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【九阳神功】2016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、(15分) 三个人独立地同时破译密码, 且三人能破译密码的概率分别为1/5, 1/3和1/4, 求此密码能够被破译的概率.
二、(15分) 从(0,1)中随机地取两个数, 求其积不小于3/16且其和不大于1的概率.
三、(15分) X ∼ N ( 0 , 1 ) , X \sim N(0,1), X∼N(0,1), 求 Y = X 2 Y=X^{2} Y=X2 的密度函数.
四、(30分) 记(0,1),(1,0),(0,0)三点围成的区域为 D , ( X , Y ) D,(X, Y) D,(X,Y)服从 D D D上的均匀分布, 求
(1)(15分) E ( X + Y ) , Var ( X + Y ) E(X+Y), \operatorname{Var}(X+Y) E(X+Y),Var(X+Y);
(2)(15分) X , Y X, Y X,Y 的相关系数.
五、(15分) 对于两个只有两个取值的随机变量 X , Y , X, Y, X,Y, 试证明 X , Y X, Y X,Y独立当且仅当 X , Y X, Y X,Y不相关.
六、(15分) 设有来自总体 X ∼ N ( μ , σ 2 ) X \sim N\left(\mu, \sigma^{2}\right) X∼N(μ,σ2)的 2 n 2 n 2n个随机样本, 试求期望
E ( ∑ i = 1 n ( X i + X n + i − 2 X ˉ ) 2 ) . E\left(\sum_{i=1}^{n}\left(X_{i}+X_{n+i}-2 \bar{X}\right)^{2}\right). E(i=1∑n(Xi+Xn+i−2Xˉ)2).
七、(15分) X 1 , X 2 , X 3 X_{1}, X_{2}, X_{3} X1,X2,X3是i.i.d.的服从 U ( 0 , 1 ) U(0,1) U(0,1)的随机变量, 求次序统计量 X ( 1 ) X_{(1)} X(1)的分布函数与期望.
八、(30分) 有来自 X ∼ f ( x ) = { θ , 0 ≤ x < 1 1 − θ , 1 ≤ x < 2 , 0 , otherwise X \sim f(x)=\left\{\begin{array}{cl}\theta, & 0 \leq x<1 \\ 1-\theta, & 1 \leq x<2, \\ 0, & \text { otherwise }\end{array}\right. X∼f(x)=⎩⎨⎧θ,1−θ,0,0≤x<11≤x<2, otherwise 的 n n n 个简单随机样本, 求
(1)(15分) θ \theta θ 的矩估计;
(2)(15分) θ \theta θ 的最大似然估计, 已知 m = ∑ i = 1 n I [ X i < 1 ] m=\sum_{i=1}^{n} I\left[X_{i}<1\right] m=∑i=1nI[Xi<1].
解析部分
一、(15分) 三个人独立地同时破译密码, 且三人能破译密码的概率分别为1/5, 1/3和1/4, 求此密码能够被破译的概率.
Solution: 根据德摩根公式, P { 至少有一人破译 } = 1 − P { 没有一个人破译 } = 1 − 4 5 × 2 3 × 3 4 = 0.6. \begin{aligned} P\{\text { 至少有一人破译 }\} &=1-P\{\text { 没有一个人破译 }\} \\ &=1-\frac{4}{5} \times \frac{2}{3} \times \frac{3}{4}=0.6 . \end{aligned} P{ 至少有一人破译 }=1−P{ 没有一个人破译 }=1−54×32×43=0.6.
二、(15分) 从(0,1)中随机地取两个数, 求其积不小于3/16且其和不大于1的概率.
Solution: 设取到的两个数是 X , Y ∼ U ( 0 , 1 ) X, Y \sim U(0,1) X,Y∼U(0,1), 其积不小于 3 16 \frac{3}{16} 163 意味着 X Y ≥ 3 16 X Y \geq \frac{3}{16} XY≥163, 其和不大于 1 意味着 X + Y ≤ 1 X+Y \leq 1 X+Y≤1.
