当前位置:网站首页>[the Nine Yang Manual] 2022 Fudan University Applied Statistics real problem + analysis
[the Nine Yang Manual] 2022 Fudan University Applied Statistics real problem + analysis
2022-07-06 13:31:00 【Elder martial brother statistics】
Catalog
The real part
One 、(20 branch ) There is... In the bag : a a a Red ball , a a a yellow ball , b b b blue ball . Yes, put it back and touch it 3 A ball , set up A = { A=\{ A={ Draw out a yellow ball and a red ball , And the red ball is taken out before the yellow ball } \} }. seek :
(1)(10 branch ) P ( A ) P(A) P(A);
(2)(10 branch ) if { \{ { I didn't touch the basketball } \} } And A A A The probability is the same , seek a : b a:b a:b.
Two 、(10 branch ) Discrete random variables X X X The distribution column of is
P ( X = a ) = P ( X = b ) = P ( X = a + 1 ) = 1 3 , P(X=a)=P(X=b)=P(X=a+1)=\frac{1}{3}, P(X=a)=P(X=b)=P(X=a+1)=31, among a < b < a + 1 a<b<a+1 a<b<a+1. Find the value range of its variance .
3、 ... and 、(20 branch ) Discrete random variables X X X Take only x , x + a x,x+a x,x+a Two values , among a > 0 a>0 a>0, And V a r ( X ) = 1 Var(X)=1 Var(X)=1, seek a a a Value range and X X X The distribution column of .
Four 、(20 branch ) With random vectors ( X Y ) \left( \begin{array}{c} X\\ Y\\ \end{array} \right) (XY), It is known that after arbitrary rotation transformation ( cos α sin α − sin α cos α ) ( X Y ) \left( \begin{matrix} \cos \alpha& \sin \alpha\\ -\sin \alpha& \cos \alpha\\ \end{matrix} \right) \left( \begin{array}{c} X\\ Y\\ \end{array} \right) (cosα−sinαsinαcosα)(XY) Still with ( X Y ) \left( \begin{array}{c} X\\ Y\\ \end{array} \right) (XY) Homodistribution , Try to solve the following problems :
(1) seek P ( 0 < Y < X ) P(0<Y<X) P(0<Y<X);
(2) seek Y X \frac{Y}{X} XY The distribution of .
5、 ... and 、(20 branch ) It is known that ( X , Y ) ∼ N ( 0 , 0 ; 1 , 1 ; 1 2 ) (X,Y)\sim N(0,0;1,1;\frac{1}{2}) (X,Y)∼N(0,0;1,1;21), seek P ( X > 0 , Y > 0 ) P(X>0,Y>0) P(X>0,Y>0).
6、 ... and 、(10 branch ) X 1 , ⋯ , X n , ⋯ X_1,\cdots,X_n,\cdots X1,⋯,Xn,⋯ yes i.i.d. There are random variables in the second moment of , Y n = ∑ i = 1 n X i Y_n = \sum_{i=1}^n X_i Yn=∑i=1nXi, ask : { Y n n 2 } \{\frac{Y_n}{n^2}\} { n2Yn} Obey the law of large numbers .
7、 ... and 、(10 branch ) X 1 , ⋯ , X n X_1,\cdots,X_n X1,⋯,Xn yes i.i.d. obey N ( μ , σ 2 ) N(\mu,\sigma^2) N(μ,σ2) Random variable of , F F F Is its distribution function , seek − 2 ∑ i = 1 n ln F ( X i ) -2\sum_{i=1}^n \ln F(X_i) −2∑i=1nlnF(Xi) The distribution of .
8、 ... and 、(10 branch ) X 1 , ⋯ , X 6 X_1,\cdots,X_6 X1,⋯,X6 yes i.i.d. Of U ( 0 , 1 ) U(0,1) U(0,1) A random variable , seek V a r ( 2 X ( 2 ) + 3 X ( 3 ) ) Var(2X_{(2)}+3X_{(3)}) Var(2X(2)+3X(3)).
Nine 、(10 branch ) X 1 , ⋯ , X n X_1,\cdots,X_n X1,⋯,Xni yes i.i.d. Of U ( 0 , θ ) U(0,\theta) U(0,θ) A random sample , set up a X ( 1 ) , b X ( 3 ) aX_{(1)},bX_{(3)} aX(1),bX(3) yes θ \theta θ Unbiased estimation of , seek a , b a,b a,b And compare which of them is more effective .
Ten 、(10 branch ) set up X 1 , ⋯ , X n X_1,\cdots,X_n X1,⋯,Xn yes i.i.d. Of N ( μ , 16 ) N(\mu,16) N(μ,16) A random sample , μ \mu μ The prior distribution of is N ( a , b 2 ) N(a,b^2) N(a,b2), Find the posterior distribution .
11、 ... and 、(10 branch ) set up X 1 , ⋯ , X n X_1,\cdots,X_n X1,⋯,Xn yes i.i.d. Of U ( 0 , θ ) U(0,\theta) U(0,θ) A random sample , Consider the hypothesis test problem
H 0 : θ ≤ 1 v s H 1 : θ > 1 H_0:\theta \le 1 \quad \mathrm{vs} \quad H_1: \theta >1 H0:θ≤1vsH1:θ>1 Construct reject fields W = { X ( n ) ≥ c } W=\{X_{(n)}\ge c \} W={ X(n)≥c}. Answer the following questions :
(1)(5 branch ) α = 0.05 \alpha = 0.05 α=0.05, seek c c c;
(2)(5 branch ) When θ = 1.5 \theta=1.5 θ=1.5, In order to make the probability of making the second kind of mistake β ≤ 0.1 \beta\le 0.1 β≤0.1, Find the minimum sample size .
