[the Nine Yang Manual] 2022 Fudan University Applied Statistics real problem + analysis

The real part

One 、(20 branch ) There is... In the bag : $a$ Red ball , $a$ yellow ball , $b$ blue ball . Yes, put it back and touch it 3 A ball , set up $A=\{$ Draw out a yellow ball and a red ball , And the red ball is taken out before the yellow ball $\}$. seek :
(1)(10 branch ) $P(A)$;
(2)(10 branch ) if $\{$ I didn't touch the basketball $\}$ And $A$ The probability is the same , seek $a:b$.

Two 、(10 branch ) Discrete random variables $X$ The distribution column of is
$$P(X=a)=P(X=b)=P(X=a+1)=\frac{1}{3},$$ among $a<b<a+1$. Find the value range of its variance .

3、 ... and 、(20 branch ) Discrete random variables $X$ Take only $x,x+a$ Two values , among $a>0$, And $Var(X)=1$, seek $a$ Value range and $X$ The distribution column of .

Four 、(20 branch ) With random vectors $\left(\begin{array}{c}X\\ Y\end{array}\right)$, It is known that after arbitrary rotation transformation $$\left(\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right)\left(\begin{array}{c}X\\ Y\end{array}\right)$$ Still with $\left(\begin{array}{c}X\\ Y\end{array}\right)$ Homodistribution , Try to solve the following problems :
(1) seek $P(0<Y<X)$;
(2) seek $\frac{Y}{X}$ The distribution of .

5、 ... and 、(20 branch ) It is known that $(X,Y)\sim N(0,0;1,1;\frac{1}{2})$, seek $P(X>0,Y>0)$.

6、 ... and 、(10 branch ) $X_1,\cdots ,X_n,\cdots$ yes i.i.d. There are random variables in the second moment of , $Y_n={\sum }_{i=1}^n{X}_i$, ask : $\{\frac{Y_n}{n^2}\}$ Obey the law of large numbers .

7、 ... and 、(10 branch ) $X_1,\cdots ,X_n$ yes i.i.d. obey $N(\mu ,{\sigma }^2)$ Random variable of , $F$ Is its distribution function , seek $-2{\sum }_{i=1}^n\ln F(X_i)$ The distribution of .

8、 ... and 、(10 branch ) $X_1,\cdots ,X_6$ yes i.i.d. Of $U(0,1)$ A random variable , seek $Var(2X_{(2)}+3X_{(3)})$.

Nine 、(10 branch ) $X_1,\cdots ,X_n$i yes i.i.d. Of $U(0,\theta )$ A random sample , set up $a{X}_{(1)},b{X}_{(3)}$ yes $\theta$ Unbiased estimation of , seek $a,b$ And compare which of them is more effective .

Ten 、(10 branch ) set up $X_1,\cdots ,X_n$ yes i.i.d. Of $N(\mu ,16)$ A random sample , $\mu$ The prior distribution of is $N(a,b^2)$, Find the posterior distribution .

11、 ... and 、(10 branch ) set up $X_1,\cdots ,X_n$ yes i.i.d. Of $U(0,\theta )$ A random sample , Consider the hypothesis test problem
$$H_0:\theta \le 1\mathrm{v}\mathrm{s}{H}_1:\theta >1$$ Construct reject fields $W=\{X_{(n)}\ge c\}$. Answer the following questions :
(1)(5 branch ) $\alpha =0.05$, seek $c$;
(2)(5 branch ) When $\theta =1.5$, In order to make the probability of making the second kind of mistake $\beta \le 0.1$, Find the minimum sample size .

The analysis part

One 、(20 branch ) There is... In the bag : $a$ Red ball , $a$ yellow ball , $b$ blue ball . Yes, put it back and touch it 3 A ball , set up $A=\{$ Draw out a yellow ball and a red ball , And the red ball is taken out before the yellow ball $\}$. seek :
(1)(10 branch ) $P(A)$;
(2)(10 branch ) if $\{$ I didn't touch the basketball $\}$ And $A$ The probability is the same , seek $a:b$.

