[the Nine Yang Manual] 2020 Fudan University Applied Statistics real problem + analysis

The real part

One 、(20 branch ) A family has two children , Find the probability of the following events :
(1)(10 branch ) The first one is known to be a girl , Find the probability that the second is a girl ;
(2)(10 branch ) One is known to be a girl , Find the probability that the other is a girl .

Two 、(15 branch ) Jiayou 21 A coin , B has 20 A coin , Both of them toss all the coins at the same time , Find the probability that the number of coins with a facing up is more than that of B .

3、 ... and 、(15 branch ) There are countless parallel lines on the plane , Every two parallel lines are spaced 2 rice , Use side length 1 An equilateral triangle of meters is thrown at the plane , Find the probability of triangle pressing to a straight line .

Four 、(15 branch ) 8 A boy 、7 Two girls sit in a row , set up $X_i=1$ It means the first one $i$ The first position is the same as $i+1$ The opposite sex sits in this position , $X_i=0$ It means the first one $i$ The first position is the same as $i+1$ Sitting in the same seat , $\xi ={\sum }_{i=1}^{14}{X}_i,$ seek $E\xi .$

5、 ... and 、(15 branch ) Cite an expectation that tends to be positive and infinite , But it converges to 0 A sequence of random variables $\left\{X_n\right\}$.

6、 ... and 、(20 branch ) Some come from the general $X\sim f(x)=\theta x^{\theta -1}I\{0<x<1\}$ Of $n$ Random sample , seek
(1)(5 branch ) $\theta$ Of $\mathrm{M}\mathrm{L}\mathrm{E},$ And verify the unbiased ;
(2)(5 branch ) verification MLE The consistency of ;
(3)(5 branch ) $\theta$ The moment estimate of ;
(4)(5 branch ) Use the sample median pair $\theta$ Estimate .

7、 ... and 、(20 branch ) $X_1,\dots ,X_n,$ i.i.d $\sim N\left(\mu ,{\sigma }^2\right),$ prove $\left[X_{(1)},X_{(n)}\right]$ yes $\mu$ The confidence level of is $1-2^{1-n}$ The confidence interval of .

8、 ... and 、(20 branch ) Some come from the general $X\sim f(x)=\frac{1}{2}{e}^{-|x-\theta |}$ Of 7 Random sample , seek $\theta$ Of MLE.

Nine 、(10 branch ) $(X_1,X_2)\sim N(0,0;1,1;0),$ seek $\frac{X_1}{X_2}$ Probability distribution of .

The analysis part

One 、(20 branch ) A family has two children , Find the probability of the following events :
(1)(10 branch ) The first one is known to be a girl , Find the probability that the second is a girl ;
(2)(10 branch ) One is known to be a girl , Find the probability that the other is a girl .

Solution:
(1) First of all, suppose ： The probability of a child being a girl without any information is $0.5$.
In the event $A_i$ It means the first one $i$ This is a girl $(i=1,2)$, be $P\left(A_2\mid A_1\right)=\frac{P\left(A_1{A}_2\right)}{P\left(A_1\right)}=\frac{0.25}{0.5}=0.5$.
(2) $P\left(A_1{A}_2\mid A_1\cup A_2\right)=\frac{P\left(A_1{A}_2\right)}{P\left(A_1\cup A_2\right)}=\frac{0.25}{0.75}=\frac{1}{3}$.

Two 、(15 branch ) Jiayou 21 A coin , B has 20 A coin , Both of them toss all the coins at the same time , Find the probability that the number of coins with a facing up is more than that of B .

Solution:
According to the symmetry, we can know , $P\{$ There are more coins with a facing up than with B $\}=P\{$ There are more coins with a side down than with B $\}$. use A random variable $X$ Indicates the number of coins with a facing up , A random variable $Y$ Indicates the number of coins with B facing up . be :
$$\begin{array}{rl}P\{X>Y\}& =P\{21-X>20-Y\}=P\{1-X>-Y\}\\ & =P\{X<Y+1\}=P\{X⩽Y\}\end{array}$$ also $P\{X>Y\}+P\{X⩽Y\}=1$, therefore $P\{X>Y\}=P\{X⩽Y\}=0.5$.