The analysis part
One 、(20 branch ) There is... In the bag : a a a Red ball , a a a yellow ball , b b b blue ball . Yes, put it back and touch it 3 A ball , set up A = { A=\{ A={ Draw out a yellow ball and a red ball , And the red ball is taken out before the yellow ball } \} }. seek :
(1)(10 branch ) P ( A ) P(A) P(A);
(2)(10 branch ) if { \{ { I didn't touch the basketball } \} } And A A A The probability is the same , seek a : b a:b a:b.
Solution:
[ notes ]: The stem of the question can be understood as A 1 = { A_1=\{ A1={ A red ball is taken out before a yellow one } \} }, It can also be interpreted as A 2 = { A_2=\{ A2={ All red balls are taken out before yellow balls } \} }.
(1) Think about it first A 1 A_1 A1, Yes
A 1 = { Red red yellow , Red, yellow, red , Red, yellow, yellow , Blue red yellow , Red blue yellow , Red, yellow, blue } , A_1=\left\{ \text{ Red red yellow }, \text{ Red, yellow, red }, \text{ Red, yellow, yellow }, \text{ Blue red yellow }, \text{ Red blue yellow }, \text{ Red, yellow, blue } \right\} , A1={ Red red yellow , Red, yellow, red , Red, yellow, yellow , Blue red yellow , Red blue yellow , Red, yellow, blue }, There are
P ( A 1 ) = a 3 + a 3 + a 3 + 3 a 2 b ( 2 a + b ) 3 = 3 a 2 ( a + b ) ( 2 a + b ) 3 . P\left( A_1 \right) =\frac{a^3+a^3+a^3+3a^2b}{\left( 2a+b \right) ^3}=\frac{3a^2\left( a+b \right)}{\left( 2a+b \right) ^3}. P(A1)=(2a+b)3a3+a3+a3+3a2b=(2a+b)33a2(a+b). Think again A 2 A_2 A2, Yes
A 2 = { Red red yellow , Red, yellow, yellow , Blue red yellow , Red blue yellow , Red, yellow, blue } , A_2=\left\{ \text{ Red red yellow }, \text{ Red, yellow, yellow }, \text{ Blue red yellow }, \text{ Red blue yellow }, \text{ Red, yellow, blue } \right\} , A2={ Red red yellow , Red, yellow, yellow , Blue red yellow , Red blue yellow , Red, yellow, blue }, There are
P ( A 2 ) = a 3 + a 3 + 3 a 2 b ( 2 a + b ) 3 = a 2 ( 2 a + 3 b ) ( 2 a + b ) 3 . P\left( A_2 \right) =\frac{a^3+a^3+3a^2b}{\left( 2a+b \right) ^3}=\frac{a^2\left( 2a+3b \right)}{\left( 2a+b \right) ^3}. P(A2)=(2a+b)3a3+a3+3a2b=(2a+b)3a2(2a+3b). (2) To calculate B = { B=\{ B={ I didn't touch the basketball } \} }, Yes
P ( B ) = ( 2 a 2 a + b ) 3 = 8 a 3 ( 2 a + b ) 3 , P\left( B \right) =\left( \frac{2a}{2a+b} \right) ^3=\frac{8a^3}{\left( 2a+b \right) ^3}, P(B)=(2a+b2a)3=(2a+b)38a3, if P ( B ) = P ( A 1 ) P(B)=P(A_1) P(B)=P(A1), namely
8 a 3 = 3 a 2 ( a + b ) * 8 a = 3 ( a + b ) * a b = 3 5 . 8a^3=3a^2\left( a+b \right) \,\,\Longrightarrow \,\,8a=3\left( a+b \right) \,\,\Longrightarrow \frac{a}{b}=\frac{3}{5}. 8a3=3a2(a+b)*8a=3(a+b)*ba=53. if P ( B ) = P ( A 2 ) P(B)=P(A_2) P(B)=P(A2), namely
8 a 3 = a 2 ( 2 a + 3 b ) * 8 a = 2 a + 3 b * a b = 1 2 . 8a^3=a^2\left( 2a+3b \right) \,\,\Longrightarrow \,\,8a=2a+3b\,\,\Longrightarrow \frac{a}{b}=\frac{1}{2}. 8a3=a2(2a+3b)*8a=2a+3b*ba=21.
Two 、(10 branch ) Discrete random variables X X X The distribution column of is
P ( X = a ) = P ( X = b ) = P ( X = a + 1 ) = 1 3 , P(X=a)=P(X=b)=P(X=a+1)=\frac{1}{3}, P(X=a)=P(X=b)=P(X=a+1)=31, among a < b < a + 1 a<b<a+1 a<b<a+1. Find the value range of its variance .