Solution:
[ notes ]: The stem of the question can be understood as $A_1=\{$ A red ball is taken out before a yellow one $\}$, It can also be interpreted as $A_2=\{$ All red balls are taken out before yellow balls $\}$.
(1) Think about it first $A_1$, Yes
$$A_1=\left\{\text{Red red yellow},\text{Red, yellow, red},\text{Red, yellow, yellow},\text{Blue red yellow},\text{Red blue yellow},\text{Red, yellow, blue}\right\},$$ There are
$$P\left(A_1\right)=\frac{a^3+a^3+a^3+3a^{2}b}{{\left(2a+b\right)}^3}=\frac{3a^2\left(a+b\right)}{{\left(2a+b\right)}^3}.$$ Think again $A_2$, Yes
$$A_2=\left\{\text{Red red yellow},\text{Red, yellow, yellow},\text{Blue red yellow},\text{Red blue yellow},\text{Red, yellow, blue}\right\},$$ There are
$$P\left(A_2\right)=\frac{a^3+a^3+3a^{2}b}{{\left(2a+b\right)}^3}=\frac{a^2\left(2a+3b\right)}{{\left(2a+b\right)}^3}.$$ (2) To calculate $B=\{$ I didn't touch the basketball $\}$, Yes
$$P\left(B\right)={\left(\frac{2a}{2a+b}\right)}^3=\frac{8a^3}{{\left(2a+b\right)}^3},$$ if $P(B)=P(A_1)$, namely
$$8a^3=3a^2\left(a+b\right)*8a=3\left(a+b\right)*\frac{a}{b}=\frac{3}{5}.$$ if $P(B)=P(A_2)$, namely
$$8a^3=a^2\left(2a+3b\right)*8a=2a+3b*\frac{a}{b}=\frac{1}{2}.$$

Two 、(10 branch ) Discrete random variables $X$ The distribution column of is
$$P(X=a)=P(X=b)=P(X=a+1)=\frac{1}{3},$$ among $a<b<a+1$. Find the value range of its variance .

Solution:
because $Var(X)=Var(X-a)$, So suppose $X$ The value is
$$P(X=0)=P(X=c)=P(X=1)=\frac{1}{3},$$ among $c=b-a\in (0,1)$, There are $EX=\frac{c+1}{3}$, and
$$E{X}^2=\frac{c^2+1}{3},Var(X)=\frac{3c^2+3-(c+1{)}^2}{9}=\frac{2}{9}\left[{\left(c-\frac{1}{2}\right)}^2+\frac{3}{4}\right]\in \left[\frac{1}{6},\frac{2}{9}\right).$$

3、 ... and 、(20 branch ) Discrete random variables $X$ Take only $x,x+a$ Two values , among $a>0$, And $Var(X)=1$, seek $a$ Value range and $X$ The distribution column of .

Solution:
The question only gives the condition of variance , and $Var(X)=Var(X-x)$, So let's assume $X$ Take only $0,a$ Two values , And
$$Var\left(X\right)=a^{2}p-a^2{p}^2=a^{2}p\left(1-p\right)=1,$$ among $p=P(X=a)$, because $p(1-p)\le \frac{1}{4}$, therefore $a\ge 2$. And when $a$ Timing , $p$ It can be solved , namely
$$p^2-p+\frac{1}{a}=0*p=\frac{1\pm \sqrt{1-\frac{4}{a}}}{2}.$$ therefore $X$ The distribution column of is
$$P\left(X=x\right)=\frac{1+\sqrt{1-\frac{4}{a}}}{2},P\left(X=x+a\right)=\frac{1-\sqrt{1-\frac{4}{a}}}{2},$$ Or is it
$$P\left(X=x\right)=\frac{1-\sqrt{1-\frac{4}{a}}}{2},P\left(X=x+a\right)=\frac{1+\sqrt{1-\frac{4}{a}}}{2}.$$

