[the Nine Yang Manual] 2019 Fudan University Applied Statistics real problem + analysis

The real part

One 、(15 branch ) There is $n-1$ A white ball 1 A black ball , There is $n$ A white ball , Take one ball from each of the two bags for Exchange , Seek exchange $N$ The probability that the black ball will still be in the armour pocket after times .

Two 、(15 branch ) $f(x,y)=A{e}^{-(2x+3y)}I[x>0,y>0],$ seek
(1)(3 branch ) $A$;
(2)(3 branch ) $P(X<2,Y<1)$;
(3)(3 branch ) $X$ Marginal density of ;
(4)(3 branch ) $P(X<3\mid Y<1)$;
(5)(3 branch ) $f(x\mid y)$.

3、 ... and 、(10 branch ) $X_1,X_2,$i.i.d $\sim N\left(\mu ,{\sigma }^2\right),$ seek $E\max \left\{X_1,X_2\right\}$.

Four 、(20 branch ) $EX=0,\mathrm{Var}(X)={\sigma }^2,$ Prove for any $\epsilon >0,$ Yes
(1)(10 branch ) $P(|X|>\epsilon )\le \frac{{\sigma }^2}{{\epsilon }^2}$;
(2)(10 branch ) $P(X>\epsilon )\le \frac{{\sigma }^2}{{\sigma }^2+{\epsilon }^2}$.

5、 ... and 、(10 branch ) $X_1,X_2,i.i.d\sim N(0,1),$ seek $\frac{X_1}{X_2}$ The distribution of .

6、 ... and 、(20 branch ) $X_1,X_2,\dots ,X_n,i.i.d\sim N\left(\mu ,{\sigma }^2\right),\mu$ It is known that , prove :
(1)(10 branch ) $\frac{1}{n}{\sum }_{i=1}^n{\left(X_i-\mu \right)}^2$ yes ${\sigma }^2$ Effective estimation of ;
(2)(10 branch ) $\frac{1}{n}\sqrt{\frac{\pi }{2}}{\sum }_{i=1}^n\left|X_i-\mu \right|$ yes $\sigma$ Unbiased estimation of , But it's not effective .

7、 ... and 、(20 branch ) The overall distribution function $F(x)$ Continuous single increase , $X_{(1)},X_{(2)},\dots ,X_{(n)}$ Is the order statistic of random samples from the population , $Y_i=F\left(X_{(i)}\right),$ seek
(1)(10 branch ) $E{Y}_i,\mathrm{Var}\left(Y_i\right)$;
(2)(10 branch ) ${\left(Y_1,Y_2,\dots ,Y_n\right)}^T$ The covariance matrix of .

8、 ... and 、(20 branch ) Some come from the general $U(\theta ,2\theta )$ A random sample of $X_1,\cdots ,X_n$, seek $\theta$ Moment estimation sum of MLE, And verify the unbiasedness and consistency .

Nine 、(20 branch ) set up $X_1,\cdots ,X_n$ Is from $N(\mu ,1)$ A random sample of , Consider the hypothesis test problem $$H_0:\mu =1\mathrm{v}\mathrm{s}{H}_1:\mu =2$$ Given denial domain $W=\{\bar{X}>1.6\}$, Answer the following questions :
(1)(10 branch ) Find the probability of making two kinds of mistakes $\alpha$, $\beta$;
(2)(10 branch ) Request the second kind of error $\beta \le 0.01$, Find the value range of the sample size .

The analysis part

One 、(15 branch ) There is $n-1$ A white ball 1 A black ball , There is $n$ A white ball , Take one ball from each of the two bags for Exchange , Seek exchange $N$ The probability that the black ball will still be in the armour pocket after times .

Solution:
set up $p_k$ For exchange $k$ The probability that the black ball will still be in the armour pocket after times , Then according to the full probability formula ,
$$p_{k+1}=p_k\cdot \frac{n-1}{n}+\left(1-p_k\right)\cdot \frac{1}{n}=\left(1-\frac{2}{n}\right)p_k+\frac{1}{n},$$ There are $$p_N-\frac{1}{2}=\left(1-\frac{2}{n}\right)\left(p_{N-1}-\frac{1}{2}\right)=\cdots ={\left(1-\frac{2}{n}\right)}^N\left(p_0-\frac{1}{2}\right),$$ namely $$p_N=\frac{1}{2}+\frac{1}{2}{\left(1-\frac{2}{n}\right)}^N\text{. }$$

Two 、(15 branch ) $f(x,y)=A{e}^{-(2x+3y)}I[x>0,y>0],$ seek
(1)(3 branch ) $A$;
(2)(3 branch ) $P(X<2,Y<1)$;
(3)(3 branch ) $X$ Marginal density of ;
(4)(3 branch ) $P(X<3\mid Y<1)$;
(5)(3 branch ) $f(x\mid y)$.