Solution:
because V a r ( X ) = V a r ( X − a ) Var(X)=Var(X-a) Var(X)=Var(X−a), So suppose X X X The value is
P ( X = 0 ) = P ( X = c ) = P ( X = 1 ) = 1 3 , P(X=0)=P(X=c)=P(X=1)=\frac{1}{3}, P(X=0)=P(X=c)=P(X=1)=31, among c = b − a ∈ ( 0 , 1 ) c=b-a\in (0,1) c=b−a∈(0,1), There are E X = c + 1 3 EX=\frac{c+1}{3} EX=3c+1, and
E X 2 = c 2 + 1 3 , V a r ( X ) = 3 c 2 + 3 − ( c + 1 ) 2 9 = 2 9 [ ( c − 1 2 ) 2 + 3 4 ] ∈ [ 1 6 , 2 9 ) . EX^2 = \frac{c^2+1}{3},\quad Var(X)=\frac{3c^2+3-(c+1)^2}{9}=\frac{2}{9}\left[ \left( c-\frac{1}{2} \right) ^2+\frac{3}{4} \right] \in \left[ \frac{1}{6},\frac{2}{9} \right) . EX2=3c2+1,Var(X)=93c2+3−(c+1)2=92[(c−21)2+43]∈[61,92).
3、 ... and 、(20 branch ) Discrete random variables X X X Take only x , x + a x,x+a x,x+a Two values , among a > 0 a>0 a>0, And V a r ( X ) = 1 Var(X)=1 Var(X)=1, seek a a a Value range and X X X The distribution column of .
Solution:
The question only gives the condition of variance , and V a r ( X ) = V a r ( X − x ) Var(X)=Var(X-x) Var(X)=Var(X−x), So let's assume X X X Take only 0 , a 0,a 0,a Two values , And
V a r ( X ) = a 2 p − a 2 p 2 = a 2 p ( 1 − p ) = 1 , Var\left( X \right) =a^2p-a^2p^2=a^2p\left( 1-p \right) =1, Var(X)=a2p−a2p2=a2p(1−p)=1, among p = P ( X = a ) p=P(X=a) p=P(X=a), because p ( 1 − p ) ≤ 1 4 p(1-p)\le \frac{1}{4} p(1−p)≤41, therefore a ≥ 2 a\ge 2 a≥2. And when a a a Timing , p p p It can be solved , namely
p 2 − p + 1 a = 0 * p = 1 ± 1 − 4 a 2 . p^2-p+\frac{1}{a}=0 \Longrightarrow p=\frac{1\pm \sqrt{1-\frac{4}{a}}}{2}. p2−p+a1=0*p=21±1−a4. therefore X X X The distribution column of is
P ( X = x ) = 1 + 1 − 4 a 2 , P ( X = x + a ) = 1 − 1 − 4 a 2 , P\left( X=x \right) =\frac{1+\sqrt{1-\frac{4}{a}}}{2},\quad P\left( X=x+a \right) =\frac{1-\sqrt{1-\frac{4}{a}}}{2}, P(X=x)=21+1−a4,P(X=x+a)=21−1−a4, Or is it
P ( X = x ) = 1 − 1 − 4 a 2 , P ( X = x + a ) = 1 + 1 − 4 a 2 . P\left( X=x \right) =\frac{1-\sqrt{1-\frac{4}{a}}}{2},\quad P\left( X=x+a \right) =\frac{1+\sqrt{1-\frac{4}{a}}}{2}. P(X=x)=21−1−a4,P(X=x+a)=21+1−a4.
Four 、(20 branch ) With random vectors ( X Y ) \left( \begin{array}{c} X\\ Y\\ \end{array} \right) (XY), It is known that after arbitrary rotation transformation ( cos α sin α − sin α cos α ) ( X Y ) \left( \begin{matrix} \cos \alpha& \sin \alpha\\ -\sin \alpha& \cos \alpha\\ \end{matrix} \right) \left( \begin{array}{c} X\\ Y\\ \end{array} \right) (cosα−sinαsinαcosα)(XY) Still with ( X Y ) \left( \begin{array}{c} X\\ Y\\ \end{array} \right) (XY) Homodistribution , Try to solve the following problems :
(1) seek P ( 0 < Y < X ) P(0<Y<X) P(0<Y<X);
(2) seek Y X \frac{Y}{X} XY The distribution of .
