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【九阳神功】2022复旦大学应用统计真题+解析
2022-07-06 09:20:00 【大师兄统计】
真题部分
一、(20分) 袋子里有: a a a红球, a a a黄球, b b b蓝球. 有放回摸3个球, 设 A = { A=\{ A={ 抽出的求有黄球也有红球, 且红球比黄球先取出 } \} }. 求:
(1)(10分) P ( A ) P(A) P(A);
(2)(10分) 若 { \{ { 没有摸到蓝球 } \} }与 A A A概率相同, 求 a : b a:b a:b.
二、(10分) 离散型随机变量 X X X的分布列是
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, 其中 a < b < a + 1 a<b<a+1 a<b<a+1. 求其方差的取值范围.
三、(20分) 离散型随机变量 X X X只取 x , x + a x,x+a x,x+a两个值, 其中 a > 0 a>0 a>0, 且 V a r ( X ) = 1 Var(X)=1 Var(X)=1, 求 a a a的取值范围及 X X X的分布列.
四、(20分) 设有随机向量 ( X Y ) \left( \begin{array}{c} X\\ Y\\ \end{array} \right) (XY), 已知它经过任意旋转变换后 ( 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)仍与 ( X Y ) \left( \begin{array}{c} X\\ Y\\ \end{array} \right) (XY)同分布, 试解决下述问题:
(1) 求 P ( 0 < Y < X ) P(0<Y<X) P(0<Y<X);
(2) 求 Y X \frac{Y}{X} XY的分布.
五、(20分) 已知 ( 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), 求 P ( X > 0 , Y > 0 ) P(X>0,Y>0) P(X>0,Y>0).
六、(10分) X 1 , ⋯ , X n , ⋯ X_1,\cdots,X_n,\cdots X1,⋯,Xn,⋯是i.i.d.的二阶矩存在随机变量, Y n = ∑ i = 1 n X i Y_n = \sum_{i=1}^n X_i Yn=∑i=1nXi, 问: { Y n n 2 } \{\frac{Y_n}{n^2}\} { n2Yn}是否服从大数定律.
七、(10分) X 1 , ⋯ , X n X_1,\cdots,X_n X1,⋯,Xn是i.i.d.服从 N ( μ , σ 2 ) N(\mu,\sigma^2) N(μ,σ2)的随机变量, F F F是其分布函数, 求 − 2 ∑ i = 1 n ln F ( X i ) -2\sum_{i=1}^n \ln F(X_i) −2∑i=1nlnF(Xi)的分布.
八、(10分) X 1 , ⋯ , X 6 X_1,\cdots,X_6 X1,⋯,X6是i.i.d.的 U ( 0 , 1 ) U(0,1) U(0,1)随机变量, 求 V a r ( 2 X ( 2 ) + 3 X ( 3 ) ) Var(2X_{(2)}+3X_{(3)}) Var(2X(2)+3X(3)).
九、(10分) X 1 , ⋯ , X n X_1,\cdots,X_n X1,⋯,Xni是i.i.d.的 U ( 0 , θ ) U(0,\theta) U(0,θ)随机样本, 设 a X ( 1 ) , b X ( 3 ) aX_{(1)},bX_{(3)} aX(1),bX(3)是 θ \theta θ的无偏估计, 求 a , b a,b a,b并比较它们的何者更有效.
十、(10分) 设 X 1 , ⋯ , X n X_1,\cdots,X_n X1,⋯,Xn是i.i.d.的 N ( μ , 16 ) N(\mu,16) N(μ,16)随机样本, μ \mu μ的先验分布是 N ( a , b 2 ) N(a,b^2) N(a,b2), 求后验分布.
十一、(10分) 设 X 1 , ⋯ , X n X_1,\cdots,X_n X1,⋯,Xn是i.i.d.的 U ( 0 , θ ) U(0,\theta) U(0,θ)随机样本, 考虑假设检验问题
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构造拒绝域 W = { X ( n ) ≥ c } W=\{X_{(n)}\ge c \} W={ X(n)≥c}. 回答下述问题:
(1)(5分) α = 0.05 \alpha = 0.05 α=0.05, 求 c c c;
(2)(5分) 当 θ = 1.5 \theta=1.5 θ=1.5, 为使得犯第二类错误的概率 β ≤ 0.1 \beta\le 0.1 β≤0.1, 求至少要多少样本量.