令 D = { ( x , y ) : x y ≥ 3 16 , x + y ≤ 1 } ∩ ( 0 , 1 ) 2 D=\left\{(x, y): x y \geq \frac{3}{16}, x+y \leq 1\right\} \cap(0,1)^{2} D={ (x,y):xy≥163,x+y≤1}∩(0,1)2, 该概率为 ∫ D f ( x , y ) d x d y = 1 8 − ∫ 1 4 3 4 ∫ 1 4 3 16 x d y d x = 1 8 − ∫ 1 4 3 4 ( 3 16 x − 1 4 ) d x = 1 4 − 3 ln 3 16 \int_{D} f(x, y) d x d y=\frac{1}{8}-\int_{\frac{1}{4}}^{\frac{3}{4}} \int_{\frac{1}{4}}^{\frac{3}{16 x}} d y d x=\frac{1}{8}-\int_{\frac{1}{4}}^{\frac{3}{4}}\left(\frac{3}{16 x}-\frac{1}{4}\right) d x=\frac{1}{4}-\frac{3 \ln 3}{16} ∫Df(x,y)dxdy=81−∫4143∫4116x3dydx=81−∫4143(16x3−41)dx=41−163ln3
三、(15分) X ∼ N ( 0 , 1 ) , X \sim N(0,1), X∼N(0,1), 求 Y = X 2 Y=X^{2} Y=X2 的密度函数.
Solution: 当 y ≥ 0 y \geq 0 y≥0 时,
F ( y ) = P { Y ≤ y } = P { X 2 ≤ > y } = P { − y ≤ X ≤ y } = 2 Φ ( y ) − 1 F(y)=P\{Y \leq y\}=P\left\{X^{2} \leq>y\right\}=P\{-\sqrt{y} \leq X \leq \sqrt{y}\}=2 \Phi(\sqrt{y})-1 F(y)=P{ Y≤y}=P{ X2≤>y}=P{ −y≤X≤y}=2Φ(y)−1, 故
f Y ( y ) = F ′ ( y ) = 2 φ ( y ) ⋅ 1 2 y = 1 2 π y e − y 2 , y ≥ 0 f_{Y}(y)=F^{\prime}(y)=2 \varphi(\sqrt{y}) \cdot \frac{1}{2 \sqrt{y}}=\frac{1}{\sqrt{2 \pi y}} e^{-\frac{y}{2}}, y \geq 0 fY(y)=F′(y)=2φ(y)⋅2y1=2πy1e−2y,y≥0
四、(30分) 记(0,1),(1,0),(0,0)三点围成的区域为 D , ( X , Y ) D,(X, Y) D,(X,Y)服从 D D D上的均匀分布, 求
(1)(15分) E ( X + Y ) , Var ( X + Y ) E(X+Y), \operatorname{Var}(X+Y) E(X+Y),Var(X+Y);
(2)(15分) X , Y X, Y X,Y 的相关系数.