Solution:
(1) set up X , Y X,Y X,Y The density function of is f X , Y ( x , y ) f_{X,Y}(x,y) fX,Y(x,y), For rotation transformation ( U , V ) = ( X , Y ) A T (U,V)=(X,Y)A^T (U,V)=(X,Y)AT, By the method of variable transformation , Yes
f U , V ( u , v ) = f X , Y ( ( u , v ) A ) ∣ A ∣ = f X , Y ( ( u , v ) A ) , f_{U,V}\left( u,v \right) =f_{X,Y}\left( \left( u,v \right) A \right) \left| A \right|=f_{X,Y}\left( \left( u,v \right) A \right) , fU,V(u,v)=fX,Y((u,v)A)∣A∣=fX,Y((u,v)A), On the other hand , because U , V U,V U,V And X , Y X,Y X,Y Homodistribution , so f U , V ( u , v ) = f X , Y ( u , v ) f_{U,V}\left( u,v \right) =f_{X,Y}\left( u,v \right) fU,V(u,v)=fX,Y(u,v), in summary , Yes
f X , Y ( x , y ) = f X , Y ( ( x , y ) A ) , f_{X,Y}\left( x,y \right) =f_{X,Y}\left( \left( x,y \right) A \right) , fX,Y(x,y)=fX,Y((x,y)A), Transform for any rotation A A A establish , This indicates that there is a function u u u bring f X , Y ( x , y ) = u ( x 2 + y 2 ) f_{X,Y}(x,y)=u(\sqrt{x^2+y^2}) fX,Y(x,y)=u(x2+y2). Make polar coordinate transformation { X = R cos Θ , Y = R sin Θ , \begin{cases} X=R\cos \Theta ,\\ Y=R\sin \Theta ,\\ \end{cases} { X=RcosΘ,Y=RsinΘ, By the method of variable transformation , Yes ( R , Θ ) (R,\Theta) (R,Θ) The distribution of is
f R , Θ ( r , θ ) = f X , Y ( r cos θ , r sin θ ) r = r u ( r ) , r ∈ ( 0 , + ∞ ) , θ ∈ ( 0 , 2 π ) , f_{R,\Theta}\left( r,\theta \right) =f_{X,Y}\left( r\cos \theta ,r\sin \theta \right) r=ru\left( r \right) ,\quad r\in \left( 0,+\infty \right) ,\theta \in \left( 0,2\pi \right) , fR,Θ(r,θ)=fX,Y(rcosθ,rsinθ)r=ru(r),r∈(0,+∞),θ∈(0,2π), Factorization , therefore R , Θ R,\Theta R,Θ Independent , And f Θ ( θ ) f_{\Theta}(\theta) fΘ(θ) Is constant , so Θ ∼ U ( 0 , 2 π ) \Theta \sim U(0,2\pi) Θ∼U(0,2π). therefore P ( 0 < Y < X ) = P ( Θ ∈ ( 0 , π 4 ) ) = 1 8 . P\left( 0<Y<X \right) =P\left( \Theta \in \left( 0,\frac{\pi}{4} \right) \right) =\frac{1}{8}. P(0<Y<X)=P(Θ∈(0,4π))=81. (2) T = Y X = tan Θ ∼ c h ( 0 , 1 ) T=\frac{Y}{X}=\tan \Theta \sim \mathrm{ch}\left( 0,1 \right) T=XY=tanΘ∼ch(0,1), Standard Cauchy distribution , The distribution function method can be used to explain : Pay attention here Θ \Theta Θ The value is ( 0 , 2 π ) (0,2\pi) (0,2π), Not on the inverse function arctan \arctan arctan The domain of definition , Discuss carefully . To any t > 0 t>0 t>0, Yes
{ T ≤ t } = { tan Θ ≤ t } = { Θ ∈ [ 0 , a r c tan t ] ∪ ( π 2 , a r c tan t + π ] ∪ ( 3 π 2 , 2 π ] } , \begin{aligned} \left\{ T\le t \right\} &=\left\{ \tan \Theta \le t \right\}\\ &=\left\{ \Theta \in \left[ 0,\mathrm{arc}\tan t \right] \cup \left( \frac{\pi}{2},\mathrm{arc}\tan t+\pi \right] \cup \left( \frac{3\pi}{2},2\pi \right] \right\}\\ \end{aligned}, { T≤t}={ tanΘ≤t}={ Θ∈[0,arctant]∪(2π,arctant+π]∪(23π,2π]}, so P ( T ≤ t ) = π + 2 arctan t 2 π = 1 2 + arctan t π P(T\le t)= \frac{\pi + 2\arctan t}{2\pi}=\frac{1}{2} + \frac{\arctan t}{\pi} P(T≤t)=2ππ+2arctant=21+πarctant, t > 0 t>0 t>0. Discuss again about any t < 0 t<0 t<0, Yes
{ T ≤ t } = { tan Θ ≤ t } = { Θ ∈ ( π 2 , a r c tan t + π ] ∪ ( 3 π 2 , a r c tan t + 2 π ) } \begin{aligned} \left\{ T\le t \right\} &=\left\{ \tan \Theta \le t \right\}\\ &=\left\{ \Theta \in \left( \frac{\pi}{2},\mathrm{arc}\tan t+\pi \right] \cup \left( \frac{3\pi}{2},\mathrm{arc}\tan t+2\pi \right) \right\}\\ \end{aligned} { T≤t}={ tanΘ≤t}={ Θ∈(2π,arctant+π]∪(23π,arctant+2π)} so P ( T ≤ t ) = π + 2 arctan t 2 π = 1 2 + arctan t π P(T\le t)= \frac{\pi + 2\arctan t}{2\pi}=\frac{1}{2} + \frac{\arctan t}{\pi} P(T≤t)=2ππ+2arctant=21+πarctant, t < 0 t<0 t<0. To sum up, there are
F T ( t ) = 1 2 + a r c tan t π , t ∈ R , F_T\left( t \right) =\frac{1}{2}+\frac{\mathrm{arc}\tan t}{\pi},\quad t\in R, FT(t)=21+πarctant,t∈R, To derive
f T ( t ) = 1 π ( 1 + t 2 ) , t ∈ R , f_T\left( t \right) =\frac{1}{\pi \left( 1+t^2 \right)},\quad t\in R, fT(t)=π(1+t2)1,t∈R, This is the standard Cauchy distribution .
