{i}\stackrel{L^2}{⟶}i(n\to \infty )$。

Consistency is the most basic requirement for an estimator, which requires that when the sample size increases infinitely, the estimator can be used to estimate parameters with arbitrarily small precision.

It can be proved that common moment estimators are consistent estimators (eg $A_k\to a_k$、$\overline{X}\to E(X)$、$S^2\to p^2$、$S\to \sigma$ ). A mean-square consistent estimator must be a consistent estimator, but not necessarily vice versa.

An important fact: Let the overall $The\; distribution\; function\; of\; X$ is$F(x)$，$(X_1,X_2,\cdots ,X_n)$ is from the overall$X$的样本，$F_n(X)=\frac{1}{n}\sum _{i=1}^n[X_i\le x]$ is an empirical distribution function, then for any fixed$x$，$F_n\left(x\right)$ isF ( x ) F(x)Unbiased estimator, consistent estimator, mean square consistent estimator of$F$$($$x$$)\; .$

Summarize

Form : $E\left(\hat{i}\right)=\theta$
Validity: The smaller the variance, the better
Consistency: Convergence according to probability (when the sample size is large enough, the gap between the estimated value and the real value can be arbitrarily small)

3. Interval estimation

1. Two-sided interval estimation

$$\underset{\Downarrow }{P\left\{{\hat{i}}_1(X_1,X_2,\cdots ,X_n)<i<{\hat{i}}_2(X_1,X_2,\cdots ,X_n)\right\}=1-}$$Random interval $\left({\hat{i}}_1,{\hat{i}}_2\right)$ is the parameter$The\; confidence\; level\; of\; \theta$ is$1-$Two-sided confidence intervalfor$\alpha$ .
${\hat{i}}_1$: lower confidence limit
${\hat{i}}_2$: Confidence upper limit
$1-\alpha$ :Confidence
$\alpha$ : range$\left({\hat{i}}_1,{\hat{i}}_2\right)$ does notcontain$Probability\; of\; \theta$ (generally small)

At confidence level $1-When\; \alpha$ is given, the length of the confidence interval$E\left({\hat{i}}_2-{\hat{i}}_1\right)$ the smaller the better.

Find the unknown parameter θ \thetaThe specific practice of the two-sided confidence interval of$\theta \; :$

(1) Find the pivot amount $Z=Z\left(X_1,X_2,\cdots ,X_n,\theta \right)$，我们需要知道$Z$的分布，并且此分布不依赖于任何未知参数，也不依赖于$\theta$。
(2) 对于给定的置信度$1-\alpha$，求出两个常数$k_1,k_2$使得 P { k 1 < Z < k 2 } = 1 − α P\{k_1<Z<k_2\}=1-\alpha P{ k1​<Z<k2​}=1−α。
(3) $k_1<Z<k_2\stackrel{\text{改写}}{⟶}{\hat{\theta }}_1<\theta <{\hat{\theta }}_2$：$\left({\hat{\theta }}_1,{\hat{\theta }}_2\right)$是置信度为$1-\alpha$的置信区间。
(4) 根据样本值计算${\hat{\theta }}_1,{\hat{\theta }}_2$的具体值。

例 设$X\text{~}N(\mu ,{\sigma }^2)$，${\sigma }^2$已知，$\mu$未知，求参数$\mu$的置信度为$1-\alpha$的置信区间。
解：取枢轴量$U=\frac{\overline{X}-\mu }{\sigma /\sqrt{n}}\text{~}N(0,1)$ , it can be seen that$N(0,1)$ Does not depend on any parameters.
Now to find$k_1,k_2$使得 P { k 1 < U < k 2 } = 1 − α P\left\{k_1<U<k_2\right\}=1-\alpha P{ k1​<U<k2​}=1−α , generally take$k_1=-u_{a/}$，$k_2=u_{a/}$。 notice− $-u_{a/}=u_{1-a/}$Let P { k 1 < U < k 2 } = 1 − P { U ≥ k 2 } − P { U ≤ k 1 } = 1 − α 2 − ( 1 − P { U > k 1 } ) = − α 2 − [ 1 − ( 1 − α 2 ) ] = α \begin{aligned} P\{k_1<U<k_2\}&=1-P\{U\ge k_2\}-P\{U\ and k_1\}\\ &=1-\frac{\alpha}{2}-\left(1-P\left\{U>k_1\right\}\right)\\ &=1-\frac{\ alpha}{2}-\left[1-\left(1-\frac{\alpha}{2}\right)\right]\\ &=\alpha\end{aligned}P{ k1​<U<k2​}​=1−P{ U≥k2​}−P{ U≤k1​}=1−2a−(1−P{ U>k1​})=1−2a−[1−(1−2a)]=a