Solution: (1) E ( X + Y ) = 2 ∫ 0 1 ∫ 0 1 − y ( x + y ) d x d y = ∫ 0 1 ( 1 − y 2 ) d y = 1 − 1 3 = 2 3 E(X+Y)=2 \int_{0}^{1} \int_{0}^{1-y}(x+y) d x d y=\int_{0}^{1}\left(1-y^{2}\right) d y=1-\frac{1}{3}=\frac{2}{3} E(X+Y)=2∫01∫01−y(x+y)dxdy=∫01(1−y2)dy=1−31=32, E ( X + Y ) 2 = 2 ∫ 0 1 ∫ 0 1 − y ( x + y ) 2 d x d y = 2 3 ∫ 0 1 ( 1 − y 3 ) d y = 2 3 ( 1 − 1 4 ) = 1 2 , E(X+Y)^{2}=2 \int_{0}^{1} \int_{0}^{1-y}(x+y)^{2} d x d y=\frac{2}{3} \int_{0}^{1}\left(1-y^{3}\right) d y=\frac{2}{3}\left(1-\frac{1}{4}\right)=\frac{1}{2}, E(X+Y)2=2∫01∫01−y(x+y)2dxdy=32∫01(1−y3)dy=32(1−41)=21, 因此
Var ( X + Y ) = 1 2 − 4 9 = 1 18 . \operatorname{Var}(X+Y)=\frac{1}{2}-\frac{4}{9}=\frac{1}{18}. Var(X+Y)=21−94=181.(2) E X Y = 2 ∫ 0 1 ∫ 0 1 − y x y d x d y = ∫ 0 1 ( y 3 − 2 y 2 + y ) d y = 1 4 − 2 3 + 1 2 = 1 12 E X Y=2 \int_{0}^{1} \int_{0}^{1-y} x y d x d y=\int_{0}^{1}\left(y^{3}-2 y^{2}+y\right) d y=\frac{1}{4}-\frac{2}{3}+\frac{1}{2}=\frac{1}{12} EXY=2∫01∫01−yxydxdy=∫01(y3−2y2+y)dy=41−32+21=121, E X = E Y = 2 ∫ 0 1 ∫ 0 1 − y x d x d y = ∫ 0 1 ( y 2 − 2 y + 1 ) d y = 1 3 − 1 + 1 = 1 3 , E X=E Y=2 \int_{0}^{1} \int_{0}^{1-y} x d x d y=\int_{0}^{1}\left(y^{2}-2 y+1\right) d y=\frac{1}{3}-1+1=\frac{1}{3}, EX=EY=2∫01∫01−yxdxdy=∫01(y2−2y+1)dy=31−1+1=31, 故
Cov ( X , Y ) = 1 12 − 1 9 = − 1 36 . \operatorname{Cov}(X, Y)=\frac{1}{12}-\frac{1}{9}=-\frac{1}{36}. Cov(X,Y)=121−91=−361. E X 2 = 2 ∫ 0 1 ∫ 0 1 − y x 2 d x d y = 2 3 ∫ 0 1 ( 1 − y ) 3 d y = 1 6 , E X^{2}=2 \int_{0}^{1} \int_{0}^{1-y} x^{2} d x d y=\frac{2}{3}\int_{0}^{1}(1-y)^{3} d y=\frac{1}{6}, EX2=2∫01∫01−yx2dxdy=32∫01(1−y)3dy=61, 故 Var ( X ) = Var ( Y ) = 1 6 − 1 9 = 1 18 , \operatorname{Var}(X)=\operatorname{Var}(Y)=\frac{1}{6}-\frac{1}{9}=\frac{1}{18}, Var(X)=Var(Y)=61−91=181,进而有 ρ X Y = − 1 2 \rho_{X Y}=-\frac{1}{2} ρXY=−21.
五、(15分) 对于两个只有两个取值的随机变量 X , Y , X, Y, X,Y, 试证明 X , Y X, Y X,Y独立当且仅当 X , Y X, Y X,Y不相关.
Solution:
X = x 1 X=x_{1} X=x1 | X = x 2 X=x_{2} X=x2 | ||
---|---|---|---|
Y = y 1 Y=y_{1} Y=y1 | p 11 p_{11} p11 | p 21 p_{21} p21 | 1 − q 1-q 1−q |
Y = y 2 Y=y_{2} Y=y2 | p 12 p_{12} p12 | p 22 p_{22} p22 | q q q |
1 − p 1-p 1−p | p p p |
Solution: 独立一定不相关, 只需证明充分性. 不妨仅讨论 X ′ = X − x 1 x 2 − x 1 X'=\frac{X-x_1}{x_2-x_1} X′=x2−x1X−x1, Y ′ = Y − y 1 y 2 − y 1 Y'=\frac{Y-y_1}{y_2-y_1} Y′=y2−y1Y−y1, X , Y X,Y X,Y不相关与 X ′ , Y ′ X',Y' X′,Y′不相关是等价的. 当 X ′ , Y ′ X',Y' X′,Y′不相关时, 说明 p 22 = E ( X ′ Y ′ ) = E ( X ′ ) E ( Y ′ ) = p q , p_{22}=E(X'Y')=E(X')E(Y')=pq, p22=E(X′Y′)=E(X′)E(Y′)=pq,进一步有 p 21 = p − p 22 = p − p q = p ( 1 − q ) , p_{21}=p-p_{22}=p-pq=p(1-q), p21=p−p22=p−pq=p(1−q), p 12 = q − p 22 = q − p q = q ( 1 − p ) , p_{12}=q-p_{22}=q-pq=q(1-p), p12=q−p22=q−pq=q(1−p), p 11 = 1 − p − p 12 = ( 1 − p ) − q ( 1 − p ) = ( 1 − p ) ( 1 − q ) . p_{11}=1-p-p_{12}=(1-p)-q(1-p)=(1-p)(1-q). p11=1−p−p12=(1−p)−q(1−p)=(1−p)(1−q). 因此 X ′ , Y ′ X',Y' X′,Y′独立, 故 X , Y X,Y X,Y也独立.