3、 ... and 、(15 branch ) There are countless parallel lines on the plane , Every two parallel lines are spaced 2 rice , Use side length 1 An equilateral triangle of meters is thrown at the plane , Find the probability of triangle pressing to a straight line .

Solution:
remember $△ABC$ The three sides of the are $a,b,c$. Then there are the following situations when a triangle intersects a parallel line :
(1) One vertex of the triangle is on the parallel line ;
(2) One side of the triangle coincides with the straight line ;
(3) Two lines of triangle Edges intersect parallel lines .
According to the geometric probability $P(1)=P(2)=0$, So just consider the situation (3). and $P(3)=P_{ab}+P_{ac}+P_{bc}$, among $P_{ab}$ edge $a、b$ Intersect with parallel lines . So , remember $P_a$ edge $a$ Intersect with parallel lines , be $P_a=P_{ac}+P_{ab}$. so $$P(3)=\frac{1}{2}\left(P_a+P_b+P_c\right),$$ Now we only need to find $P_a、P_b、P_c$. This is a Buffon Injection model , The probability is $P_a=\frac{2a}{d\pi }$, among $a$ Is the edge $a$ The length of , $d$ It's parallel Spacing between lines , Substituting data can be calculated $P_a=\frac{2}{2\pi }=\frac{1}{\pi }$. Empathy $P_b=P_c=\frac{1}{\pi }$. so
$$P\{\text{ Triangle pressed to a straight line }\}=P(3)=\frac{1}{2}\left(P_a+P_b+P_c\right)=\frac{3}{2\pi }.$$

Four 、(15 branch ) 8 A boy 、7 Two girls sit in a row , set up $X_i=1$ It means the first one $i$ The first position is the same as $i+1$ The opposite sex sits in this position , $X_i=0$ It means the first one $i$ The first position is the same as $i+1$ Sitting in the same seat , $\xi ={\sum }_{i=1}^{14}{X}_i,$ seek $E\xi .$

Solution:
$E\xi =E\left({\sum }_{i=1}^{14}{X}_i\right)={\sum }_{i=1}^{14}E{X}_i$, Considering all $X_i$ It's identically distributed , Now $E{X}_1$.
$$E{X}_1=P\left(X_1=1\right)=\frac{C_8^1{C}_7^1}{C_{15}^2}=\frac{8\times 7}{\frac{15\times 14}{2\times 1}}=\frac{8}{15}$$ therefore $E\xi ={\sum }_{i=1}^{14}E{X}_i=14E{X}_1=\frac{112}{15}$.

5、 ... and 、(15 branch ) Cite an expectation that tends to be positive and infinite , But it converges to 0 A sequence of random variables $\left\{X_n\right\}$.

Solution:
Give such a sequence of random variables : $P\left(X_n=0\right)=1-\frac{1}{n},P\left(X_n=n^2\right)=\frac{1}{n}$. $E{X}_n=n\to +\infty$. and $P\left(X_n\ne 0\right)=\frac{1}{n}$, be $X_n\stackrel{P}{\to }0$.

6、 ... and 、(20 branch ) Some come from the general $X\sim f(x)=\theta x^{\theta -1}I\{0<x<1\}$ Of $n$ Random sample , seek
(1)(5 branch ) $\theta$ Of $\mathrm{M}\mathrm{L}\mathrm{E},$ And verify the unbiased ;
(2)(5 branch ) verification MLE The consistency of ;
(3)(5 branch ) $\theta$ The moment estimate of ;
(4)(5 branch ) Use the sample median pair $\theta$ Estimate .