5、 ... and 、(20 branch ) It is known that ( X , Y ) ∼ N ( 0 , 0 ; 1 , 1 ; 1 2 ) (X,Y)\sim N(0,0;1,1;\frac{1}{2}) (X,Y)∼N(0,0;1,1;21), seek P ( X > 0 , Y > 0 ) P(X>0,Y>0) P(X>0,Y>0).
Solution:
Make W = 2 3 ( Y − 1 2 X ) W=\frac{2}{\sqrt{3}}(Y-\frac{1}{2}X) W=32(Y−21X), Then there are E W = 0 EW=0 EW=0, V a r ( W ) = 1 Var(W)=1 Var(W)=1, And C o v ( X , W ) = 0 Cov(X,W)=0 Cov(X,W)=0, There are X , W X,W X,W The independent identity obeys the standard normal distribution . Further consider
P ( X > 0 , Y > 0 ) = P ( − X > 0 , − Y > 0 ) = P ( X < 0 , Y < 0 ) , P\left( X>0,Y>0 \right) =P\left( -X>0,-Y>0 \right) =P\left( X<0,Y<0 \right) , P(X>0,Y>0)=P(−X>0,−Y>0)=P(X<0,Y<0), Find out
P ( X > 0 , Y > 0 ) = 1 2 P ( X Y > 0 ) = 1 2 P ( Y X > 0 ) , P\left( X>0,Y>0 \right) =\frac{1}{2}P\left( XY>0 \right) =\frac{1}{2}P\left( \frac{Y}{X}>0 \right) , P(X>0,Y>0)=21P(XY>0)=21P(XY>0), recycling W W W To deal with , namely
{ Y X > 0 } = { 3 W 2 + 1 2 X X > 0 } = { W X > − 1 3 } . \left\{ \frac{Y}{X}>0 \right\} =\left\{ \frac{\frac{\sqrt{3}W}{2}+\frac{1}{2}X}{X}>0 \right\} =\left\{ \frac{W}{X}>-\frac{1}{\sqrt{3}}\right\} . { XY>0}={ X23W+21X>0}={ XW>−31}. utilize W X \frac{W}{X} XW Obey the standard Cauchy distribution , Yes P ( X > 0 , Y > 0 ) = 1 2 ∫ − 3 3 + ∞ 1 π ( 1 + t 2 ) d t = 1 3 . P\left( X>0,Y>0 \right) =\frac{1}{2}\int_{-\frac{\sqrt{3}}{3}}^{+\infty}{\frac{1}{\pi \left( 1+t^2 \right)}dt}=\frac{1}{3}. P(X>0,Y>0)=21∫−33+∞π(1+t2)1dt=31.
6、 ... and 、(10 branch ) X 1 , ⋯ , X n , ⋯ X_1,\cdots,X_n,\cdots X1,⋯,Xn,⋯ yes i.i.d. There are random variables in the second moment of , Y n = ∑ i = 1 n X i Y_n = \sum_{i=1}^n X_i Yn=∑i=1nXi, ask : { Y n n 2 } \{\frac{Y_n}{n^2}\} { n2Yn} Obey the law of large numbers .
Solution:
[ Law 1 ]: Make Z n = Y n n 2 Z_n = \frac{Y_n}{n^2} Zn=n2Yn, Calculate covariance directly , First of all, there is
C o v ( Y k , Y k + l ) = C o v ( ∑ j = 1 k X j , ∑ j = 1 k + l X j ) = k V a r ( X 1 ) , Cov\left( Y_k,Y_{k+l} \right) =Cov\left( \sum_{j=1}^k{X_j},\sum_{j=1}^{k+l}{X_j} \right) =kVar\left( X_1 \right) , Cov(Yk,Yk+l)=Cov(j=1∑kXj,j=1∑k+lXj)=kVar(X1), Further, there are , When l → ∞ l\rightarrow \infty l→∞, Yes
C o v ( Z k , Z k + l ) = 1 k 2 ( k + l ) 2 C o v ( Y k , Y l ) = V a r ( X 1 ) k ( k + l ) 2 → 0 , Cov\left( Z_k,Z_{k+l} \right) =\frac{1}{k^2\left( k+l \right) ^2}Cov\left( Y_k,Y_l \right) =\frac{Var\left( X_1 \right)}{k\left( k+l \right) ^2}\rightarrow 0, Cov(Zk,Zk+l)=k2(k+l)21Cov(Yk,Yl)=k(k+l)2Var(X1)→0, By Bernstein condition , { Y n n 2 } \{\frac{Y_n}{n^2}\} { n2Yn} Obey the law of large numbers .
[ Law two ]: By strong law of numbers , Yes Z n = 1 n ⋅ Y n n → 0 ⋅ E X 1 = 0 Z_n=\frac{1}{n}\cdot \frac{Y_n}{n}\rightarrow 0\cdot EX_1=0 Zn=n1⋅nYn→0⋅EX1=0, a.s., from stolz Theorem , Yes lim n → ∞ ∑ k = 1 n Z k n = lim n → ∞ Z n = 0 , a.s. \underset{n\rightarrow \infty}{\lim}\frac{\sum_{k=1}^n{Z_k}}{n}=\underset{n\rightarrow \infty}{\lim}Z_n=0, \text{a.s.} n→∞limn∑k=1nZk=n→∞limZn=0,a.s.