If $P\left\{-u_{a/}<\frac{\overline{X}-}{s/\sqrt{n}}<u_{a/}\right\}=1-\alpha$ , then$$\begin{array}{rl}P\left\{-\frac{\sqrt{n}}{p}{u}_{a/}<\overline{X}-m<\frac{\sqrt{n}}{p}{u}_{a/}\right\}& =1-\alpha \\ P\left\{\overline{X}-\frac{\sqrt{n}}{p}{u}_{a/}<m<\overline{X}+\frac{\sqrt{n}}{p}{u}_{a/}\right\}& =1-\alpha \end{array}$$于是得$\mu$的置信度为$1-\alpha$的置信区间为$\left(\overline{X}-\frac{\sqrt{n}}{\sigma }{u}_{\alpha /2},\overline{X}+\frac{\sqrt{n}}{\sigma }{u}_{\alpha /2}\right)$。

其实，选取枢轴量的过程就是从$X$的分布中剔除参数$\theta$的影响的过程。$X$的分布受$\theta$影响，我们就需要消除这种影响，所以我们提出统计量$Z$，它的分布是完全确定的，只有这样我们才能确定参数$k_1,k_2$。如果$X$的分布不是确定的，那么我们很难求出置信区间。

2. One-sided interval estimation

P { θ ‾ ( X 1 , X 2 , ⋯   , X n ) < θ } = 1 − α    ⟹    ( θ ‾ , + ∞ ) P\left\{\underline{\theta}(X_1,X_2,\cdots,X_n)<\theta\right\}=1-\alpha\implies\left(\underline{\theta},+\infty\right) P{ i(X1​,X2​,⋯,Xn​)<i }=1−a⟹(i,+ ∞ ) current$The\; confidence\; level\; of\; \theta$ is$1-$One-sided confidence intervalfor$\alpha$ ,$\underline{}$Let P { θ < θ ‾ ( X 1 , X 2 , ⋯ , X n ) } = 1 − α ⟹ ( − ∞ , θ ‾ ) P\left\{\theta<\overline{\theta } (
$P\left\{i<\overline{i}(X_1,X_2,\cdots ,X_n)\right\}=1-a⟹\left(-\infty ,\overline{i}\right)$是$The\; confidence\; level\; of\; \theta$ is$1-$One-sided confidence intervalfor$\alpha$ ,$\overline{i}$is the upper bound of confidence .

即：$\left(\underline{},+\infty \right)$ contains$The\; probability\; of\; \theta$ is$1-\alpha$，$\left(-\infty ,\overline{i}\right)$ contains$The\; probability\; of\; \theta$ is$1-a$ .

At confidence level $1-When\; \alpha$ is given, the larger the confidence lower bound is, the better, and the smaller the confidence upper bound is, the better.

4. Interval Estimation of Normal Population Parameters

For the case of a single population, we set $X\text{~}N(\mu ,p^2)$; for the case of two populations, we assume$X\text{~}N(m_1,p_1^2)$，$Y\text{~}N(m_2,p_2^2)$。$X$ has a sample size of$n$ , the sample variance is$S_X^2$；$The\; sample\; size\; of\; Y$ is$m$ , the sample variance is$S_Y^2$。

The following is the case of a single population ( $X\text{~}N(\mu ,p^2)$）。

$p^2\text{known, examine}\mu$

枢長量$U=\frac{\sqrt{n}(\overline{X}-m}{p}\text{~}N(0,1)$

注意$P\left\{-u_{a/}<U<u_{a/}\right\}=1-\alpha$
P { U < u α } = 1 − α P\left\{U<u_\alpha\right\}=1-\alpha P{ U<ua}=1−α
$P\left\{U>-u_{}\right\}=1-a$

$p^2\text{unknown, investigate}\mu$

枢轴量$T=\frac{\sqrt{n}(\overline{X}-m}{S}\text{~}t(n-1)$

注意$P\left\{-t_{a/}<T<t_{a/}\right\}=1-\alpha$
P { T < t α } = 1 − α P\left\{T<t_\alpha\right\}=1-\alpha P{ T<ta}=1−α
$P\left\{T>-t_{}\right\}=1-a$

$\mu \text{is known, examine}{\sigma }^2$

Definition $h^2=\frac{\sum _{i=1}^n{\left(X_i-m\right)}^2}{p^2}\text{~}{h}^2(n)$

$\mu \text{unknown, examine}{\sigma }^2$

Definition $h^2=\frac{\sum _{i=1}^n{\left(X_i-\overline{X}\right)}^2}{p^2}=\frac{(n-1)S^2}{p^2}\text{~}{h}^2(n-1)$