Four 、(20 branch ) With random vectors $\left(\begin{array}{c}X\\ Y\end{array}\right)$, It is known that after arbitrary rotation transformation $$\left(\begin{array}{cc}\cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{array}\right)\left(\begin{array}{c}X\\ Y\end{array}\right)$$ Still with $\left(\begin{array}{c}X\\ Y\end{array}\right)$ Homodistribution , Try to solve the following problems :
(1) seek $P(0<Y<X)$;
(2) seek $\frac{Y}{X}$ The distribution of .

Solution:
(1) set up $X,Y$ The density function of is $f_{X,Y}(x,y)$, For rotation transformation $(U,V)=(X,Y)A^T$, By the method of variable transformation , Yes
$$f_{U,V}\left(u,v\right)=f_{X,Y}\left(\left(u,v\right)A\right)\left|A\right|=f_{X,Y}\left(\left(u,v\right)A\right),$$ On the other hand , because $U,V$ And $X,Y$ Homodistribution , so $f_{U,V}\left(u,v\right)=f_{X,Y}\left(u,v\right)$, in summary , Yes
$$f_{X,Y}\left(x,y\right)=f_{X,Y}\left(\left(x,y\right)A\right),$$ Transform for any rotation $A$ establish , This indicates that there is a function $u$ bring $f_{X,Y}(x,y)=u(\sqrt{x^2+y^2})$. Make polar coordinate transformation $$\{\begin{array}{l}X=R\cos \Theta ,\\ Y=R\sin \Theta ,\end{array}$$ By the method of variable transformation , Yes $(R,\Theta )$ The distribution of is
$$f_{R,\Theta }\left(r,\theta \right)=f_{X,Y}\left(r\cos \theta ,r\sin \theta \right)r=ru\left(r\right),r\in \left(0,+\infty \right),\theta \in \left(0,2\pi \right),$$ Factorization , therefore $R,\Theta$ Independent , And $f_{\Theta }(\theta )$ Is constant , so $\Theta \sim U(0,2\pi )$. therefore $$P\left(0<Y<X\right)=P\left(\Theta \in \left(0,\frac{\pi }{4}\right)\right)=\frac{1}{8}.$$ (2) $T=\frac{Y}{X}=\tan \Theta \sim \mathrm{c}\mathrm{h}\left(0,1\right)$, Standard Cauchy distribution , The distribution function method can be used to explain : Pay attention here $\Theta$ The value is $(0,2\pi )$, Not on the inverse function $\arctan$ The domain of definition , Discuss carefully . To any $t>0$, Yes
$$\begin{array}{rl}\left\{T\le t\right\}& =\left\{\tan \Theta \le t\right\}\\ & =\left\{\Theta \in \left[0,\mathrm{a}\mathrm{r}\mathrm{c}\tan t\right]\cup \left(\frac{\pi }{2},\mathrm{a}\mathrm{r}\mathrm{c}\tan t+\pi \right]\cup \left(\frac{3\pi }{2},2\pi \right]\right\}\end{array},$$ so $P(T\le t)=\frac{\pi +2\arctan t}{2\pi }=\frac{1}{2}+\frac{\arctan t}{\pi }$, $t>0$. Discuss again about any $t<0$, Yes
$$\begin{array}{rl}\left\{T\le t\right\}& =\left\{\tan \Theta \le t\right\}\\ & =\left\{\Theta \in \left(\frac{\pi }{2},\mathrm{a}\mathrm{r}\mathrm{c}\tan t+\pi \right]\cup \left(\frac{3\pi }{2},\mathrm{a}\mathrm{r}\mathrm{c}\tan t+2\pi \right)\right\}\end{array}$$ so $P(T\le t)=\frac{\pi +2\arctan t}{2\pi }=\frac{1}{2}+\frac{\arctan t}{\pi }$, $t<0$. To sum up, there are
$$F_T\left(t\right)=\frac{1}{2}+\frac{\mathrm{a}\mathrm{r}\mathrm{c}\tan t}{\pi },t\in R,$$ To derive
$$f_T\left(t\right)=\frac{1}{\pi \left(1+t^2\right)},t\in R,$$ This is the standard Cauchy distribution .

