[the Nine Yang Manual] 2016 Fudan University Applied Statistics real problem + analysis

The real part

One 、(15 branch ) Three people decipher the code independently at the same time , And the probability that three people can decipher the password is 1/5, 1/3 and 1/4, Find the probability that this password can be decoded .

Two 、(15 branch ) from (0,1) Take two numbers randomly in , Find that the product is not less than 3/16 And its sum is not greater than 1 Probability .

3、 ... and 、(15 branch ) $X\sim N(0,1),$ seek $Y=X^2$ Density function of .

Four 、(30 branch ) remember (0,1),(1,0),(0,0) The area enclosed by three points is $D,(X,Y)$ obey $D$ Even distribution on , seek

(1)(15 branch ) $E(X+Y),\mathrm{Var}(X+Y)$;

(2)(15 branch ) $X,Y$ The correlation coefficient of .

5、 ... and 、(15 branch ) For two random variables with only two values $X,Y,$ Try to prove $X,Y$ Independent if and only if $X,Y$ Unrelated .

6、 ... and 、(15 branch ) Set from the overall $X\sim N\left(\mu ,{\sigma }^2\right)$ Of $2n$ Random sample , Try to expect
$$E\left(\sum _{i=1}^n{\left(X_i+X_{n+i}-2\bar{X}\right)}^2\right).$$

7、 ... and 、(15 branch ) $X_1,X_2,X_3$ yes i.i.d. To obey $U(0,1)$ Random variable of , Find order statistics $X_{(1)}$ Distribution function and expectation of .

8、 ... and 、(30 branch ) From $X\sim f(x)=\{\begin{array}{cl}\theta ,& 0\le x<1\\ 1-\theta ,& 1\le x<2,\\ 0,& \text{ otherwise }\end{array}$ Of $n$ A simple random sample , seek

(1)(15 branch ) $\theta$ The moment estimate of ;

(2)(15 branch ) $\theta$ Maximum likelihood estimation of , It is known that $m={\sum }_{i=1}^{n}I\left[X_i<1\right]$.

The analysis part

One 、(15 branch ) Three people decipher the code independently at the same time , And the probability that three people can decipher the password is 1/5, 1/3 and 1/4, Find the probability that this password can be decoded .

Solution: According to de Morgan formula , $$\begin{array}{rl}P\{\text{ At least one person decoded }\}& =1-P\{\text{ No one decoded }\}\\ & =1-\frac{4}{5}\times \frac{2}{3}\times \frac{3}{4}=0.6.\end{array}$$

Two 、(15 branch ) from (0,1) Take two numbers randomly in , Find that the product is not less than 3/16 And its sum is not greater than 1 Probability .

Solution: Let the two numbers obtained be $X,Y\sim U(0,1)$, Its product is not less than $\frac{3}{16}$ signify $XY\ge \frac{3}{16}$, Its sum is not greater than 1 signify $X+Y\le 1$.
Make $D=\left\{(x,y):xy\ge \frac{3}{16},x+y\le 1\right\}\cap (0,1{)}^2$, The probability is $${\int }_{D}f(x,y)dxdy=\frac{1}{8}-{\int }_{\frac{1}{4}}^{\frac{3}{4}}{\int }_{\frac{1}{4}}^{\frac{3}{16x}}dydx=\frac{1}{8}-{\int }_{\frac{1}{4}}^{\frac{3}{4}}\left(\frac{3}{16x}-\frac{1}{4}\right)dx=\frac{1}{4}-\frac{3\ln 3}{16}$$

3、 ... and 、(15 branch ) $X\sim N(0,1),$ seek $Y=X^2$ Density function of .

Solution: When $y\ge 0$ when ,
$F(y)=P\{Y\le y\}=P\left\{X^2\le >y\right\}=P\{-\sqrt{y}\le X\le \sqrt{y}\}=2\Phi (\sqrt{y})-1$, so
$$f_Y(y)=F^{\prime }(y)=2\phi (\sqrt{y})\cdot \frac{1}{2\sqrt{y}}=\frac{1}{\sqrt{2\pi y}}{e}^{-\frac{y}{2}},y\ge 0$$

Four 、(30 branch ) remember (0,1),(1,0),(0,0) The area enclosed by three points is $D,(X,Y)$ obey $D$ Even distribution on , seek

(1)(15 branch ) $E(X+Y),\mathrm{Var}(X+Y)$;

(2)(15 branch ) $X,Y$ The correlation coefficient of .

Solution: (1) $E(X+Y)=2{\int }_0^1{\int }_0^{1-y}(x+y)dxdy={\int }_0^1\left(1-y^2\right)dy=1-\frac{1}{3}=\frac{2}{3}$, $$E(X+Y{)}^2=2{\int }_0^1{\int }_0^{1-y}(x+y{)}^{2}dxdy=\frac{2}{3}{\int }_0^1\left(1-y^3\right)dy=\frac{2}{3}\left(1-\frac{1}{4}\right)=\frac{1}{2},$$ therefore
$$\mathrm{Var}(X+Y)=\frac{1}{2}-\frac{4}{9}=\frac{1}{18}.$$

(2) $EXY=2{\int }_0^1{\int }_0^{1-y}xydxdy={\int }_0^1\left(y^3-2y^2+y\right)dy=\frac{1}{4}-\frac{2}{3}+\frac{1}{2}=\frac{1}{12}$, $$EX=EY=2{\int }_0^1{\int }_0^{1-y}xdxdy={\int }_0^1\left(y^2-2y+1\right)dy=\frac{1}{3}-1+1=\frac{1}{3},$$ so
$$\mathrm{Cov}(X,Y)=\frac{1}{12}-\frac{1}{9}=-\frac{1}{36}.$$$$E{X}^2=2{\int }_0^1{\int }_0^{1-y}{x}^{2}dxdy=\frac{2}{3}{\int }_0^1(1-y{)}^{3}dy=\frac{1}{6},$$ so $$\mathrm{Var}(X)=\mathrm{Var}(Y)=\frac{1}{6}-\frac{1}{9}=\frac{1}{18},$$ And then there are ${\rho }_{XY}=-\frac{1}{2}$.

