[the Nine Yang Manual] 2017 Fudan University Applied Statistics real problem + analysis

The real part

One 、(30 branch ) A term is used to explain

(1)(5 branch ) Sample mean 、 Sample variance ;

(2)(5 branch ) statistic ;

(3)(5 branch ) Order statistics ;

(4)(5 branch ) Median 、 Sample median ;

(5)(5 branch ) Empirical distribution function ;

(6)(5 branch ) Unbiased estimate .

Two 、(20 branch ) $X_1,X_2,$ i.i.d $\sim \mathrm{Exp}(1),$ seek

(1)(10 branch ) $\frac{X_1}{X_1+X_2}$ Density function of ;

(2)(10 branch ) $X_{(2)}-X_{(1)}$ Density function of .

3、 ... and 、(20 branch ) $X_1,X_2$ i.i.d.$\sim N(0,1),$ seek $\frac{X_1}{X_2}$ Probability distribution of .

Four 、(20 branch ) $X_1,X_2,\dots ,X_n$ i.i.d.$\sim F(x),$ remember $Y_n(x)={\sum }_{i=1}^{n}I\left[X_i\le x\right],$ seek ${\lim }_{n\to \infty }\frac{Y_n(x)}{n}$.

5、 ... and 、(20 branch ) $X_0,X_1,\cdots ,X_{2n},$ i.i.d $\sim U(0,1),X_{(0)},X_{(1)},\cdots ,X_{(2n)}$ Is the corresponding order statistic , Try to prove $X_{(n)}\stackrel{P}{\to }\frac{1}{2}$.

6、 ... and 、(20 branch ) Continuous random variables are known $X$ The expectations of the $EX$ There is , $f(t)=E|X-t|$ stay $t=m$ Take the minimum value , prove $P(X\le m)=\frac{1}{2}$.

7、 ... and 、(20 branch ) $X_1,X_2,X_3$ i.i.d $\sim N(0,1),$ seek

(1)(10 branch ) $P\left(X_1>X_2>X_3\right)$;

(2)(10 branch ) $P\left(X_1>X_2,X_1>X_3\right)$.

The analysis part

One 、(30 branch ) A term is used to explain

(1)(5 branch ) Sample mean 、 Sample variance ;

(2)(5 branch ) statistic ;

(3)(5 branch ) Order statistics ;

(4)(5 branch ) Median 、 Sample median ;

(5)(5 branch ) Empirical distribution function ;

(6)(5 branch ) Unbiased estimate .

Solution: (1) The sample mean is taken from a specific population $n$ Independent random samples , Calculated average , Write it down as $\bar{X}=\frac{1}{n}{\sum }_{i=1}^n{X}_i$, If the overall expectation exists , be $\bar{X}$ It is a strong consistent estimate of the overall expectation . The sample variance is based on this $n$ The square of the corrected mean deviation calculated from independent random samples , Correction means that the average is divided by its degrees of freedom $n-1$ Not the number of data $n$, The sample variance is generally recorded as $S^2=\frac{1}{n-1}{\sum }_{i=1}^n{\left(X_i-\bar{X}\right)}^2$, If the variance of the population exists , be $S^2$ It is also a strong consistent estimate of the population variance .

(2) A statistic is a function whose expression contains only samples but no unknown parameters , It is essentially a random variable ( vector ), Of course, when the value of random samples is given , Statistics can also be regarded as a known constant ( Constant vector ), At this time, the statistic is “ Observations of Statistics ” For short .

(3) Order statistics refers to random samples $X_1,X_2,\dots ,X_n$ Rearrange the statistics from small to large , Generally, it is written from small to large $X_{(1)},X_{(2)},\dots ,X_{(n)}$.

(4) If $x_{0.5}$ Satisfy $P\left\{X\le x_{0.5}\right\}=0.5$ said $x_{0.5}$ Is the median . And the median of the sample $m_{0.5}$ It refers to the number in the middle of the random sample , If expressed in order statistics, it is $$m_{0.5}=\{\begin{array}{ll}{x}_{\left(\frac{n+1}{2}\right)},& n=\text{ Odd number , }\\ \frac{x_{\left(\frac{n}{2}\right)}+x_{\left(\frac{n}{2}+1\right)}}{2},& n=\text{ even numbers . }\end{array}$$
(5) Empirical distribution function is an estimate of the overall distribution function based on sample information , Write it down as $$F_n(x)=\frac{1}{n}\sum _{i=1}^{n}I\left[X_i\le x\right],$$ according to Glivenko-Cantelli Theorem , The empirical distribution function is a uniformly strong consistent estimate of the population distribution function , namely $$\underset{x}{\sup }\left|F_n(x)-F(x)\right|\stackrel{a.s.}{*}0.$$ (6) If $\hat{g}\left(X_1,\dots ,X_n\right)$ Satisfy $E\hat{g}\left(X_1,\dots ,X_n\right)=g(\theta )$, said $\hat{g}$ yes $g$ Unbiased estimation of , Unbiasedness is a A very important good standard , But it's not necessary , As if $EX$ There is , A random sample $X_1$ Always an unbiased estimate of overall expectations , But you can hardly say that it is a good estimate .