Proportion$P\left\{x^2>h_{a/2}^2(n-1)\right\}=\frac{}{2}$, $P\left\{x^2>h_{1-a/2}^2(n-1)\right\}=1-\frac{}{2}$，故$P\left\{x_{1-a/2}^2\right(n-1)<h^2<h_{a/2}^2(n-1)\}=1-\alpha$。
$P\left\{x^2<h_a^2(n-1)\right\}=1-\alpha$
$P\left\{x^2>h_{1-a}^2(n-1)\right\}=1-a$

The following is the case of two populations ( $X\text{~}N(m_1,p_1^2)$，$Y\text{~}N(m_2,p_2^2)$）。

$p_1^2,p_2^2\text{Known, examine}{\mu }_1-m_2$

枢长量$U=\frac{\left(\overline{X}-\overline{Y}\right)-(m_1-m_2)}{\sqrt{\frac{p_1^2}{n}+\frac{p_2^2}{m}}}\text{~}N(0,1)$

Note that $D\left(\overline{X}-\overline{Y}\right)=\frac{p_1^2}{n}+\frac{p_2^2}{m}$。

$p_1^2=p_2^2\text{Unknown, investigate}{\mu }_1-m_2$

枢轴量$T=\frac{\left(\overline{X}-\overline{Y}\right)-({\mu }_1-{\mu }_2)}{S_W\sqrt{\frac{1}{n}+\frac{1}{m}}}\text{~}t(n+m-2)$，其中$S_W=\sqrt{\frac{(n-1)S_X^2+(m-1)S_Y^2}{n+m-2}}$

NoteU $U=\frac{\left(\overline{X}-\overline{Y}\right)-(m_1-m_2)}{p\sqrt{\frac{1}{n}+\frac{1}{m}}}\text{~}N(0,1)$，$V=\frac{(n-1)S_X^2+(m-1)S_Y^2}{p^2}\text{~}{h}^2(n+m-2)$。

${\mu }_1,{\mu }_2\text{已知，考察}\frac{{\sigma }_1^2}{{\sigma }_2^2}$

枢轴量$F=\frac{\sum _{i=1}^n\frac{{(X_i-m_1)}^2}{p_1^2}/n}{\sum _{j=1}^m\frac{{(Y_j-m_2)}^2}{{\sigma }_2^2}/m}=\frac{p_2^2}{p_1^2}\frac{m\sum _{i=1}^n{(X_i-m_1)}^2}{n\sum _{j=1}^m{(Y_j-m_2)}^2}\text{~}F(n,m)$

$m_1,m_2\text{unknown, investigation}\frac{p_1^2}{p_2^2}$

Definition $F=\frac{p_2^2}{p_1^2}\frac{S_X^2}{S_Y^2}\text{~}F(n-1,m-1)$

注意$P\left\{F>F_{\alpha /2}(n-1,m-1)\right\}=\frac{\alpha }{2}$，$P\left\{F>F_{1-\alpha /2}(n-1,m-1)\right\}=1-\frac{\alpha }{2}$，故 P { F 1 − α / 2 ( n − 1 , m − 1 ) < F < F α / 2 ( n − 1 , m − 1 ) } = 1 − α P\{F_{1-\alpha/2}(n-1,m-1)<F<F_{\alpha/2}(n-1,m-1)\}=1-\alpha P{ F1−α/2​(n−1,m−1)<F<Fα/2​(n−1,m−1)}=1−α。
P { F < F α ( n − 1 , m − 1 ) } = 1 − α P\left\{F<F_{\alpha}(n-1,m-1)\right\}=1-\alpha P{ F<Fa(n−1,m−1)}=1−α
$P\left\{F>F_{1-}(n-1,m-1)\right\}=1-a$

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$t$ distribution and standard normal distribution$N(0,1)$ Similarly, probability density curves are all about$x=0$ symmetric，$u_{1-}=-u_{}$，$t_{1-}(n)=-t_{}(n)$；
$F$ distribution and$h^2$ distribution is similar, the probability density is only at$x>0$ is positive.

Regardless of XXAny distribution$X$ obeys satisfies P { q 1 − α / 2 < X < q α / 2 } = 1 − α P\{q_{1-\alpha/2}<X<q_{\alpha/2}\ }=1-\alphaP{ q1 − a / 2<X<qa / 2}=1−α , where$q_{}$means XXThe upper side of the distribution$X$$alpha$ quantile.

Regarding the degree of freedom, it can be considered as follows: If the overall mean $\mu$ is known, then the degree of freedom is the sample size; if the population mean$\mu$ is unknown, and with the sample mean$\overline{X}$Instead, one degree of freedom is lost. For the tt used when testing the sample mean difference$t$ distribution, its degrees of freedom are the sum of the degrees of freedom of the two.