5、 ... and 、(20 branch ) It is known that $(X,Y)\sim N(0,0;1,1;\frac{1}{2})$, seek $P(X>0,Y>0)$.

Solution:
Make $W=\frac{2}{\sqrt{3}}(Y-\frac{1}{2}X)$, Then there are $EW=0$, $Var(W)=1$, And $Cov(X,W)=0$, There are $X,W$ The independent identity obeys the standard normal distribution . Further consider
$$P\left(X>0,Y>0\right)=P\left(-X>0,-Y>0\right)=P\left(X<0,Y<0\right),$$ Find out
$$P\left(X>0,Y>0\right)=\frac{1}{2}P\left(XY>0\right)=\frac{1}{2}P\left(\frac{Y}{X}>0\right),$$ recycling $W$ To deal with , namely
$$\left\{\frac{Y}{X}>0\right\}=\left\{\frac{\frac{\sqrt{3}W}{2}+\frac{1}{2}X}{X}>0\right\}=\left\{\frac{W}{X}>-\frac{1}{\sqrt{3}}\right\}.$$ utilize $\frac{W}{X}$ Obey the standard Cauchy distribution , Yes $$P\left(X>0,Y>0\right)=\frac{1}{2}{\int }_{-\frac{\sqrt{3}}{3}}^{+\infty }\frac{1}{\pi \left(1+t^2\right)}dt=\frac{1}{3}.$$

6、 ... and 、(10 branch ) $X_1,\cdots ,X_n,\cdots$ yes i.i.d. There are random variables in the second moment of , $Y_n={\sum }_{i=1}^n{X}_i$, ask : $\{\frac{Y_n}{n^2}\}$ Obey the law of large numbers .

Solution:
[ Law 1 ]: Make $Z_n=\frac{Y_n}{n^2}$, Calculate covariance directly , First of all, there is
$$Cov\left(Y_k,Y_{k+l}\right)=Cov\left(\sum _{j=1}^k{X}_j,\sum _{j=1}^{k+l}{X}_j\right)=kVar\left(X_1\right),$$ Further, there are , When $l\to \infty$, Yes
$$Cov\left(Z_k,Z_{k+l}\right)=\frac{1}{k^2{\left(k+l\right)}^2}Cov\left(Y_k,Y_l\right)=\frac{Var\left(X_1\right)}{k{\left(k+l\right)}^2}\to 0,$$ By Bernstein condition , $\{\frac{Y_n}{n^2}\}$ Obey the law of large numbers .
[ Law two ]: By strong law of numbers , Yes $Z_n=\frac{1}{n}\cdot \frac{Y_n}{n}\to 0\cdot E{X}_1=0$, a.s., from stolz Theorem , Yes $$\underset{n\to \infty }{\lim }\frac{{\sum }_{k=1}^n{Z}_k}{n}=\underset{n\to \infty }{\lim }{Z}_n=0,\text{a.s.}$$

7、 ... and 、(10 branch ) $X_1,\cdots ,X_n$ yes i.i.d. obey $N(\mu ,{\sigma }^2)$ Random variable of , $F$ Is its distribution function , seek $-2{\sum }_{i=1}^n\ln F(X_i)$ The distribution of .