7、 ... and 、(10 branch ) X 1 , ⋯ , X n X_1,\cdots,X_n X1,⋯,Xn yes i.i.d. obey N ( μ , σ 2 ) N(\mu,\sigma^2) N(μ,σ2) Random variable of , F F F Is its distribution function , seek − 2 ∑ i = 1 n ln F ( X i ) -2\sum_{i=1}^n \ln F(X_i) −2∑i=1nlnF(Xi) The distribution of .
Solution:
First of all Y i = F ( X i ) ∼ U ( 0 , 1 ) Y_i = F(X_i)\sim U(0,1) Yi=F(Xi)∼U(0,1), Just calculate Z 1 = − 2 Y 1 Z_1=-2Y_1 Z1=−2Y1 The distribution of , By the distribution function method , Yes z > 0 z>0 z>0, Yes
P ( Z 1 ≤ z ) = P ( − 2 ln Y 1 ≤ z ) = P ( Y 1 ≥ e − 2 z ) = 1 − e − 2 z , P\left( Z_1\le z \right) =P\left( -2\ln Y_1\le z \right) =P\left( Y_1\ge e^{-2z} \right) =1-e^{-2z}, P(Z1≤z)=P(−2lnY1≤z)=P(Y1≥e−2z)=1−e−2z, This is an average of 1 / 2 1/2 1/2 The exponential distribution of , It's also χ 2 ( 2 ) \chi^2(2) χ2(2) Distribution , By additivity , have to
− 2 ∑ i = 1 n ln F ( X i ) ∼ χ 2 ( 2 n ) . -2\sum_{i=1}^n \ln F(X_i)\sim \chi^2(2n). −2i=1∑nlnF(Xi)∼χ2(2n).
8、 ... and 、(10 branch ) X 1 , ⋯ , X 6 X_1,\cdots,X_6 X1,⋯,X6 yes i.i.d. Of U ( 0 , 1 ) U(0,1) U(0,1) A random variable , seek V a r ( 2 X ( 2 ) + 3 X ( 3 ) ) Var(2X_{(2)}+3X_{(3)}) Var(2X(2)+3X(3)).
Solution:
Direct calculation , Yes
V a r ( 2 X ( 2 ) + 3 X ( 3 ) ) = 4 V a r ( X ( 2 ) ) + 9 V a r ( X ( 3 ) ) + 12 C o v ( X ( 2 ) , X ( 3 ) ) , Var\left( 2X_{\left( 2 \right)}+3X_{\left( 3 \right)} \right) =4Var\left( X_{\left( 2 \right)} \right) +9Var\left( X_{\left( 3 \right)} \right) +12Cov\left( X_{\left( 2 \right)},X_{\left( 3 \right)} \right) , Var(2X(2)+3X(3))=4Var(X(2))+9Var(X(3))+12Cov(X(2),X(3)), And marginal distribution X ( 2 ) ∼ B e t a ( 2 , 5 ) X_{(2)}\sim Beta(2,5) X(2)∼Beta(2,5), X ( 3 ) ∼ B e t a ( 3 , 4 ) X_{(3)}\sim Beta(3,4) X(3)∼Beta(3,4), Therefore, the two variance terms can be calculated directly , namely
4 V a r ( X ( 2 ) ) = 4 ⋅ 10 7 2 ⋅ 8 = 10 98 , 9 V a r ( X ( 3 ) ) = 9 ⋅ 12 7 2 ⋅ 8 = 27 98 , 4Var\left( X_{\left( 2 \right)} \right) =\frac{4\cdot 10}{7^2\cdot 8}=\frac{10}{98},\quad 9Var\left( X_{\left( 3 \right)} \right) =\frac{9\cdot 12}{7^2\cdot 8}=\frac{27}{98}, 4Var(X(2))=72⋅84⋅10=9810,9Var(X(3))=72⋅89⋅12=9827, Covariance term can be written in formula i ( n + 1 − j ) ( n + 1 ) 2 ( n + 2 ) \frac{i(n+1-j)}{(n+1)^2(n+2)} (n+1)2(n+2)i(n+1−j)(2019 Fudan Yingtong seventh question ), You can also write the joint density first
g 2 , 3 ( x , y ) = 6 ! 1 ! 1 ! 1 ! 3 ! x ⋅ ( 1 − y ) 3 = 6 ! 1 ! 3 ! x ( 1 − y ) 3 , 0 < x < y < 1 , g_{2,3}\left( x,y \right) =\frac{6!}{1!1!1!3!}x\cdot \left( 1-y \right) ^3=\frac{6!}{1!3!}x\left( 1-y \right) ^3,\quad 0<x<y<1, g2,3(x,y)=1!1!1!3!6!x⋅(1−y)3=1!3!6!x(1−y)3,0<x<y<1, Calculate the mixing moment , namely
E ( X ( 2 ) X ( 3 ) ) = 6 ! 3 ! ∫ 0 1 ∫ 0 y x 2 y ( 1 − y ) 3 d x d y = 6 ! 3 ! 3 ∫ 0 1 y 4 ( 1 − y ) 3 d y = 6 ! 3 ! 4 ! 3 ! 8 ! 3 = 8 56 . \begin{aligned} E\left( X_{\left( 2 \right)}X_{\left( 3 \right)} \right) &=\frac{6!}{3!}\int_0^1{\int_0^y{x^2y\left( 1-y \right) ^3dx}dy}\\ &=\frac{6!}{3!3}\int_0^1{y^4\left( 1-y \right) ^3dy}\\ &=\frac{6!3!4!}{3!8!3}=\frac{8}{56}.\\ \end{aligned} E(X(2)X(3))=3!6!∫01∫0yx2y(1−y)3dxdy=3!36!∫01y4(1−y)3dy=3!8!36!3!4!=568. So the covariance is C o v ( X ( 2 ) , X ( 3 ) ) = 8 56 − 6 49 = 2 98 . Cov\left( X_{\left( 2 \right)},X_{\left( 3 \right)} \right) =\frac{8}{56}-\frac{6}{49}=\frac{2}{98}. Cov(X(2),X(3))=568−496=982. Sum up all the calculation results to
V a r ( 2 X ( 2 ) + 3 X ( 3 ) ) = 61 98 . Var\left( 2X_{\left( 2 \right)}+3X_{\left( 3 \right)} \right) =\frac{61}{98}. Var(2X(2)+3X(3))=9861.