Solution:
(1) By the regularity of probability , $1={\int }_{R^2}f(x,y)dxdy=A{\int }_0^{+\infty }{e}^{-2x}dx{\int }_0^{+\infty }{e}^{-3y}dy=\frac{A}{6}\Rightarrow A=6$.
(2) $P(X<2,Y<1)=6{\int }_0^2{e}^{-2x}dx{\int }_0^1{e}^{-3y}dy=6\left(1-e^{-4}\right)\left(1-e^{-3}\right)$.
(3) $f_X(x)={\int }_0^{+\infty }6e^{-2x-3y}dy=2e^{-2x},x>0$.
(4) because $f_X(x)=2e^{-2x},f_Y(y)=3e^{-3y}$, so $X,Y$ Are independent of each other , therefore $$P(X<3\mid Y<1)=P(X<3)=1-e^{-6}.$$(5) $f(x\mid y)=\frac{f(x,y)}{f_Y(y)}=2e^{-2x},x>0,y>0$.

3、 ... and 、(10 branch ) $X_1,X_2,$i.i.d $\sim N\left(\mu ,{\sigma }^2\right),$ seek $E\max \left\{X_1,X_2\right\}$.

Solution:
Make $Y_i=\frac{X_i-\mu }{\sigma },i=1,2$, so $\max \left\{Y_1,Y_2\right\}=\frac{\max \left\{X_1,X_2\right\}-\mu }{\sigma }$.
and $$E\max \left\{Y_1,Y_2\right\}=E{Y}_1{I}_{\left[Y_1\ge Y_2\right]}+E{Y}_2{I}_{\left[Y_1<Y_2\right]}=2E{Y}_1{I}_{\left[Y_1\ge Y_2\right]},$$$$E[Y_1{I}_{\left[Y_1\ge Y_2\right]}]=\frac{1}{2\pi }{\int }_{-\infty }^{+\infty }{\int }_x^{+\infty }y{e}^{-\frac{x^2+y^2}{2}}dxdy=\frac{1}{2\pi }{\int }_{\frac{\pi }{4}}^{\frac{5\pi }{4}}\sin \theta d\theta {\int }_0^{+\infty }{r}^2{e}^{-\frac{r^2}{2}}dr$$
among ${\int }_0^{+\infty }{r}^2{e}^{-\frac{r^2}{2}}dr=\sqrt{2}{\int }_0^{+\infty }\sqrt{\frac{r^2}{2}}{e}^{-\frac{r^2}{2}}d\left(\frac{r^2}{2}\right)=\sqrt{2}\Gamma \left(\frac{3}{2}\right)=\sqrt{\frac{\pi }{2}}$, so $E\max \left\{Y_1,Y_2\right\}=\frac{1}{\sqrt{\pi }},E\max \left\{X_1,X_2\right\}=\mu +\frac{\sigma }{\sqrt{\pi }}$.

Four 、(20 branch ) $EX=0,\mathrm{Var}(X)={\sigma }^2,$ Prove for any $\epsilon >0,$ Yes
(1)(10 branch ) $P(|X|>\epsilon )\le \frac{{\sigma }^2}{{\epsilon }^2}$;
(2)(10 branch ) $P(X>\epsilon )\le \frac{{\sigma }^2}{{\sigma }^2+{\epsilon }^2}$.

Solution:
(1) remember $I_{[|X|>\epsilon ]}=\{\begin{array}{ll}1,& |X|>\epsilon ,\\ 0,& |X|\le \epsilon .\end{array}$ It's a collection $\{\omega :|X(\omega )|>\epsilon \}$ The indicative function of , It can be seen that : $$I_{[|X|>\epsilon ]}\le \frac{|X{|}^2}{{\epsilon }^2}$$ so $P(|X|>\epsilon )=E{I}_{[|X|>\epsilon ]}\le \frac{E|X{|}^2}{{\epsilon }^2}=\frac{{\sigma }^2}{{\epsilon }^2}$.
(2) By Markov inequality , Yes $$P(X>\epsilon )=P(X+a>\epsilon +a)\le \frac{E(X+a{)}^2}{(\epsilon +a{)}^2}=\frac{{\sigma }^2+a^2}{(\epsilon +a{)}^2},$$ take $a=\frac{{\sigma }^2}{\epsilon }$, Then there happens to be $$\frac{{\sigma }^2+a^2}{(\epsilon +a{)}^2}=\frac{{\sigma }^2+\frac{{\sigma }^4}{{\epsilon }^2}}{{\left(\epsilon +\frac{{\sigma }^2}{\epsilon }\right)}^2}=\frac{{\sigma }^2}{{\sigma }^2+{\epsilon }^2},$$ therefore $P(X>\epsilon )\le \frac{{\sigma }^2}{{\sigma }^2+{\epsilon }^2}$.

