[the Nine Yang Manual] 2018 Fudan University Applied Statistics real problem + analysis

The real part

One 、(20 branch ) from 1-10 Don't put it back 3 A digital , Find the following probability
(1)(5 branch ) The minimum number is 5;
(2)(5 branch ) The biggest number is 5;
(3)(5 branch ) At least one is less than 6;
(4)(5 branch ) One less than 5, One is equal to 5, One is greater than 5.

Two 、(15 branch ) You are trying again. The probability of success is $p$ Events , Don't stop until there are two consecutive successes or two failures , Please take the probability of two successful stops .

3、 ... and 、(15 branch ) Find binomial distribution , $(a,b)$ Evenly distributed , Expectation and variance of gamma distribution .

Four 、(20 branch ) prove $E\left(X^2\right)<\infty$ The necessary and sufficient condition of is the series $\sum nP(|X|>n)$ convergence .

5、 ... and 、(20 branch ) $X_1,X_2,X_3$ Is taken from the expectation that $\alpha$ Random samples of exponential distribution , Find the probability $P\left(X_1<X_2<X_3\right)$ as well as $X_{(1)}$ Probability density of .

6、 ... and 、(20 branch ) $P\left(X_i=-0.3\right)=P\left(X_i=0.4\right)=\frac{1}{2},i=1,2,\dots ,n,$ Are independent of each other , Construct a sequence of random variables $Y_n={\prod }_{i=1}^n\left(X_i+1\right),$ seek $Y_n$ And prove $Y_n$ Expectations tend to be infinite .

7、 ... and 、(20 branch ) There is a pile of balls : 2 red , 3 black , 4 white . Touch a ball randomly , If it's black, remember that you win , If it's other colors , Some people put it back and continue to touch the ball , Until the color or black appears repeatedly , If there is the color you touch for the first time , Then you win , Otherwise you lose . Please ask the probability of winning .

8、 ... and 、(20 branch )
(1)(10 branch ) Explain consistency estimates ;
(2)(10 branch ) $X_1,\dots ,X_n$ It is a sample from the same population , Write a consistent estimate of the median , And explain why .

The analysis part

One 、(20 branch ) from 1-10 Don't put it back 3 A digital , Find the following probability
(1)(5 branch ) The minimum number is 5;
(2)(5 branch ) The biggest number is 5;
(3)(5 branch ) At least one is less than 6;
(4)(5 branch ) One less than 5, One is equal to 5, One is greater than 5.

Solution:

(1) $\#\Omega =C_{10}^3=120$, #A $A_1=1\cdot C_5^2=10,P\left(A_1\right)=\frac{\#A_1}{\#\Omega }=\frac{1}{12}$.

(2) $\#A_2=1\cdot C_4^2=6,P\left(A_2\right)=\frac{\#A_2}{\#\Omega }=\frac{1}{20}$.

(3) $\#\overline{A_3}=C_5^3=10,P\left(A_3\right)=1-\frac{\#\overline{A_3}}{\#\Omega }=\frac{11}{12}$.

(4) $\#A_4=4\cdot 1\cdot 5=20,P\left(A_4\right)=\frac{\#A_4}{\#\Omega }=\frac{1}{6}$.

Two 、(15 branch ) You are trying again. The probability of success is $p$ Events , Don't stop until there are two consecutive successes or two failures , Please With the probability of two successful stops .

Solution:

set up $A$ “ Stop successfully twice ”", $p_0=P(A),p_1=P(A\mid$ For the first time $),p_{-1}=P$ ($A\mid$ The first failure $)$, According to the full probability formula : $$\{\begin{array}{l}{p}_0=p_1\cdot p+p_{-1}\cdot (1-p)\\ p_1=p+p_{-1}\cdot (1-p)\\ p_{-1}=p_1\cdot p\end{array}$$ Solution $p_0=\frac{p^2(2-p)}{1-p(1-p)}$.

3、 ... and 、(15 branch ) Find binomial distribution , $(a,b)$ Evenly distributed , Expectation and variance of gamma distribution .