解析部分
一、(20分) 袋子里有: a a a红球, a a a黄球, b b b蓝球. 有放回摸3个球, 设 A = { A=\{ A={ 抽出的求有黄球也有红球, 且红球比黄球先取出 } \} }. 求:
(1)(10分) P ( A ) P(A) P(A);
(2)(10分) 若 { \{ { 没有摸到蓝球 } \} }与 A A A概率相同, 求 a : b a:b a:b.
Solution:
[注]: 题干可以理解成 A 1 = { A_1=\{ A1={ 有一个红球比黄球先取出 } \} }, 也可以理解成 A 2 = { A_2=\{ A2={ 所有红球比黄球先取出 } \} }.
(1) 先考虑 A 1 A_1 A1, 有
A 1 = { 红红黄 , 红黄红 , 红黄黄 , 蓝红黄 , 红蓝黄 , 红黄蓝 } , A_1=\left\{ \text{红红黄}, \text{红黄红}, \text{红黄黄}, \text{蓝红黄}, \text{红蓝黄}, \text{红黄蓝} \right\} , A1={ 红红黄,红黄红,红黄黄,蓝红黄,红蓝黄,红黄蓝},故有
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). 再考虑 A 2 A_2 A2, 有
A 2 = { 红红黄 , 红黄黄 , 蓝红黄 , 红蓝黄 , 红黄蓝 } , A_2=\left\{ \text{红红黄}, \text{红黄黄}, \text{蓝红黄}, \text{红蓝黄}, \text{红黄蓝} \right\} , A2={ 红红黄,红黄黄,蓝红黄,红蓝黄,红黄蓝}, 故有
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) 先计算 B = { B=\{ B={ 没有摸到蓝球 } \} }, 有
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, 若 P ( B ) = P ( A 1 ) P(B)=P(A_1) P(B)=P(A1), 即
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. 若 P ( B ) = P ( A 2 ) P(B)=P(A_2) P(B)=P(A2), 即
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.
二、(10分) 离散型随机变量 X X X的分布列是
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, 其中 a < b < a + 1 a<b<a+1 a<b<a+1. 求其方差的取值范围.
Solution:
由于 V a r ( X ) = V a r ( X − a ) Var(X)=Var(X-a) Var(X)=Var(X−a), 故不妨假设 X X X的取值是
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, 其中 c = b − a ∈ ( 0 , 1 ) c=b-a\in (0,1) c=b−a∈(0,1), 故有 E X = c + 1 3 EX=\frac{c+1}{3} EX=3c+1, 而
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).
三、(20分) 离散型随机变量 X X X只取 x , x + a x,x+a x,x+a两个值, 其中 a > 0 a>0 a>0, 且 V a r ( X ) = 1 Var(X)=1 Var(X)=1, 求 a a a的取值范围及 X X X的分布列.
Solution:
题目只给了方差的条件, 而 V a r ( X ) = V a r ( X − x ) Var(X)=Var(X-x) Var(X)=Var(X−x), 故不妨先假设 X X X只取 0 , a 0,a 0,a两个值, 且
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, 其中 p = P ( X = a ) p=P(X=a) p=P(X=a), 由于 p ( 1 − p ) ≤ 1 4 p(1-p)\le \frac{1}{4} p(1−p)≤41, 因此 a ≥ 2 a\ge 2 a≥2. 且当 a a a给定时, p p p是可以解出的, 即
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. 所以 X X X的分布列是
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, 或者是
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.
四、(20分) 设有随机向量 ( X Y ) \left( \begin{array}{c} X\\ Y\\ \end{array} \right) (XY), 已知它经过任意旋转变换后 ( 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)仍与 ( X Y ) \left( \begin{array}{c} X\\ Y\\ \end{array} \right) (XY)同分布, 试解决下述问题:
(1) 求 P ( 0 < Y < X ) P(0<Y<X) P(0<Y<X);
(2) 求 Y X \frac{Y}{X} XY的分布.