Two 、(20 branch ) $X_1,X_2,$ i.i.d $\sim \mathrm{Exp}(1),$ seek

(1)(10 branch ) $\frac{X_1}{X_1+X_2}$ Density function of ;

(2)(10 branch ) $X_{(2)}-X_{(1)}$ Density function of .

Solution: (1) According to the meaning , $2X_1\sim {\chi }^2(2),2X_2\sim {\chi }^2(2)$ And independent of each other , so $\frac{X_1}{X_1+X_2}=\frac{2X_1}{2X_1+2X_2}\sim \mathrm{Beta}(1,1)$, The density function is $f(x)=1,0<x<1$, namely $U(0,1)$.

(2) It's easy to see $X_{(2)}-X_{(1)}=\left|X_1-X_2\right|$, According to the convolution formula $Y=X_1-X_2$ The density function of is $$f(y)=\{\begin{array}{ll}\frac{1}{2}{e}^{-y},& y\ge 0\\ \frac{1}{2}{e}^y,& y<0\end{array}$$ so $|Y|$ The density function of is $f(x)=e^{-y},y\ge 0$.

[ notes ] if $X\sim {\chi }^2(m),Y\sim {\chi }^2(n)$, And independent of each other , be $$\frac{X}{X+Y}\sim \mathrm{Beta}\left(\frac{m}{2},\frac{n}{2}\right).$$ The analysis is as follows : remember $U=\frac{X}{X+Y},V=X+Y$, After inverse solution $x=uv,y=v-uv$, $$|J|=\left|\frac{\partial (x,y)}{\partial (u,v)}\right|=\left|\begin{array}{cc}v& u\\ -v& 1-u\end{array}\right|=v,$$ therefore $f_{U,V}(u,v)=v{f}_{X,Y}(uv,(1-u)v)=\frac{u^{\frac{m}{2}-1}(1-u{)}^{\frac{n}{2}-1}}{\Gamma \left(\frac{m}{2}\right)\Gamma \left(\frac{n}{2}\right)}\cdot {\left(\frac{1}{2}\right)}^{\frac{m+n}{2}}{v}^{\frac{m+n}{2}-1}{e}^{-\frac{v}{2}}$, so $$f_U(u)={\int }_0^{+\infty }{f}_{U,V}(u,v)dv=\frac{\Gamma \left(\frac{m+n}{2}\right)}{\Gamma \left(\frac{m}{2}\right)\Gamma \left(\frac{n}{2}\right)}{u}^{\frac{m}{2}-1}(1-u{)}^{\frac{n}{2}-1}\sim \mathrm{Beta}\left(\frac{m}{2},\frac{n}{2}\right).$$

3、 ... and 、(20 branch ) $X_1,X_2$ i.i.d $\sim N(0,1),$ 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 $$\frac{N(0,1)}{\sqrt{\frac{{\chi }^2(1)}{1}}}$$ Is a degree of freedom 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 .

Four 、(20 branch ) $X_1,X_2,\dots ,X_n$ i.i.d $\sim F(x),$ remember $Y_n(x)={\sum }_{i=1}^{n}I\left[X_i\le x\right],$ seek ${\lim }_{n\to \infty }\frac{Y_n(x)}{n}$.

Solution: According to the strong law of numbers , ${\lim }_{n\to \infty }\frac{Y_n(x)}{n}=EI\left[X_1\le x\right]=P\left(X_1\le x\right)=F(x)$, a.s.

5、 ... and 、(20 branch ) $X_0,X_1,\cdots ,X_{2n},$ i.i.d $\sim U(0,1),X_{(0)},X_{(1)},\cdots ,X_{(2n)}$ Is the corresponding order statistic , Try to prove $X_{(n)}\stackrel{P}{\to }\frac{1}{2}$.