Solution:
First of all $Y_i=F(X_i)\sim U(0,1)$, Just calculate $Z_1=-2Y_1$ The distribution of , By the distribution function method , Yes $z>0$, Yes
$$P\left(Z_1\le z\right)=P\left(-2\ln Y_1\le z\right)=P\left(Y_1\ge e^{-2z}\right)=1-e^{-2z},$$ This is an average of $1/2$ The exponential distribution of , It's also ${\chi }^2(2)$ Distribution , By additivity , have to
$$-2\sum _{i=1}^n\ln F(X_i)\sim {\chi }^2(2n).$$

8、 ... and 、(10 branch ) $X_1,\cdots ,X_6$ yes i.i.d. Of $U(0,1)$ A random variable , seek $Var(2X_{(2)}+3X_{(3)})$.

Solution:
Direct calculation , Yes
$$Var\left(2X_{\left(2\right)}+3X_{\left(3\right)}\right)=4Var\left(X_{\left(2\right)}\right)+9Var\left(X_{\left(3\right)}\right)+12Cov\left(X_{\left(2\right)},X_{\left(3\right)}\right),$$ And marginal distribution $X_{(2)}\sim Beta(2,5)$, $X_{(3)}\sim Beta(3,4)$, Therefore, the two variance terms can be calculated directly , namely
$$4Var\left(X_{\left(2\right)}\right)=\frac{4\cdot 10}{7^2\cdot 8}=\frac{10}{98},9Var\left(X_{\left(3\right)}\right)=\frac{9\cdot 12}{7^2\cdot 8}=\frac{27}{98},$$ Covariance term can be written in formula $\frac{i(n+1-j)}{(n+1{)}^2(n+2)}$(2019 Fudan Yingtong seventh question ), You can also write the joint density first
$$g_{2,3}\left(x,y\right)=\frac{6!}{1!1!1!3!}x\cdot {\left(1-y\right)}^3=\frac{6!}{1!3!}x{\left(1-y\right)}^3,0<x<y<1,$$ Calculate the mixing moment , namely
$$\begin{array}{rl}E\left(X_{\left(2\right)}{X}_{\left(3\right)}\right)& =\frac{6!}{3!}{\int }_0^1{\int }_0^y{x}^{2}y{\left(1-y\right)}^{3}dxdy\\ & =\frac{6!}{3!3}{\int }_0^1{y}^4{\left(1-y\right)}^{3}dy\\ & =\frac{6!3!4!}{3!8!3}=\frac{8}{56}.\end{array}$$ So the covariance is $Cov\left(X_{\left(2\right)},X_{\left(3\right)}\right)=\frac{8}{56}-\frac{6}{49}=\frac{2}{98}.$ Sum up all the calculation results to
$$Var\left(2X_{\left(2\right)}+3X_{\left(3\right)}\right)=\frac{61}{98}.$$

Nine 、(10 branch ) $X_1,\cdots ,X_n$ yes i.i.d. Of $U(0,\theta )$ A random sample , set up $a{X}_{(1)},b{X}_{(3)}$ yes $\theta$ Unbiased estimation of , seek $a,b$ And compare which of them is more effective .

Solution:
because $\frac{X_{(1)}}{\theta }\sim Beta(1,3)$, so $E{X}_{(1)}=\frac{1}{4}\theta$, $Var(X_{(1)})=\frac{3}{80}{\theta }^2$, so $a=4$, And $Var(a{X}_{(1)})=\frac{3}{5}{\theta }^2$. Empathy $X_{(3)}\sim Beta(3,1)$, so $E{X}_{(3)}=\frac{3}{4}\theta$, $Var(X_{(3)})=\frac{3}{80}{\theta }^2$, so $b=\frac{4}{3}$, And $Var(b{X}_{(3)})=\frac{1}{60}{\theta }^2$. It can be seen that $b{X}_{(3)}$ More effective .

Ten 、(10 branch ) set up $X_1,\cdots ,X_n$ yes i.i.d. Of $N(\mu ,16)$ A random sample , $\mu$ The prior distribution of is $N(a,b^2)$, Find the posterior distribution .

