[the Nine Yang Manual] 2021 Fudan University Applied Statistics real problem + analysis

The real part

One 、(15 branch ) There are countless lines with a distance of 2 Equidistant parallel lines of , First throw one side as 1 An equilateral triangle , Find the probability that the triangle intersects the parallel line .

Two 、(15 branch ) Party A and Party B toss coins , If the front is up, a wins 1 element , Reverse up B wins 1 element . Together with 20 round , In the end, neither of them will lose or win . It is known that a has no money at first , Ask the probability that a doesn't owe money in the whole process .

3、 ... and 、(15 branch ) $X_0,\cdots ,X_n,\cdots$ yes i.i.d. To obey $U(0,1)$ Random variable of , remember
$$N=\inf \{n\ge 1:X_n>X_0\},$$ seek $N$ The distribution law of .

Four 、(15 branch ) $X_1,\cdots ,X_n$ yes i.i.d. Random variable of , $\mathrm{P}(X=1)=0.4,\mathrm{P}(X=-0.5)=0.6$. Make $S_n=(1+X_1)(1+X_2)\cdots (1+X_n)$, ask $\mathrm{E}(S_n)$ And $S_n$ Convergence or not ? If convergence , Find its limit .

5、 ... and 、(15 branch ) Toss a coin , If the same surface appears three times in a row, it will stop , Record the number of throws as $N$, seek $\mathrm{E}N$.

6、 ... and 、(15 branch ) $X_1,X_2$ Independence and obedience $N(0,1)$, seek $\frac{X_1}{X_2}$ And $\frac{X_1^2}{X_2^2}$ The distribution of .

7、 ... and 、(10 branch ) Short answer :
(1)(5 branch ) Describe the definition of sufficient statistics .
(2)(5 branch ) Narration C-R inequality .

8、 ... and 、(10 branch ) Some come from the general $f(x)=2x,0<x<1$ Of 10 Random sample $X_1,\cdots ,X_{10}$, ask $\frac{X_{(3)}}{X_{(6)}}$ And $X_{(6)}$ Is it independent , Please give reasons .

Nine 、(10 branch ) Some come from the general $N(\mu ,1)$ Of $n$ Random sample $X_1,\cdots ,X_n$, prove :$\bar{X}$ It's full statistics .

Ten 、(10 branch ) Some come from the general $U(\theta ,2\theta )$ Of $n$ Random sample $X_1,\cdots ,X_n$, seek $\theta$ Maximum likelihood estimation of , And judge its unbiasedness and consistency .

11、 ... and 、(10 branch ) Some come from the general $N(\mu ,{\sigma }^2)$ Of $n$ Random sample $X_1,\cdots ,X_n$, among $\mu$ It is known that , Make
$$\hat{\sigma }=\frac{1}{n}\sqrt{\frac{\pi }{2}}\sum _{i=1}^n\left|X_i-\mu \right|,$$ verification $\hat{\sigma }$ yes $\sigma$ Unbiased estimation of , But it is not an effective estimate .

Twelve 、(10 branch ) Some come from the general $\mathcal{P}(\theta )$ Of $n$ Random sample , Consider the hypothesis test problem :
$${\mathrm{H}}_0:\theta =2\mathrm{v}\mathrm{s}{\mathrm{H}}_1:\theta =3,$$ There are denial domains $W=\{\bar{x}\ge 2.8\}$, ask
(1)(5 branch ) $n=5$ when , What is the probability of making two kinds of mistakes ?
(2)(5 branch ) $n$ Towards infinity , What happens to the probability of making two kinds of mistakes ? Please explain .

The analysis part

One 、(15 branch ) There are countless lines with a distance of 2 Equidistant parallel lines of , First throw one side as 1 An equilateral triangle , Find the probability that the triangle intersects the parallel 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
$$P\{\text{ Triangle pressed to a straight line }\}=P(3)=\frac{1}{2}\left(P_a+P_b+P_c\right)=\frac{3}{2\pi }.$$

Two 、(15 branch ) Party A and Party B toss coins , If the front is up, a wins 1 element , Reverse up B wins 1 element . Together with 20 round , In the end, neither of them will lose or win . It is known that a has no money at first , Ask the probability that a doesn't owe money in the whole process .