Nine 、(10 branch ) X 1 , ⋯ , X n X_1,\cdots,X_n X1,⋯,Xn yes i.i.d. Of U ( 0 , θ ) U(0,\theta) U(0,θ) A random sample , set up a X ( 1 ) , b X ( 3 ) aX_{(1)},bX_{(3)} aX(1),bX(3) yes θ \theta θ Unbiased estimation of , seek a , b a,b a,b And compare which of them is more effective .
Solution:
because X ( 1 ) θ ∼ B e t a ( 1 , 3 ) \frac{X_{(1)}}{\theta}\sim Beta(1,3) θX(1)∼Beta(1,3), so E X ( 1 ) = 1 4 θ EX_{(1)}=\frac{1}{4}\theta EX(1)=41θ, V a r ( X ( 1 ) ) = 3 80 θ 2 Var(X_{(1)})=\frac{3}{80}\theta^2 Var(X(1))=803θ2, so a = 4 a=4 a=4, And V a r ( a X ( 1 ) ) = 3 5 θ 2 Var(aX_{(1)})=\frac{3}{5}\theta^2 Var(aX(1))=53θ2. Empathy X ( 3 ) ∼ B e t a ( 3 , 1 ) X_{(3)}\sim Beta(3,1) X(3)∼Beta(3,1), so E X ( 3 ) = 3 4 θ EX_{(3)}=\frac{3}{4}\theta EX(3)=43θ, V a r ( X ( 3 ) ) = 3 80 θ 2 Var(X_{(3)})=\frac{3}{80}\theta^2 Var(X(3))=803θ2, so b = 4 3 b=\frac{4}{3} b=34, And V a r ( b X ( 3 ) ) = 1 60 θ 2 Var(bX_{(3)})=\frac{1}{60}\theta^2 Var(bX(3))=601θ2. It can be seen that b X ( 3 ) bX_{(3)} bX(3) More effective .
Ten 、(10 branch ) set up X 1 , ⋯ , X n X_1,\cdots,X_n X1,⋯,Xn yes i.i.d. Of N ( μ , 16 ) N(\mu,16) N(μ,16) A random sample , μ \mu μ The prior distribution of is N ( a , b 2 ) N(a,b^2) N(a,b2), Find the posterior distribution .
Solution:
Consider sufficient statistics X ˉ ∣ μ ∼ N ( μ , 16 n ) \bar{X}|\mu \sim N(\mu,\frac{16}{n}) Xˉ∣μ∼N(μ,n16), The joint density is
p ( x , μ ) = p ( x ∣ μ ) π ( π ) = C ⋅ e − ( x − μ ) 2 2 ⋅ 16 n ⋅ e − ( μ − a ) 2 2 b 2 = C ⋅ e − ( x − μ ) 2 2 ⋅ 16 n ⋅ e − ( μ − a ) 2 2 b 2 = C e − b 2 ( x − μ ) 2 + 16 n ( μ − a ) 2 2 ⋅ 16 n b 2 = C e − b 2 ( x − μ ) 2 + 16 n ( μ − a ) 2 2 ⋅ 16 n b 2 = C e − ( 16 n + b 2 ) μ 2 − 2 ( b 2 x + 16 n a ) μ + ( b 2 x 2 + 16 n a 2 ) 2 ⋅ 16 n b 2 = C 1 ( x ) e − ( μ − b 2 x + 16 n a b 2 + 16 n ) 2 2 ⋅ 16 n b 2 16 n + b 2 , \begin{aligned} p\left( x,\mu \right) &=p\left( x|\mu \right) \pi \left( \pi \right) =C\cdot e^{-\frac{\left( x-\mu \right) ^2}{2\cdot \frac{16}{n}}}\cdot e^{-\frac{\left( \mu -a \right) ^2}{2b^2}}\\ &=C\cdot e^{-\frac{\left( x-\mu \right) ^2}{2\cdot \frac{16}{n}}}\cdot e^{-\frac{\left( \mu -a \right) ^2}{2b^2}}\\ &=Ce^{-\frac{b^2\left( x-\mu \right) ^2+\frac{16}{n}\left( \mu -a \right) ^2}{2\cdot \frac{16}{n}b^2}}\\ &=Ce^{-\frac{b^2\left( x-\mu \right) ^2+\frac{16}{n}\left( \mu -a \right) ^2}{2\cdot \frac{16}{n}b^2}}\\ &=Ce^{-\frac{\left( \frac{16}{n}+b^2 \right) \mu ^2-2\left( b^2x+\frac{16}{n}a \right) \mu +\left( b^2x^2+\frac{16}{n}a^2 \right)}{2\cdot \frac{16}{n}b^2}}\\ &=C_1(x)e^{-\frac{\left( \mu -\frac{b^2x+\frac{16}{n}a}{b^2+\frac{16}{n}} \right) ^2}{2\cdot \frac{\frac{16}{n}b^2}{\frac{16}{n}+b^2}}},\\ \end{aligned} p(x,μ)=p(x∣μ)π(π)=C⋅e−2⋅n16(x−μ)2⋅e−2b2(μ−a)2=C⋅e−2⋅n16(x−μ)2⋅e−2b2(μ−a)2=Ce−2⋅n16b2b2(x−μ)2+n16(μ−a)2=Ce−2⋅n16b2b2(x−μ)2+n16(μ−a)2=Ce−2⋅n16b2(n16+b2)μ2−2(b2x+n16a)μ+(b2x2+n16a2)=C1(x)e−2⋅n16+b2n16b2(μ−b2+n16b2x+n16a)2, It is found that there is a normally distributed kernel , So the posterior distribution is