Solution:
(1) 设 X , Y X,Y X,Y的密度函数是 f X , Y ( x , y ) f_{X,Y}(x,y) fX,Y(x,y), 对旋转变换 ( U , V ) = ( X , Y ) A T (U,V)=(X,Y)A^T (U,V)=(X,Y)AT, 由变量变换法, 有
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),另一方面, 由于 U , V U,V U,V与 X , Y X,Y X,Y同分布, 故 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), 综上所述, 有
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),对任意旋转变换 A A A成立, 这说明存在函数 u u u使得 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). 作极坐标变换 { X = R cos Θ , Y = R sin Θ , \begin{cases} X=R\cos \Theta ,\\ Y=R\sin \Theta ,\\ \end{cases} { X=RcosΘ,Y=RsinΘ, 由变量变换法, 有 ( R , Θ ) (R,\Theta) (R,Θ)的分布是
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), 可因式分解, 因此 R , Θ R,\Theta R,Θ独立, 且 f Θ ( θ ) f_{\Theta}(\theta) fΘ(θ)是常数, 故 Θ ∼ U ( 0 , 2 π ) \Theta \sim U(0,2\pi) Θ∼U(0,2π). 因此 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), 标准柯西分布, 可利用分布函数法说明: 这里要注意 Θ \Theta Θ取值是 ( 0 , 2 π ) (0,2\pi) (0,2π), 对不上反函数 arctan \arctan arctan的定义域, 要仔细讨论. 对任意 t > 0 t>0 t>0, 有
{ 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]}, 故 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. 再讨论对任意 t < 0 t<0 t<0, 有
{ 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)} 故 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. 综上所述有
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,求导得
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, 这是标准柯西分布.
五、(20分) 已知 ( 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), 求 P ( X > 0 , Y > 0 ) P(X>0,Y>0) P(X>0,Y>0).
Solution:
令 W = 2 3 ( Y − 1 2 X ) W=\frac{2}{\sqrt{3}}(Y-\frac{1}{2}X) W=32(Y−21X), 则有 E W = 0 EW=0 EW=0, V a r ( W ) = 1 Var(W)=1 Var(W)=1, 且 C o v ( X , W ) = 0 Cov(X,W)=0 Cov(X,W)=0, 故有 X , W X,W X,W独立同服从标准正态分布. 进一步考虑到
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), 发现
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), 再利用 W W W作处理, 即
{ 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}. 利用 W X \frac{W}{X} XW服从标准柯西分布, 有 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.
六、(10分) X 1 , ⋯ , X n , ⋯ X_1,\cdots,X_n,\cdots X1,⋯,Xn,⋯是i.i.d.的二阶矩存在随机变量, Y n = ∑ i = 1 n X i Y_n = \sum_{i=1}^n X_i Yn=∑i=1nXi, 问: { Y n n 2 } \{\frac{Y_n}{n^2}\} { n2Yn}是否服从大数定律.
Solution:
[法一]: 令 Z n = Y n n 2 Z_n = \frac{Y_n}{n^2} Zn=n2Yn, 直接计算协方差, 首先有
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), 进一步有, 当 l → ∞ l\rightarrow \infty l→∞, 有
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,由伯恩斯坦条件, { Y n n 2 } \{\frac{Y_n}{n^2}\} { n2Yn}服从大数定律.
[法二]: 由强大数律, 有 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., 由stolz定理, 有 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.
七、(10分) X 1 , ⋯ , X n X_1,\cdots,X_n X1,⋯,Xn是i.i.d.服从 N ( μ , σ 2 ) N(\mu,\sigma^2) N(μ,σ2)的随机变量, F F F是其分布函数, 求 − 2 ∑ i = 1 n ln F ( X i ) -2\sum_{i=1}^n \ln F(X_i) −2∑i=1nlnF(Xi)的分布.