Solution:

(1) The binomial distribution : $X\sim B(n,p),P(X=k)=C_n^k{p}^k(1-p{)}^{n-k},k=0,1,\dots ,n$,
$$EX=\sum _{k=0}^{n}k\frac{n!}{k!(n-k)!}{p}^k(1-p{)}^{n-k}=np\sum _{k=1}^n\frac{(n-1)!}{(k-1)!(n-k)!}{p}^{k-1}(1-p{)}^{n-k}=np,$$
$$EX(X-1)=\sum _{k=0}^{n}k(k-1)C_n^k{p}^k(1-p{)}^{n-k}=n(n-1)p^2\sum _{k=2}^n{C}_{n-2}^{k-2}{p}^{k-2}(1-p{)}^{n-k}=n(n-1)p^2,$$
therefore $E{X}^2=n(n-1)p^2+np,DX=E{X}^2-(EX{)}^2=n(n-1)p^2+np-n^2{p}^2=np(1-p)$.

(2) Uniform distribution : $X\sim U(a,b),f(x)=\frac{1}{b-a},a<x<b$,
$$EX={\int }_a^b\frac{x}{b-a}dx=\frac{a+b}{2},DX={\int }_a^b{\left(x-\frac{a+b}{2}\right)}^2\frac{1}{b-a}dx=\frac{(b-a{)}^2}{12}.$$
(3) Gamma distribution : $X\sim Ga(\alpha ,\lambda ),f(x)=\frac{{\lambda }^{\alpha }}{\Gamma (\alpha )}{x}^{\alpha -1}{e}^{-\lambda x},x>0$,
$$EX={\int }_0^{+\infty }\frac{{\lambda }^{\alpha }}{\Gamma (\alpha )}{x}^{\alpha }{e}^{-\lambda x}dx=\frac{1}{\lambda \Gamma (\alpha )}{\int }_0^{+\infty }(\lambda x{)}^{\alpha }{e}^{-\lambda x}d(\lambda x)=\frac{\Gamma (\alpha +1)}{\lambda \Gamma (\alpha )}=\frac{\alpha }{\lambda },$$
$E{X}^2={\int }_0^{+\infty }\frac{{\lambda }^{\alpha }}{\Gamma (\alpha )}{x}^{\alpha +1}{e}^{-\lambda x}dx=\frac{1}{{\lambda }^2\Gamma (\alpha )}{\int }_0^{+\infty }(\lambda x{)}^{\alpha +1}{e}^{-\lambda x}d(\lambda x)=\frac{\Gamma (\alpha +2)}{{\lambda }^2\Gamma (\alpha )}=\frac{\alpha (\alpha +1)}{\lambda }$,
so $DX=E{X}^2-(EX{)}^2=\frac{\alpha (\alpha +1)}{{\lambda }^2}-\frac{{\alpha }^2}{{\lambda }^2}=\frac{\alpha }{{\lambda }^2}.$

Four 、(20 branch ) prove $E\left(X^2\right)<\infty$ The necessary and sufficient condition of is the series $\sum nP(|X|>n)$ convergence .

Solution:
(1) First explain $E|X|<+\infty$ The necessary and sufficient condition of is the series ${\sum }_{n=1}^{\infty }P(|X|>n)$ convergence ： because $$\begin{array}{rl}\sum _{n=1}^{\infty }P(|X|>n)& =\sum _{n=1}^{\infty }\sum _{k=n}^{\infty }P(k<|X|\le k+1)\\ & =\sum _{k=1}^{\infty }\sum _{n=1}^{k}P(k<|X|\le k+1)\\ & =\sum _{k=1}^{\infty }kP(k<|X|\le k+1)\end{array}$$ At the same time, by the comparison and discrimination of positive series , Superior number and ${\sum }_{k=1}^{\infty }(k+1)P(k<|X|\le k+1)$ It is also convergent and scattered , in consideration of $E|X|={\int }_0^{+\infty }xd{F}_{|X|}(x)$, On the one hand $$\begin{array}{rl}{\int }_0^{+\infty }xd{F}_{|X|}(x)& =\sum _{k=0}^{\infty }{\int }_k^{k+1}xd{F}_{|X|}(x)\\ & \le \sum _{k=0}^{\infty }{\int }_k^{k+1}(k+1)d{F}_{|X|}(x)\\ & =\sum _{k=0}^{\infty }(k+1)P(k<|X|\le k+1),\end{array}$$ On the other hand, there are
$$\begin{array}{rl}{\int }_0^{+\infty }xd{F}_{|X|}(x)& =\sum _{k=0}^{\infty }{\int }_k^{k+1}xd{F}_{|X|}(x)\\ & \ge \sum _{k=0}^{\infty }{\int }_k^{k+1}kd{F}_{|X|}(x)\\ & =\sum _{k=0}^{\infty }kP(k<|X|\le k+1),\end{array}$$ in summary , $E|X|<+\infty$ The necessary and sufficient condition of is the series ${\sum }_{n=1}^{\infty }P(|X|>n)$ convergence .