μ ∣ X ˉ ∼ N ( b 2 X ˉ + 16 n a b 2 + 16 n , 16 n b 2 16 n + b 2 ) = N ( n 16 X ˉ + 1 b 2 a n 16 + 1 b 2 , 1 n 16 + 1 b 2 ) . \mu |\bar{X}\sim N\left( \frac{b^2\bar{X}+\frac{16}{n}a}{b^2+\frac{16}{n}},\frac{\frac{16}{n}b^2}{\frac{16}{n}+b^2} \right) =N\left( \frac{\frac{n}{16}\bar{X}+\frac{1}{b^2}a}{\frac{n}{16}+\frac{1}{b^2}},\frac{1}{\frac{n}{16}+\frac{1}{b^2}} \right) . μ∣Xˉ∼N(b2+n16b2Xˉ+n16a,n16+b2n16b2)=N(16n+b2116nXˉ+b21a,16n+b211). It can be seen that the posterior mean is the weighted average of sample information and prior information .
11、 ... and 、(10 branch ) set up X 1 , ⋯ , X n X_1,\cdots,X_n X1,⋯,Xn yes i.i.d. Of U ( 0 , θ ) U(0,\theta) U(0,θ) A random sample , Consider the hypothesis test problem
H 0 : θ ≤ 1 v s H 1 : θ > 1 H_0:\theta \le 1 \quad \mathrm{vs} \quad H_1: \theta >1 H0:θ≤1vsH1:θ>1 Construct reject fields W = { X ( n ) ≥ c } W=\{X_{(n)}\ge c \} W={ X(n)≥c}. Answer the following questions :
(1)(5 branch ) α = 0.05 \alpha = 0.05 α=0.05, seek c c c;
(2)(5 branch ) When θ = 1.5 \theta=1.5 θ=1.5, In order to make the probability of making the second kind of mistake β ≤ 0.1 \beta\le 0.1 β≤0.1, Find the minimum sample size .
Solution:
(1) In order to make the significance level 0.05 0.05 0.05, Yes
0.05 = s u p θ ≤ 1 P θ ( X ( n ) ≥ c ) = P θ = 1 ( X ( n ) ≥ c ) = ( 1 − c ) n , 0.05 = \underset{\theta \le 1}{\mathrm{sup}}P_{\theta}\left( X_{\left( n \right)}\ge c \right) =P_{\theta =1}\left( X_{\left( n \right)}\ge c \right) =\left( 1-c \right) ^n, 0.05=θ≤1supPθ(X(n)≥c)=Pθ=1(X(n)≥c)=(1−c)n,
Solution c = 0.9 5 1 n c=0.95^{\frac{1}{n}} c=0.95n1.
(2) The probability of making the second kind of mistake is
β ( 1.5 ) = P θ = 1.5 ( X ( n ) < 0.9 5 1 n ) = ( 0.9 5 1 n 1.5 ) n = 0.95 1. 5 n , \beta \left( 1.5 \right) =P_{\theta =1.5}\left( X_{\left( n \right)}<0.95^{\frac{1}{n}} \right) =\left( \frac{0.95^{\frac{1}{n}}}{1.5} \right) ^n=\frac{0.95}{1.5^n}, β(1.5)=Pθ=1.5(X(n)<0.95n1)=(1.50.95n1)n=1.5n0.95, Make it less than or equal to 0.1 0.1 0.1, have to
0.95 1. 5 n ≤ 0.1 * n ≥ ln 9.5 ln 1.5 * n ≥ 6. \frac{0.95}{1.5^n}\le 0.1 \Longrightarrow \,\,n\ge \frac{\ln 9.5}{\ln 1.5}\,\,\Longrightarrow \,\,n\ge 6. 1.5n0.95≤0.1*n≥ln1.5ln9.5*n≥6.
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