Solution:
首先记 Y i = F ( X i ) ∼ U ( 0 , 1 ) Y_i = F(X_i)\sim U(0,1) Yi=F(Xi)∼U(0,1), 只需计算 Z 1 = − 2 Y 1 Z_1=-2Y_1 Z1=−2Y1的分布, 由分布函数法, 对 z > 0 z>0 z>0, 有
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, 这是均值为 1 / 2 1/2 1/2的指数分布, 也是 χ 2 ( 2 ) \chi^2(2) χ2(2)分布, 由可加性, 得
− 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).
八、(10分) X 1 , ⋯ , X 6 X_1,\cdots,X_6 X1,⋯,X6是i.i.d.的 U ( 0 , 1 ) U(0,1) U(0,1)随机变量, 求 V a r ( 2 X ( 2 ) + 3 X ( 3 ) ) Var(2X_{(2)}+3X_{(3)}) Var(2X(2)+3X(3)).
Solution:
直接计算, 有
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)), 而边际分布 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), 故两个方差项可以直接计算, 即
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, 协方差项可以记公式 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复旦应统第七题), 也可以先写出联合密度
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, 计算混合矩, 即
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. 因此协方差为 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. 将所有计算结果汇总得
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.
九、(10分) X 1 , ⋯ , X n X_1,\cdots,X_n X1,⋯,Xn是i.i.d.的 U ( 0 , θ ) U(0,\theta) U(0,θ)随机样本, 设 a X ( 1 ) , b X ( 3 ) aX_{(1)},bX_{(3)} aX(1),bX(3)是 θ \theta θ的无偏估计, 求 a , b a,b a,b并比较它们的何者更有效.
Solution:
由于 X ( 1 ) θ ∼ B e t a ( 1 , 3 ) \frac{X_{(1)}}{\theta}\sim Beta(1,3) θX(1)∼Beta(1,3), 故 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, 故 a = 4 a=4 a=4, 且 V a r ( a X ( 1 ) ) = 3 5 θ 2 Var(aX_{(1)})=\frac{3}{5}\theta^2 Var(aX(1))=53θ2. 同理 X ( 3 ) ∼ B e t a ( 3 , 1 ) X_{(3)}\sim Beta(3,1) X(3)∼Beta(3,1), 故 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, 故 b = 4 3 b=\frac{4}{3} b=34, 且 V a r ( b X ( 3 ) ) = 1 60 θ 2 Var(bX_{(3)})=\frac{1}{60}\theta^2 Var(bX(3))=601θ2. 可以看出 b X ( 3 ) bX_{(3)} bX(3)更有效.
十、(10分) 设 X 1 , ⋯ , X n X_1,\cdots,X_n X1,⋯,Xn是i.i.d.的 N ( μ , 16 ) N(\mu,16) N(μ,16)随机样本, μ \mu μ的先验分布是 N ( a , b 2 ) N(a,b^2) N(a,b2), 求后验分布.
Solution:
考虑充分统计量 X ˉ ∣ μ ∼ N ( μ , 16 n ) \bar{X}|\mu \sim N(\mu,\frac{16}{n}) Xˉ∣μ∼N(μ,n16), 有联合密度是
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, 发现有一个正态分布的核, 故后验分布是
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). 可以看出后验均值是样本信息与先验信息的加权平均.
十一、(10分) 设 X 1 , ⋯ , X n X_1,\cdots,X_n X1,⋯,Xn是i.i.d.的 U ( 0 , θ ) U(0,\theta) U(0,θ)随机样本, 考虑假设检验问题
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构造拒绝域 W = { X ( n ) ≥ c } W=\{X_{(n)}\ge c \} W={ X(n)≥c}. 回答下述问题:
(1)(5分) α = 0.05 \alpha = 0.05 α=0.05, 求 c c c;
(2)(5分) 当 θ = 1.5 \theta=1.5 θ=1.5, 为使得犯第二类错误的概率 β ≤ 0.1 \beta\le 0.1 β≤0.1, 求至少要多少样本量.
Solution:
(1) 为使显著性水平为 0.05 0.05 0.05, 有
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,
解得 c = 0.9 5 1 n c=0.95^{\frac{1}{n}} c=0.95n1.
(2) 犯第二类错误的概率是
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,令其小于等于 0.1 0.1 0.1, 得
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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