(2) Further explanation $E{X}^2<+\infty$ The necessary and sufficient condition of is the series ${\sum }_{n=1}^{+\infty }nP(|X|>n)$ convergence : because $$\begin{array}{rl}\sum _{n=1}^{\infty }nP(|X|>n)& =\sum _{n=1}^{\infty }\sum _{k=n}^{\infty }nP(k<|X|\le k+1)\\ & =\sum _{k=1}^{\infty }\sum _{n=1}^{\infty }nP(k<|X|\le k+1)\\ & =\sum _{k=1}^{\infty }\frac{k(k+1)}{2}P(k<|X|\le k+1)\end{array}$$ At the same time, by the comparison and discrimination of positive series , The convergence and divergence of the above formula is obviously equivalent to ${\sum }_{n=1}^{\infty }{n}^{2}P(n<|X|\le n+1)$ Convergence and divergence of , It's also equivalent to ${\sum }_{n=1}^{\infty }(n+1{)}^{2}P(n<|X|\le n+1)$ Convergence and divergence of , Also with the help of second-order moment determination Semantic formula $E{X}^2={\int }_0^{+\infty }{x}^{2}d{F}_{|X|}(x)$, On the one hand $$\begin{array}{rl}{\int }_0^{+\infty }{x}^{2}d{F}_{|X|}(x)& =\sum _{n=0}^{\infty }{\int }_n^{n+1}{x}^{2}d{F}_{|X|}(x)\\ & \le \sum _{n=0}^{\infty }{\int }_n^{n+1}(n+1{)}^{2}d{F}_{|X|}(x)\\ & =\sum _{n=0}^{\infty }(n+1{)}^{2}P(n<|X|\le n+1)\end{array}$$
On the other hand, there are $$\begin{array}{rl}{\int }_0^{+\infty }{x}^{2}d{F}_{|X|}(x)& =\sum _{n=0}^{\infty }{\int }_n^{n+1}{x}^{2}d{F}_{|X|}(x)\\ & \le \sum _{n=0}^{\infty }{\int }_n^{n+1}{n}^{2}d{F}_{|X|}(x)\\ & =\sum _{n=0}^{\infty }{n}^{2}P(n<|X|\le n+1),\end{array}$$ in summary , $E{X}^2<+\infty$ The necessary and sufficient condition of is the series ${\sum }_{n=1}^{+\infty }nP(|X|>n)$ convergence .

5、 ... and 、(20 branch ) $X_1,X_2,X_3$ Is taken from the expectation that $\alpha$ Random samples of exponential distribution , Find the probability $P\left(X_1<X_2<X_3\right)$ as well as $X_{(1)}$ Probability density of .

Solution:
According to the rotation symmetry , $P\left(X_1<X_2<X_3\right)=\frac{1}{6}$. Make $Y=X_{(1)}$, Then when $y>0$ when , $1-F(y)=P\{Y>y\}=P^3\left\{X_1>y\right\}=e^{-\frac{3}{\alpha }y}$, so $f(y)=\frac{3}{\alpha }{e}^{-\frac{3}{\alpha }y},y>0$. This happens to be $\mathrm{Exp}\left(\frac{3}{\alpha }\right)$.

6、 ... and 、(20 branch ) $P\left(X_i=-0.3\right)=P\left(X_i=0.4\right)=\frac{1}{2},i=1,2,\dots ,n,$ Are independent of each other , Construct a sequence of random variables $Y_n={\prod }_{i=1}^n\left(X_i+1\right),$ seek $Y_n$ And prove $Y_n$ Expectations tend to be infinite .