5、 ... and 、(10 branch ) $X_1,X_2$, i.i.d. $\sim N(0,1),$ seek $\frac{X_1}{X_2}$ The distribution of .

Solution:
Due to denominator $X_2$ The distribution of is about 0 symmetry , therefore $\frac{X_1}{|X_2|}$ And $\frac{X_1}{X_2}$ Homodistribution , And obviously $\frac{N(0,1)}{\sqrt{\frac{{\chi }^2(1)}{1}}}$ It's a The degree of freedom is 1 Of $t$ Distribution , therefore $\frac{X_1}{|X_2|}$ Also, the degree of freedom is 1 Of $t$ Distribution , Its probability density is $$f(x)=\frac{\Gamma (1)}{\sqrt{\pi }\Gamma \left(\frac{1}{2}\right)}{\left(x^2+1\right)}^{-1}=\frac{1}{\pi }\cdot \frac{1}{1+x^2},-\infty <x<+\infty ,$$ The standard Cauchy distribution .

6、 ... and 、(20 branch ) $X_1,X_2,\dots ,X_n,i.i.d\sim N\left(\mu ,{\sigma }^2\right),\mu$ It is known that , prove :
(1)(10 branch ) $\frac{1}{n}{\sum }_{i=1}^n{\left(X_i-\mu \right)}^2$ yes ${\sigma }^2$ Effective estimation of ;
(2)(10 branch ) $\frac{1}{n}\sqrt{\frac{\pi }{2}}{\sum }_{i=1}^n\left|X_i-\mu \right|$ yes $\sigma$ Unbiased estimation of , But it's not effective .

Solution:
(1) To calculate ${\sigma }^2$ Of Fisher The amount of information , According to the definition $$I\left({\sigma }^2\right)=E{\left[\frac{\partial \ln f\left(X;{\sigma }^2\right)}{\partial {\sigma }^2}\right]}^2=\frac{1}{4{\sigma }^4}E{\left[{\left(\frac{X-\mu }{\sigma }\right)}^2-1\right]}^2,$$ just $E{\left[{\left(\frac{X-\mu }{\sigma }\right)}^2-1\right]}^2$ yes ${\chi }^2$ (1) The variance of , so $I\left({\sigma }^2\right)=\frac{1}{2{\sigma }^4}$. therefore , ${\sigma }^2$ Of C-R The lower bound is $\frac{1}{nI\left({\sigma }^2\right)}=\frac{2}{n}{\sigma }^4.$ Calculate again $\frac{1}{n}{\sum }_{i=1}^n{\left(X_i-\mu \right)}^2$ The expected variance of , because $\frac{1}{{\sigma }^2}{\sum }_{i=1}^n{\left(X_i-\mu \right)}^2\sim {\chi }^2(n)$, Expectation is $n$, The variance is $2n$, therefore $$E\left[\frac{1}{n}\sum _{i=1}^n{\left(X_i-\mu \right)}^2\right]={\sigma }^2,\mathrm{Var}\left[\frac{1}{n}\sum _{i=1}^n{\left(X_i-\mu \right)}^2\right]=\frac{2}{n}{\sigma }^4,$$ It is an unbiased estimate , The variance just reaches C-R Lower bound , Therefore, it is an effective estimate .
(2) First , Make $g(x)=\sqrt{x}$, be $\sigma$ Of $\mathrm{C}-\mathrm{R}$ The lower bound is $\frac{{\left[g^{\prime }\left({\sigma }^2\right)\right]}^2}{nI\left({\sigma }^2\right)}=\frac{1}{2n}{\sigma }^2$, Let's calculate $\frac{1}{n}\sqrt{\frac{\pi }{2}}{\sum }_{i=1}^n\left|X_i-\mu \right|$ The expected variance of :$$\frac{1}{\sigma }E\left|X_1-\mu \right|={\int }_{-\infty }^{+\infty }|x|\frac{1}{\sqrt{2\pi }}{e}^{-\frac{x^2}{2}}dx=\sqrt{\frac{2}{\pi }}{\int }_0^{+\infty }x{e}^{-\frac{x^2}{2}}dx=\sqrt{\frac{2}{\pi }}\Rightarrow E\left|X_1-\mu \right|=\sqrt{\frac{2}{\pi }}\sigma ,$$$$E{\left|X_1-\mu \right|}^2={\sigma }^2\text{, so }\mathrm{Var}\left(\left|X_1-\mu \right|\right)={\sigma }^2-\frac{2}{\pi }{\sigma }^2=\left(1-\frac{2}{\pi }\right){\sigma }^2\text{, }$$ so $E\left[\frac{1}{n}\sqrt{\frac{\pi }{2}}{\sum }_{i=1}^n\left|X_i-\mu \right|\right]=\sigma ,\mathrm{Var}\left[\frac{1}{n}\sqrt{\frac{\pi }{2}}{\sum }_{i=1}^n\left|X_i-\mu \right|\right]=\frac{\left(\frac{\pi }{2}-1\right)}{n}{\sigma }^2$, It is an unbiased estimate , but Don't reach C-R Lower bound , Not a valid estimate .