Solution:
Use the broken line method , The problem is : From the point of $(0,0)$ Random walk to point $(20,0)$, Do not touch the straight line during $y=-1$. First , From the point of $(0,0)$ Random walk to point $(20,0)$ A total of $C_{20}^{10}$ Kind of (20 Next time , A win 10 Time ). From the point of $(0,0)$ Random walk to and $(20,0)$ About $y=-1$ Symmetrical points $(20,-2)$ There are kinds in total $C_{20}^9$ Kind of . so Suppose the probability of the event is :
$$\mathrm{P}(A)=1-\frac{C_{20}^9}{C_{20}^{10}}=\frac{1}{11}.$$

3、 ... and 、(15 branch ) $X_0,\cdots ,X_n,\cdots$ yes i.i.d. To obey $U(0,1)$ Random variable of , remember
$$N=\inf \{n\ge 1:X_n>X_0\},$$ seek $N$ The distribution law of .

Solution:
First calculate the given $X_0=x$ Conditional distribution of time , It is similar to geometric distribution :
$$\begin{array}{rl}\mathrm{P}\left(N=k\mid X_0=x\right)& ={\mathrm{P}}^{k-1}\left(X_1\le x\right)\mathrm{P}\left(X_k>x\right)\\ & =x^{k-1}(1-x)\end{array}$$ By the formula of total probability in continuous occasions , Yes $\mathrm{P}(N=k)={\int }_0^1{x}^{k-1}(1-x)dx=\frac{1}{k(k+1)}$.

Four 、(15 branch ) $X_1,\cdots ,X_n$ yes i.i.d. Random variable of , $\mathrm{P}(X=1)=0.4,\mathrm{P}(X=-0.5)=0.6$. Make $S_n=(1+X_1)(1+X_2)\cdots (1+X_n)$, ask $\mathrm{E}(S_n)$ And $S_n$ Convergence or not ? If convergence , Find its limit .

Solution:
$\mathrm{E}\left(S_n\right)={\prod }_{i=1}^n\mathrm{E}\left(1+X_i\right)=(1.1{)}^n\to +\infty$, Expectation does not converge ; By strong law of numbers
$\frac{1}{n}\ln S_n=\frac{1}{n}{\sum }_{i=1}^n\ln \left(1+X_i\right)\stackrel{\text{ a.s. }}{*}\mathrm{Eln}\left(1+X_1\right)=-\frac{1}{5}\ln 2<0$,
so $S_n\stackrel{\text{ a.s. }}{*}{\lim }_{n\to \infty }\exp \left\{-\frac{n}{5}\ln 2\right\}=0.$

5、 ... and 、(15 branch ) Toss a coin , If the same surface appears three times in a row, it will stop , Record the number of throws as $N$, seek $\mathrm{E}N$.

Solution:
set up $A_{-2},A_{-1},A_0,A_1,A_2$ They are states “ Just cast continuously 2 The second reverse ”、“ Just cast continuously 1 The second reverse ”、 The initial state ”、“ Just cast continuously 1 Suboptimal ”、“ Just cast continuously 2 Suboptimal ”. set up $X$ Indicates starting from the current state , The number of times to stop on the same side for three consecutive times , Then there are :
$$\begin{array}{rl}\mathrm{E}\left[X\mid A_{-2}\right]& =\frac{1}{2}+\frac{1}{2}\left(\mathrm{E}\left[X\mid A_1\right]+1\right)\\ \mathrm{E}\left[X\mid A_{-1}\right]& =\frac{1}{2}\left(\mathrm{E}\left[X\mid A_{-2}\right]+1\right)+\frac{1}{2}\left(\mathrm{E}\left[X\mid A_1\right]+1\right)\\ E\left[X\mid A_0\right]& =\frac{1}{2}\left(\mathrm{E}\left[X\mid A_{-1}\right]+1\right)+\frac{1}{2}\left(\mathrm{E}\left[X\mid A_1\right]+1\right)\\ \mathrm{E}\left[X\mid A_1\right]& =\frac{1}{2}\left(\mathrm{E}\left[X\mid A_{-1}\right]+1\right)+\frac{1}{2}\left(\mathrm{E}\left[X\mid A_2\right]+1\right)\\ \mathrm{E}\left[X\mid A_2\right]& =\frac{1}{2}+\frac{1}{2}\left(\mathrm{E}\left[X\mid A_{-1}\right]+1\right)\end{array}$$ Obviously, the problem has symmetry , set up
$$x=E\left[X\mid A_{-2}\right]=E\left[X\mid A_2\right],y=E\left[X\mid A_{-1}\right]=E\left[X\mid A_1\right],z=E\left[X\mid A_0\right],$$
There are
$$\{\begin{array}{l}x=\frac{1}{2}+\frac{1}{2}\left(y+1\right),\\ y=\frac{1}{2}\left(x+1\right)+\frac{1}{2}\left(y+1\right),\\ z=\frac{1}{2}\left(y+1\right)+\frac{1}{2}\left(y+1\right),\end{array}*\{\begin{array}{l}x-\frac{1}{2}y=1,\\ x-y=-2,\\ z=y+1,\end{array}*\{\begin{array}{l}x=4,\\ y=6,\\ z=7.\end{array}$$ so $\mathrm{E}N=\mathrm{E}\left[X\mid A_0\right]=7$.