Solution:
Consider sufficient statistics $\bar{X}|\mu \sim N(\mu ,\frac{16}{n})$, The joint density is
$$\begin{array}{rl}p\left(x,\mu \right)& =p\left(x|\mu \right)\pi \left(\pi \right)=C\cdot e^{-\frac{{\left(x-\mu \right)}^2}{2\cdot \frac{16}{n}}}\cdot e^{-\frac{{\left(\mu -a\right)}^2}{2b^2}}\\ & =C\cdot e^{-\frac{{\left(x-\mu \right)}^2}{2\cdot \frac{16}{n}}}\cdot e^{-\frac{{\left(\mu -a\right)}^2}{2b^2}}\\ & =C{e}^{-\frac{b^2{\left(x-\mu \right)}^2+\frac{16}{n}{\left(\mu -a\right)}^2}{2\cdot \frac{16}{n}{b}^2}}\\ & =C{e}^{-\frac{b^2{\left(x-\mu \right)}^2+\frac{16}{n}{\left(\mu -a\right)}^2}{2\cdot \frac{16}{n}{b}^2}}\\ & =C{e}^{-\frac{\left(\frac{16}{n}+b^2\right){\mu }^2-2\left(b^{2}x+\frac{16}{n}a\right)\mu +\left(b^2{x}^2+\frac{16}{n}{a}^2\right)}{2\cdot \frac{16}{n}{b}^2}}\\ & =C_1(x)e^{-\frac{{\left(\mu -\frac{b^{2}x+\frac{16}{n}a}{b^2+\frac{16}{n}}\right)}^2}{2\cdot \frac{\frac{16}{n}{b}^2}{\frac{16}{n}+b^2}}},\end{array}$$ It is found that there is a normally distributed kernel , So the posterior distribution is
$$\mu |\bar{X}\sim N\left(\frac{b^2\bar{X}+\frac{16}{n}a}{b^2+\frac{16}{n}},\frac{\frac{16}{n}{b}^2}{\frac{16}{n}+b^2}\right)=N\left(\frac{\frac{n}{16}\bar{X}+\frac{1}{b^2}a}{\frac{n}{16}+\frac{1}{b^2}},\frac{1}{\frac{n}{16}+\frac{1}{b^2}}\right).$$ It can be seen that the posterior mean is the weighted average of sample information and prior information .

11、 ... and 、(10 branch ) set up $X_1,\cdots ,X_n$ yes i.i.d. Of $U(0,\theta )$ A random sample , Consider the hypothesis test problem
$$H_0:\theta \le 1\mathrm{v}\mathrm{s}{H}_1:\theta >1$$ Construct reject fields $W=\{X_{(n)}\ge c\}$. Answer the following questions :
(1)(5 branch ) $\alpha =0.05$, seek $c$;
(2)(5 branch ) When $\theta =1.5$, In order to make the probability of making the second kind of mistake $\beta \le 0.1$, Find the minimum sample size .

Solution:
(1) In order to make the significance level $0.05$, Yes
$$0.05=\underset{\theta \le 1}{\sup }{P}_{\theta }\left(X_{\left(n\right)}\ge c\right)=P_{\theta =1}\left(X_{\left(n\right)}\ge c\right)={\left(1-c\right)}^n,$$
Solution $c=0.95^{\frac{1}{n}}$.
(2) The probability of making the second kind of mistake is
$$\beta \left(1.5\right)=P_{\theta =1.5}\left(X_{\left(n\right)}<0.95^{\frac{1}{n}}\right)={\left(\frac{0.95^{\frac{1}{n}}}{1.5}\right)}^n=\frac{0.95}{1.5^n},$$ Make it less than or equal to $0.1$, have to
$$\frac{0.95}{1.5^n}\le 0.1*n\ge \frac{\ln 9.5}{\ln 1.5}*n\ge 6.$$