7、 ... and 、(20 branch ) The overall distribution function $F(x)$ Continuous single increase , $X_{(1)},X_{(2)},\dots ,X_{(n)}$ Is the order statistic of random samples from the population , $Y_i=F\left(X_{(i)}\right),$ seek
(1)(10 branch ) $E{Y}_i,\mathrm{Var}\left(Y_i\right)$;
(2)(10 branch ) ${\left(Y_1,Y_2,\dots ,Y_n\right)}^T$ The covariance matrix of .

Solution:
(1) because $U_1=F\left(X_1\right),U_2=F\left(X_2\right),\cdots ,U_n=F\left(X_n\right)$ Independence and obedience $[0,1]$ Evenly distributed , so $Y_i=U_{(i)}$ It happens to be the order statistics of uniform distribution , According to the definition of order statistics , Yes :
$$\begin{array}{c}{f}_i(y)=\frac{n!}{(i-1)!(n-i)!}{f}_U(y)P^{i-1}\{U\le y\}{P}^{n-i}\{U\ge y\},\text{ Insert one by one , Yes }\\ f_i(y)=\frac{n!}{(i-1)!(n-i)!}{y}^{i-1}(1-y{)}^{n-i},0\le y\le 1\text{, Exactly }\mathrm{Beta}(i,n-i+1).\end{array}$$ According to the nature of beta distribution $E{Y}_i=\frac{i}{n+1},\mathrm{Var}\left(Y_i\right)=\frac{i(n-i+1)}{(n+1{)}^2(n+2)}$.
(2) We consider the $\left(Y_i,Y_j\right),i<j$ The joint distribution of , According to the definition of order statistics , Yes
$$f_{i,j}(x,y)=\frac{n!}{(i-1)!(j-i-1)!(n-j)!}{f}_U(x)f_U(y)P^{i-1}\{U\le x\}{P}^{j-i-1}\{x\le U\le y\}{P}^{n-j}\{U\ge y\}$$ namely $$f_{i,j}(x,y)=\frac{n!}{(i-1)!(j-i-1)!(n-j)!}{x}^{i-1}(y-x{)}^{j-i-1}(1-y{)}^{n-j},0\le x\le y\le 1.$$
The covariance problem is reduced to the calculation of integral ${\int }_0^1{\int }_0^{y}xy\cdot x^{i-1}(y-x{)}^{j-i-1}(1-y{)}^{n-j}dxdy$, And if I We put $y$ Regard as $y=x+(y-x)$, Then this integral can be reduced to two parts :$$\begin{array}{c}{\int }_0^1{\int }_0^y{x}^{i+1}(y-x{)}^{j-i-1}(1-y{)}^{n-j}dxdy=\frac{(i+1)!(j-i-1)!(n-j)!}{(n+2)!},\\ {\int }_0^1{\int }_0^y{x}^i(y-x{)}^{j-i}(1-y{)}^{n-j}dxdy=\frac{i!(j-i)!(n-j)!}{(n+2)!},\end{array}$$( First of all, look at $f_{i,j}(x,y)$ The expression of , Will find
$$\frac{(n+2)!}{(i+1)!(j-i-1)!(n-j)!}{x}^{i+1}(y-x{)}^{j-i-1}(1-y{)}^{n-j}$$ It is also a density function , Therefore, the above two integral values can be easily obtained through the regularity of probability ) There are $E[Y_i{Y}_j]=\frac{i(j+1)}{(n+1)(n+2)}$, so $\mathrm{Cov}\left(Y_i,Y_j\right)=\frac{i(n+1-j)}{(n+1{)}^2(n+2)}$. So the covariance matrix is $\Sigma ={\left(a_{ij}\right)}_{n\times n}$, among $a_{ij}=\frac{i(n+1-j)}{(n+1{)}^2(n+2)},i<j$.