5、 ... and 、(15 branch ) For two random variables with only two values $X,Y,$ Try to prove $X,Y$ Independent if and only if $X,Y$ Unrelated .

Solution:

	$X=x_1$	$X=x_2$
$Y=y_1$	$p_{11}$	$p_{21}$	$1-q$
$Y=y_2$	$p_{12}$	$p_{22}$	$q$
	$1-p$	$p$

Solution: Independence must not be relevant , Just prove the sufficiency . Just discuss $X^{\prime }=\frac{X-x_1}{x_2-x_1}$, $Y^{\prime }=\frac{Y-y_1}{y_2-y_1}$, $X,Y$ Irrelevant and $X^{\prime },Y^{\prime }$ Irrelevance is equivalent . When $X^{\prime },Y^{\prime }$ When it's not relevant , explain $$p_{22}=E(X^{\prime }{Y}^{\prime })=E(X^{\prime })E(Y^{\prime })=pq,$$ Further, there are $$p_{21}=p-p_{22}=p-pq=p(1-q),$$$$p_{12}=q-p_{22}=q-pq=q(1-p),$$$$p_{11}=1-p-p_{12}=(1-p)-q(1-p)=(1-p)(1-q).$$ therefore $X^{\prime },Y^{\prime }$ Independent , so $X,Y$ Also independent .

6、 ... and 、(15 branch ) Set from the overall $X\sim N\left(\mu ,{\sigma }^2\right)$ Of $2n$ Random sample , Try to expect
$$E\left(\sum _{i=1}^n{\left(X_i+X_{n+i}-2\bar{X}\right)}^2\right).$$

Solution: remember $Y_i=X_i+X_{n+i},i=1,2,\dots ,n$, such $Y_1,Y_2,\dots ,Y_n$ It comes from the whole $N\left(2\mu ,2{\sigma }^2\right)$ Of $n$ Random sample , $\bar{Y}=2\bar{X}$, According to the definition and nature of sample variance , We have $$\frac{1}{2{\sigma }^2}\sum _{i=1}^n{\left(X_i+X_{n+i}-2\bar{X}\right)}^2=\frac{1}{2{\sigma }^2}\sum _{i=1}^n{\left(Y_i-\bar{Y}\right)}^2\sim {\chi }^2(n-1)$$ therefore
$$E\sum _{i=1}^n{\left(X_i+X_{n+i}-2\bar{X}\right)}^2=2(n-1){\sigma }^2.$$

7、 ... and 、(15 branch ) $X_1,X_2,X_3$, yes i.i.d. To obey $U(0,1)$ Random variable of , Find order statistics $X_{(1)}$ Distribution function and expectation of .

Solution: set up $Y=X_{(1)}$, When $y\in (0,1)$ when , $$P\{Y\ge y\}=P\left(\min \left\{X_1,X_2,X_3\right\}\ge y\right)=P^3\left\{X_1\ge y\right\}=(1-y{)}^3$$ So the distribution function is
$$F(y)=\{\begin{array}{lc}0,& y<0,\\ 1-(1-y{)}^3,& 0\le y<1,\\ 1,& y\ge 1.\end{array}$$ This happens to be $\mathrm{Beta}(1,3)$ Distribution function of . $$EY={\int }_0^{1}ydF(y)=1\cdot F(1)-{\int }_0^1\left[1-(1-y{)}^3\right]dy=\frac{1}{4}$$

8、 ... and 、(30 branch ) From $X\sim f(x)=\{\begin{array}{cl}\theta ,& 0\le x<1\\ 1-\theta ,& 1\le x<2,\\ 0,& \text{ otherwise }\end{array}$ Of $n$ A simple random sample , seek

(1)(15 branch ) $\theta$ The moment estimate of ;

(2)(15 branch ) $\theta$ Maximum likelihood estimation of , It is known that $m={\sum }_{i=1}^{n}I\left[X_i<1\right]$.

Solution: (1) $EX={\int }_0^{1}x\theta dx+{\int }_1^{2}x(1-\theta )dx=\frac{3}{2}-\theta$, therefore $${\hat{\theta }}_1=\frac{3}{2}-\bar{X}.$$(2) Likelihood function $L\left(x_1,\dots ,x_n;\theta \right)={\theta }^m(1-\theta {)}^{n-m},$
$\ln L=m\ln \theta +(n-m)\ln (1-\theta )$, Yes $\theta$ Finding partial derivatives , have to $$\frac{\partial \ln L}{\partial \theta }=\frac{m}{\theta }-\frac{n-m}{1-\theta }\stackrel{let}{=}0,$$ Solution ${\hat{\theta }}_2=\frac{m}{n}$.