Solution:
(1) Likelihood function $L(\mathbf{X};\theta )={\theta }^n{\left({\prod }_{i=1}^n{x}_i\right)}^{\theta -1}$, Log likelihood function $\ln L=n\ln \theta +(\theta -1){\sum }_{i=1}^n\ln x_i$. Make $\frac{\partial \ln L}{\partial \theta }=\frac{n}{\theta }+{\sum }_{i=1}^n\ln x_i=0$, Solution ${\hat{\theta }}_L=\frac{n}{{\sum }_{i=1}^n\left(-\ln x_i\right)}$. And because the whole obeys beta distribution , Belong to Exponential family distribution , The stationary point of its log likelihood function must be maximum likelihood estimation . therefore ${\hat{\theta }}_L=\frac{n}{{\sum }_{i=1}^n\left(-\ln x_i\right)}$ yes $\theta$ Maximum likelihood estimation of . If order $Y_i=-\ln X_i\sim \mathrm{Exp}(\theta )$, And from the additivity of gamma distribution $T={\sum }_{i=1}^n{y}_i\sim Ga(n,\theta )$, Zeji Large likelihood estimation can be written as ${\hat{\theta }}_L=\frac{n}{T}$.
$$E{\hat{\theta }}_L=E\frac{n}{T}={\int }_0^{+\infty }\frac{n}{t}\frac{{\theta }^n}{\Gamma (n)}{t}^{n-1}{e}^{-\theta t}dt=\frac{n\theta }{\Gamma (n)}{\int }_0^{+\infty }(\theta t{)}^{n-2}{e}^{-\theta t}d(\theta t)=\frac{n\theta }{\Gamma (n)}\Gamma (n-1)=\frac{n}{n-1}\theta$$
therefore ${\hat{\theta }}_L$ No $\theta$ Unbiased estimation of , But it is gradual and unbiased .
(2) In the last question, we have calculated $E{\hat{\theta }}_L=\frac{n}{n-1}\theta \to \theta$, Now consider its consistency .
$$E{\hat{\theta }}_L^2=E\frac{n^2}{T^2}=\frac{n^2{\theta }^2}{\Gamma (n)}{\int }_0^{+\infty }(\theta t{)}^{n-3}{e}^{-\theta t}d(\theta t)=\frac{n^2{\theta }^2}{\Gamma (n)}\Gamma (n-2)=\frac{n^2}{(n-1)(n-2)}{\theta }^2,$$
be $\mathrm{Var}\left({\hat{\theta }}_L\right)=E{\hat{\theta }}_L^2-{\left(E{\hat{\theta }}_L\right)}^2=\frac{n^2{\theta }^2}{(n-1)(n-2)}-\frac{n^2{\theta }^2}{(n-1{)}^2}=\frac{n^2}{(n-1{)}^2(n-2)}{\theta }^2$.
$$P\left(\left|{\hat{\theta }}_L-\theta \right|\ge \epsilon \right)=P\left(\left|{\hat{\theta }}_L-\frac{n}{n-1}\theta +\frac{n}{n-1}\theta -\theta \right|\ge \epsilon \right)$$$$\le P\left(\left|{\hat{\theta }}_L-\frac{n}{n-1}\theta \right|\ge \frac{\epsilon }{2}\right)+P\left(\left|\frac{n}{n-1}\theta -\theta \right|\ge \frac{\epsilon }{2}\right)$$ According to Chebyshev inequality $P\left(\left|{\hat{\theta }}_L-\frac{n}{n-1}\theta \right|\ge \frac{\epsilon }{2}\right)\le \frac{4\mathrm{Var}\left({\hat{\theta }}_L\right)}{{\epsilon }^2}=\frac{4}{{\epsilon }^2}\frac{n^2}{(n-1{)}^2(n-2)}{\theta }^2\to 0$ And for the larger $n,P\left(\left|\frac{n}{n-1}\theta -\theta \right|\ge \frac{\epsilon }{2}\right)=0$. therefore $P\left(\left|{\hat{\theta }}_L-\theta \right|\ge \epsilon \right)\to 0$, in other words ${\hat{\theta }}_L$ yes $\theta$ A consistent estimate of .
(3) Overall obedience $\mathrm{Beta}(\theta ,1)$, Digital features distributed by beta , We know $EX=\frac{\theta }{\theta +1}$. According to the inverse solution $\theta$ The moment estimate of ${\hat{\theta }}_M=\frac{\bar{x}}{1-\bar{x}}$.
(4) The overall distribution function is $F(x)=\{\begin{array}{cc}0,& x<0\\ x^{\theta },& 0\le x<1\\ 1,& x\ge 1\end{array}$ Make $F(x)=\frac{1}{2}$, Solution $x_{0.5}={\left(\frac{1}{2}\right)}^{\frac{1}{\theta }}$, Use the median of the sample $m_{0.5}$ Replace the overall median $x_{0.5}$, And inverse solution Out $\hat{\theta }=\frac{1}{{\log }_{\frac{1}{2}}{m}_{0.5}}=\frac{{\log }_{\frac{1}{2}}\frac{1}{2}}{{\log }_{\frac{1}{2}}{m}_{0.5}}={\log }_{m_{0.5}}\frac{1}{2}$ It is based on the median pair of samples $\theta$ Estimation .