Solution: So let's calculate $Y=X_{(n)}$ Density function of , Thought is : from $X_0,X_1,\cdots ,X_{2n}$ Choose one of them as $Y$, be left over There should be $n$ Ratio $Y$ Small , $n$ Ratio $Y$ Big , Of course, there are $\frac{(2n+1)!}{n!\cdot 1!\cdot n!}$ Kind of , therefore

$$\begin{array}{c}f(y)=\frac{(2n+1)!}{n!n!}P\left\{X_0\le y,X_1\le y,\dots ,X_{n-1}\le y\right\}{f}_{X_n}(y)P\left\{X_{n+1}>y,\dots ,X_{2n}>y\right\}\\ P\left\{X_0\le y,X_1\le y,\dots ,X_{n-1}\le y\right\}=P^n\left\{X_0\le y\right\}=y^n,\\ P\left\{X_{n+1}>y,\dots ,X_{2n}>y\right\}=P^n\left\{X_{2n}>y\right\}=(1-y{)}^n,\end{array}$$ so $f(y)=\frac{\Gamma (2n+2)}{\Gamma (n+1)\Gamma (n+1)}{y}^n(1-y{)}^n,0<y<1$, This is a $\mathrm{Beta}(n+1,n+1)$ Distribution function of . according to Beta The nature of the distribution , $$EY=\frac{n+1}{2n+2}=\frac{1}{2},\mathrm{Var}(Y)=\frac{(n+1{)}^2}{(2n+2{)}^2(2n+3)}\to 0,$$ Using Chebyshev inequality , $P\left\{\left|Y-\frac{1}{2}\right|>\epsilon \right\}\le \frac{\mathrm{Var}(Y)}{{\epsilon }^2}\to 0$, therefore $X_{(n)}\stackrel{P}{\to }\frac{1}{2}$.

6、 ... and 、(20 branch ) Continuous random variables are known $X$ The expectations of the $EX$ There is , $f(t)=E|X-t|$ stay $t=m$ Take the minimum value , prove $P(X\le m)=\frac{1}{2}$.

Solution: set up $X$ The distribution function of $F(x)$, The MI degree function is $p(x)$, be $$f(t)={\int }_{-\infty }^t(t-x)dF(x)+{\int }_t^{+\infty }(x-t)dF(x)=tF(t)-{\int }_{-\infty }^{t}xdF(x)+{\int }_t^{+\infty }xdF(x)-t(1-F(t)),$$ Yes $t$ Derivation , have to $$f^{\prime }(t)=F(t)+tp(t)-tp(t)-tp(t)-1+F(t)+tp(t)=2F(t)-1,$$ Make $f^{\prime }(t)=0$, Solution $t=x_{0.5}(x_a$ Indicates that it is satisfied $F\left(x_a\right)=a)$. The second derivative $f^{\prime \prime }(t)=2p(t)\ge 0$, Therefore, the stagnation point is the minimum point . According to the meaning $m=x_{0.5}$, so $$P(X\le m)=\frac{1}{2}.$$

7、 ... and 、(20 branch ) $X_1,X_2,X_3,$ i.i.d $\sim N(0,1),$ seek

(1)(10 branch ) $P\left(X_1>X_2>X_3\right)$;

(2)(10 branch ) $P\left(X_1>X_2,X_1>X_3\right)$.

Solution: (1) According to the rotation symmetry , $$\begin{array}{rl}& P\left(X_1>X_2>X_3\right)=P\left(X_1>X_3>X_2\right)=P\left(X_2>X_1>X_3\right)\\ =& P\left(X_2>X_3>X_1\right)=P\left(X_3>X_1>X_2\right)=P\left(X_3>X_2>X_1\right)\end{array}$$ These events are disjoint , And the probability of these events is 1 Events , so $$6P\left(X_1>X_2>X_3\right)=1\Rightarrow P\left(X_1>X_2>X_3\right)=\frac{1}{6}.$$ (2) We found that :
$$P\left(X_1>X_2,X_1>X_3\right)=P\left(X_1=\max \left\{X_1,X_2,X_3\right\}\right),$$ According to the rotation symmetry , $$P\left(X_1=\max \left\{X_1,X_2,X_3\right\}\right)=P\left(X_2=\max \left\{X_1,X_2,X_3\right\}\right)=P\left(X_3=\max \left\{X_1,X_2,X_3\right\}\right),$$

so $P\left(X_1>X_2,X_1>X_3\right)=\frac{1}{3}$.