6、 ... and 、(15 branch ) $X_1,X_2$ Independence and obedience $N(0,1)$, seek $\frac{X_1}{X_2}$ And $\frac{X_1^2}{X_2^2}$ The distribution of .

Solution:
$Y=\frac{X_1}{X_2}\sim \mathrm{Cau}(0,1)$, Make $Z=Y^2$,
$$\begin{array}{rl}\mathrm{P}(Z\le z)& =\mathrm{P}(-\sqrt{z}\le Y\le \sqrt{z})\\ & =2{\int }_0^{\sqrt{z}}\frac{1}{\pi \left(1+x^2\right)}dx\\ & =\frac{2}{\pi }\arctan \sqrt{z}\end{array}$$ After derivation , There's a density function
$$f_Z(z)=\frac{1}{\pi (1+z)\sqrt{z}},z>0$$ let me put it another way , because $Y\sim t(1)$, so $Z=Y^2\sim F(1,1)$.

7、 ... and 、(10 branch ) Short answer :
(1)(5 branch ) Describe the definition of sufficient statistics .
(2)(5 branch ) Narration C-R inequality .

Solution:
(1) Sufficient statistics are statistics that fully contain the information of sample characterization parameters , When sufficient statistics $T$ Timing , Conditional distribution of samples $f\left(x_1,\cdots ,x_n\mid T=t\right)$ It has nothing to do with parameters .
(2) C-R Inequality refers to a kind of population ( Satisfy the regularity condition : Integral derivation can be ordered in a different order ) When considering the unbiased estimation of the function of parameters, the variance will have a lower bound , If yes $g(\theta )$ Make an estimate , C-R The lower bound is $\frac{{\left[g^{\prime }(\theta )\right]}^2}{nI(\theta )}$, among $I(\theta )=\mathrm{E}{\left[\frac{\partial \ln f(X;\theta )}{\partial \theta }\right]}^2$ It's Fisher information .

8、 ... and 、(10 branch ) Some come from the general $f(x)=2x,0<x<1$ Of 10 Random sample $X_1,\cdots ,X_{10}$, ask $\frac{X_{(3)}}{X_{(6)}}$ And $X_{(6)}$ Is it independent , Please give reasons .

Solution:
$\left(U=\frac{X_{(3)}}{X_{(6)}},V=X_{(6)}\right)$ The joint distribution of is :
$$\begin{array}{rl}f(u,v)& =v{f}_{X_{(3)},X_{(6)}}(uv,v)\\ & =C{v}^{10}{\left(1-v^2\right)}^4{u}^5{\left(1-u^2\right)}^2\end{array}$$ among , $u\in (0,1),v\in (0,1)$, The definition field is rectangular and can be factorized , so $U,V$ Independent .

Nine 、(10 branch ) Some come from the general $N(\mu ,1)$ Of $n$ Random sample $X_1,\cdots ,X_n$, prove :$\bar{X}$ It's full statistics .

Solution:
The joint density function can be written as
$$\begin{array}{rl}f\left(x_1,\cdots ,x_n;\mu \right)& =C\exp \left\{-\frac{{\sum }_{i=1}^n{\left(x_i-\mu \right)}^2}{2}\right\}\\ & =C\exp \left\{-\frac{-2\mu {\sum }_{i=1}^n{x}_i+n{\mu }^2+{\sum }_{i=1}^n{x}_i^2}{2}\right\}\end{array}$$ By the Factorization Theorem , $\bar{X}$ It's full statistics .

Ten 、(10 branch ) Some come from the general $U(\theta ,2\theta )$ Of $n$ Random sample $X_1,\cdots ,X_n$, seek $\theta$ Maximum likelihood estimation of , And judge its unbiasedness and consistency .