7、 ... and 、(20 branch ) $X_1,\dots ,X_n,$ i.i.d $\sim N\left(\mu ,{\sigma }^2\right),$ prove $\left[X_{(1)},X_{(n)}\right]$ yes $\mu$ The confidence level of is $1-2^{1-n}$ The confidence interval of .

Solution:
First consider asking $U=x_{(1)}$ The distribution of , According to the calculation formula of the minimum value distribution
$$F_U(u)=1-{\left[1-\Phi \left(\frac{u-\mu }{\sigma }\right)\right]}^n$$ be $P\left(\mu <x_{(1)}\right)=1-F_U(\mu )=[1-\Phi (0){]}^n={\left(\frac{1}{2}\right)}^n=2^{-n}$.
According to the symmetry $P\left(\mu >x_{(n)}\right)=2^{-n}$. therefore $P\left(x_{(1)}⩽\mu ⩽x_{(n)}\right)=1-2\cdot 2^{-n}=1-2^{1-n}$

8、 ... and 、(20 branch ) Some come from the general $X\sim f(x)=\frac{1}{2}{e}^{-|x-\theta |}$ Of 7 Random sample , seek $\theta$ Of MLE.

Solution:
Likelihood function $L(\mathbf{X};\theta )={\left(\frac{1}{2}\right)}^7{e}^{-{\sum }_{i=1}^7|x_i-\theta |}={\left(\frac{1}{2}\right)}^7{e}^{-{\sum }_{i=1}^7|x_{(i)}-\theta |}$. here $x_{(i)}$ It means No $i$ Order statistics . In order to make the likelihood function as large as possible , Should make $e^{-{\sum }_{i=1}^7|x_{(0)}-\theta |}$ As big as possible , That is to make ${\sum }_{i=1}^7\left|x_{(i)}-\theta \right|$ As small as possible . The following research ${\sum }_{i=1}^7\left|x_{(i)}-\theta \right|$ The state of being :
$$\sum _{i=1}^7\left|x_{(i)}-\theta \right|=\sum _{i=1}^3\left(\left|x_{(i)}-\theta \right|+\left|x_{(7-i+1)}-\theta \right|\right)+\left|x_{(4)}-\theta \right|$$ ( The above formula will $x_{(1)},x_{(7)}$ Divided into one group , $x_{(2)},x_{(6)}$ Divided into one group , $x_{(3)},x_{(5)}$ Divided into one group , $x_{(4)}$ A single group )
among $\left|x_{(i)}-\theta \right|+\left|x_{(7-i+1)}-\theta \right|$ stay $\theta \in \left[x_{(i)},x_{(7-i+1)}\right]$ Take the minimum value when ; $\left|x_{(4)}-\theta \right|$ stay $\theta =x_{(4)}$ Take the minimum value when . and $\left({\bigcap }_{i=1}^3\left[x_{(i)},x_{(7-i+1)}\right]\right)\cap \left\{x_{(4)}\right\}=\left\{x_{(4)}\right\}$, therefore $\hat{\theta }=x_{(4)}$ yes $\theta$ Of MLE.

Nine 、(10 branch ) $(X_1,X_2)\sim N(0,0;1,1;0),$ seek $\frac{X_1}{X_2}$ Probability 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 .