Solution:
By strong law of numbers , $\frac{1}{n}\ln Y_n=\frac{1}{n}{\sum }_{i=1}^n\ln \left(X_i+1\right)\stackrel{\text{ a.s. }}{*}E\ln \left(X_1+1\right)=\frac{1}{2}\ln 0.98<0$, so $\ln Y_n\stackrel{\text{ a.s. }}{*}-\infty ,Y_n\stackrel{\text{ a.s. }}{*}0$, therefore $Y_n$ The limit of is the single point distribution , With probability 1 take 0 . and $$E{Y}_n=\prod _{i=1}^{n}E\left(X_i+1\right)=\prod _{i=1}^n\left(\frac{0.7+1.4}{2}\right)=1.05^n\to +\infty .$$

7、 ... and 、(20 branch ) There is a pile of balls : 2 red , 3 black , 4 white . Touch a ball randomly , If it's black, remember that you win , If it's other colors , Some people put it back and continue to touch the ball , Until the color or black appears repeatedly , If there is the color you touch for the first time , Then you win , Otherwise you lose . Please ask the probability of winning .

Solution:
set up $A_k$ For the first time $k$ The probability of winning the first time , The result to be found is $P(A)=P({\bigcup }_{k=1}^{\infty }{A}_k)={\sum }_{k=1}^{\infty }P(A_k)$.
(i) The probability of winning for the first time $P(A_1)=\frac{1}{3}$.
(ii) The first $k$ Second win ($k>1$) indicate : I didn't touch the black ball for the first time , Follow up $2,3,...,k-1$ I didn't touch black for the first time, and I didn't touch the color you touched for the first time , Last $k$ I touched the original color for the first time , This probability must be related to the color you first touch , Consider the full probability formula $P(A_k)=P(R)P(A_k|R)+P(W)P(A_k|W)$, among $R$ It means touching red , $W$ It means feeling white , Yes
$$P\left(A_k|R\right)={\left(\frac{4}{9}\right)}^{k-2}\frac{2}{9},P\left(A_k\mid W\right)={\left(\frac{2}{9}\right)}^{k-2}\frac{4}{9}.$$ therefore $$P\left(A_k\right)=P\left(R\right)\frac{2}{9}{\left(\frac{4}{9}\right)}^{k-2}+P\left(W\right)\frac{4}{9}{\left(\frac{2}{9}\right)}^{k-2}=\frac{4}{81}{\left(\frac{4}{9}\right)}^{k-2}+\frac{16}{81}{\left(\frac{2}{9}\right)}^{k-2}.$$ In sum , Yes
$$P\left(A\right)=P\left(A_1\right)+\frac{4}{81\left(1-\frac{4}{9}\right)}+\frac{16}{81\left(1-\frac{2}{9}\right)}=\frac{1}{3}+\frac{4}{45}+\frac{16}{63}=\frac{71}{105}.$$

8、 ... and 、(20 branch )
(1)(10 branch ) Explain consistency estimates ;
(2)(10 branch ) $X_1,\dots ,X_n$ It is a sample from the same population , Write a consistent estimate of the median , And explain why .

Solution:
(1) $\hat{g}$ yes $g$ The consistent estimate of means $\hat{g}\stackrel{p}{\to }g$, This illustrates the $\hat{g}$ Is a good estimate , At least in the sample size $n$ large , It deviates $g$ The probability is very small . Further more , if $\hat{g}$ yes $g$ The strong consistency estimate of means $\hat{g}\stackrel{\text{ a.s. }}{*}g$.

(2) Sample median $X_{\left[\frac{n}{2}\right]}$ Is the overall median $x_{0.5}$ The consistent estimator of , Let the overall density function be $f(x)$, Asymptotically normal distribution with sample median $$X_{\left[\frac{n}{2}\right]}\sim N\left(x_{0.5},\frac{1}{4n{f}^2\left(x_{0.5}\right)}\right),$$ From its asymptotic normal distribution, we can get $$P\left(\left|X_{\left[\frac{n}{2}\right]}-x_{0.5}\right|<\epsilon \right)=P\left(2\sqrt{n}f\left(x_{0.5}\right)\left|X_{\left[\frac{n}{2}\right]}-x_{0.5}\right|<2\sqrt{n}f\left(x_{0.5}\right)\epsilon \right)\sim \Phi \left(2\sqrt{n}f\left(x_{0.5}\right)\epsilon \right)-\Phi \left(-2\sqrt{n}f\left(x_{0.5}\right)\epsilon \right)\to 1.$$ The above formula shows the median of the sample $X_{\left[\frac{n}{2}\right]}$ Is the overall median $x_{0.5}$ The consistent estimator of .