Solution:
The likelihood function is
$$L(\theta )=\frac{1}{{\theta }^n}{I}_{\left\{x_{(1)}>\theta >\frac{x_{(w)}}{2}\right\}},$$ The previous part is about $\theta$ Monotonic decline , Description of indicative function $\theta$ The minimum is $\frac{x_{(n)}}{2}$, So the maximum likelihood estimation is $\hat{\theta }=\frac{X_{(n)}}{2}$. Make $Y=\frac{X-\theta }{\theta }\sim U(0,1)$, so $\mathrm{E}{Y}_{(n)}=\frac{n}{n+1},\mathrm{E}{Y}_{(n)}\stackrel{\text{ a.s. }}{*}1$, There are
$$\mathrm{E}\hat{\theta }=\frac{2n+1}{2n+2}\theta \ne \theta ,\hat{\theta }\stackrel{\text{ a.s. }}{*}\theta \text{. }$$

11、 ... and 、(10 branch ) Some come from the general $N(\mu ,{\sigma }^2)$ Of $n$ Random sample $X_1,\cdots ,X_n$, among $\mu$ It is known that , Make
$$\hat{\sigma }=\frac{1}{n}\sqrt{\frac{\pi }{2}}\sum _{i=1}^n\left|X_i-\mu \right|,$$ verification $\hat{\sigma }$ yes $\sigma$ Unbiased estimation of , But it is not an effective estimate .

Solution:
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 $I\left({\sigma }^2\right)=\frac{1}{2{\sigma }^4}$. Make $g(x)=\sqrt{x}$, be $\sigma$ Of C-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 :
$$\begin{array}{c}\frac{1}{\sigma }E\left|X_1-\mu \right|={\int }_{-\infty }^{+\infty }|x|\frac{1}{\sqrt{2\pi }}{e}^{-\frac{x^2}{2}}dx=\sqrt{\frac{2}{\pi }}{\int }_0^{+\infty }x{e}^{-\frac{x^2}{2}}dx=\sqrt{\frac{2}{\pi }}\Rightarrow E\left|X_1-\mu \right|=\sqrt{\frac{2}{\pi }}\sigma ,\\ E{\left|X_1-\mu \right|}^2={\sigma }^2,\text{ so }\mathrm{Var}\left(\left|X_1-\mu \right|\right)={\sigma }^2-\frac{2}{\pi }{\sigma }^2=\left(1-\frac{2}{\pi }\right){\sigma }^2,\end{array}$$ 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 it didn't reach C-R Lower bound , Not a valid estimate .

Twelve 、(10 branch ) Some come from the general $\mathcal{P}(\theta )$ Of $n$ Random sample , Consider the hypothesis test problem :
$${\mathrm{H}}_0:\theta =2\mathrm{v}\mathrm{s}{\mathrm{H}}_1:\theta =3,$$ There are denial domains $W=\{\bar{x}\ge 2.8\}$, ask
(1)(5 branch ) $n=5$ when , What is the probability of making two kinds of mistakes ?
(2)(5 branch ) $n$ Towards infinity , What happens to the probability of making two kinds of mistakes ? Please explain .

Solution:
(1) The first type of error is :
$$\begin{array}{rl}\alpha & ={\mathrm{P}}_{\theta =2}\{\bar{X}\ge 2.8\}\\ & ={\mathrm{P}}_{\theta =2}\left\{\sum _{i=1}^5{X}_i\ge 14\right\}\\ & =\sum _{k=14}^{\infty }\frac{10^k}{k!}{e}^{-10}\end{array}$$ The second type of error is :
$$\begin{array}{rl}\beta & ={\mathrm{P}}_{\theta =3}\{\bar{X}<2.8\}\\ & ={\mathrm{P}}_{\theta =3}\left\{\sum _{i=1}^5{X}_i<14\right\}\\ & =\sum _{k=0}^{13}\frac{15^k}{k!}{e}^{-15}\end{array}$$(2) From the law of large numbers :
$$\begin{array}{rl}\alpha & ={\mathrm{P}}_{\theta =2}\{\bar{X}\ge 2.8\}\\ & \le {\mathrm{P}}_{\theta =2}\{|\bar{X}-2|\ge 0.8\}*0\\ \beta & ={\mathrm{P}}_{\theta =3}\{\bar{X}<2.8\}\\ & \le {\mathrm{P}}_{\theta =3}\{|\bar{X}-3|>0.2\}*0\end{array}$$ So when the sample size tends to infinity , Both types of errors tend to 0 .