8、 ... and 、(20 branch ) Some come from the general $U(\theta ,2\theta )$ A random sample of $X_1,\cdots ,X_n$, seek $\theta$ Moment estimation sum of MLE, And verify the unbiasedness and consistency .

Solution:
First find the moment estimation , Expect $E{X}_1=\frac{3\theta }{2}$, From the principle of substitution ${\hat{\theta }}_M=\frac{2}{3}\bar{X}$. The expectation is easy to see that it is unbiased , It is known from the strong law of large numbers that it is a strongly consistent estimate .
Ask again MLE, Write the likelihood function as
$$L\left(\theta \right)=\frac{1}{{\theta }^n}{I}_{\left\{X_{\left(n\right)}<2\theta \right\}}{I}_{\left\{X_{\left(1\right)}>\theta \right\}}=\frac{I_{\left\{\frac{X_{\left(n\right)}}{2}<\theta <X_{\left(1\right)}\right\}}}{{\theta }^n},$$ It can be seen that $\frac{1}{{\theta }^n}$ About $\theta$ Monotonic decline , so $\theta$ The minimum value is MLE, namely ${\hat{\theta }}_L=\frac{X_{(n)}}{2}$. Use transformation $Y_i=\frac{X_i-\theta }{\theta }\sim U(0,1)$, know $Y_{(n)}=\frac{X_{(n)}-\theta }{\theta }\sim Beta(n,1)$, There are
$$E\left(Y_{\left(n\right)}\right)=\frac{n}{n+1},E\left({\hat{\theta }}_L\right)=\frac{1}{2}E\left(X_{\left(n\right)}\right)=\frac{1}{2}\left[\theta E\left(Y_{\left(n\right)}\right)+\theta \right]=\frac{2n+1}{2n+2}\theta .$$ therefore ${\hat{\theta }}_L$ Not without bias , But asymptotically unbiased , Then look at consistency , Yes
$$\begin{array}{rl}P\left(\left|{\hat{\theta }}_L-\theta \right|>{\epsilon }_1\right)& =P\left(\left|X_{\left(n\right)}-2\theta \right|>{\epsilon }_2\right)\\ & =P\left(\left|Y_{\left(n\right)}-1\right|>{\epsilon }_3\right)\\ & =P\left(Y_{\left(n\right)}<1-{\epsilon }_3\right)\\ & ={\left(1-{\epsilon }_3\right)}^n,\end{array}$$ Convergence of series , So it is a strongly consistent estimate .

Nine 、(20 branch ) set up $X_1,\cdots ,X_n$ Is from $N(\mu ,1)$ A random sample of , Consider the hypothesis test problem $$H_0:\mu =1\mathrm{v}\mathrm{s}{H}_1:\mu =2$$ Given denial domain $W=\{\bar{X}>1.6\}$, Answer the following questions :
(1)(10 branch ) $n=10$, Find the probability of making two kinds of mistakes $\alpha$, $\beta$;
(2)(10 branch ) Request the second kind of error $\beta \le 0.01$, Find the value range of the sample size .

Solution:
(1) First count the first kind of errors , When the original assumption comes true $\bar{X}\sim N(1,\frac{1}{n})$, so
$$\begin{array}{rl}\alpha & =P_{\mu =1}\left(\bar{X}>1.6\right)\\ & =P_{\mu =1}\left(\sqrt{n}\left(\bar{X}-1\right)>0.6\sqrt{n}\right)\\ & =1-\Phi \left(0.6\sqrt{n}\right)\\ & =1-\Phi \left(0.6\sqrt{10}\right).\end{array}$$ Calculate the second kind of error , When alternative assumptions come true $\bar{X}\sim N(2,\frac{1}{n})$, so
$$\begin{array}{rl}\beta & =P_{\mu =2}\left(\bar{X}\le 1.6\right)\\ & =P_{\mu =1}\left(\sqrt{n}\left(\bar{X}-2\right)\le -0.4\sqrt{n}\right)\\ & =\Phi \left(-0.4\sqrt{n}\right)\\ & =\Phi \left(-0.4\sqrt{10}\right).\end{array}$$ (2) According to the first (1) Ask calculation , Make $\Phi \left(-0.4\sqrt{n}\right)\le 0.01$, have to
$$-0.4\sqrt{n}\le -2.33*n\ge \frac{2.33^2}{0.4^2}*n\ge 34.$$