六、(15分) 设有来自总体 X ∼ N ( μ , σ 2 ) X \sim N\left(\mu, \sigma^{2}\right) X∼N(μ,σ2)的 2 n 2 n 2n个随机样本, 试求期望
E ( ∑ i = 1 n ( X i + X n + i − 2 X ˉ ) 2 ) . E\left(\sum_{i=1}^{n}\left(X_{i}+X_{n+i}-2 \bar{X}\right)^{2}\right). E(i=1∑n(Xi+Xn+i−2Xˉ)2).
Solution: 记 Y i = X i + X n + i , i = 1 , 2 , … , n Y_{i}=X_{i}+X_{n+i}, i=1,2, \ldots, n Yi=Xi+Xn+i,i=1,2,…,n, 这样 Y 1 , Y 2 , … , Y n Y_{1}, Y_{2}, \ldots, Y_{n} Y1,Y2,…,Yn 就是来自于总体 N ( 2 μ , 2 σ 2 ) N\left(2 \mu, 2 \sigma^{2}\right) N(2μ,2σ2) 的 n n n 个随机样本, Y ˉ = 2 X ˉ \bar{Y}=2 \bar{X} Yˉ=2Xˉ, 根据样本方差的定义和性质, 我们有 1 2 σ 2 ∑ i = 1 n ( X i + X n + i − 2 X ˉ ) 2 = 1 2 σ 2 ∑ i = 1 n ( Y i − Y ˉ ) 2 ∼ χ 2 ( n − 1 ) \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) 2σ21i=1∑n(Xi+Xn+i−2Xˉ)2=2σ21i=1∑n(Yi−Yˉ)2∼χ2(n−1) 因此
E ∑ i = 1 n ( X i + X n + i − 2 X ˉ ) 2 = 2 ( n − 1 ) σ 2 . E \sum_{i=1}^{n}\left(X_{i}+X_{n+i}-2 \bar{X}\right)^{2}=2(n-1)\sigma^{2}. Ei=1∑n(Xi+Xn+i−2Xˉ)2=2(n−1)σ2.
七、(15分) X 1 , X 2 , X 3 X_{1}, X_{2}, X_{3} X1,X2,X3, 是i.i.d.的服从 U ( 0 , 1 ) U(0,1) U(0,1)的随机变量, 求次序统计量 X ( 1 ) X_{(1)} X(1)的分布函数与期望.
Solution: 设 Y = X ( 1 ) Y=X_{(1)} Y=X(1), 当 y ∈ ( 0 , 1 ) y \in(0,1) y∈(0,1) 时, P { Y ≥ y } = P ( min { X 1 , X 2 , X 3 } ≥ y ) = P 3 { X 1 ≥ y } = ( 1 − y ) 3 P\{Y \geq y\}=P\left(\min \left\{X_{1}, X_{2}, X_{3}\right\} \geq y\right)=P^{3}\left\{X_{1} \geq y\right\}=(1-y)^{3} P{ Y≥y}=P(min{ X1,X2,X3}≥y)=P3{ X1≥y}=(1−y)3 所以分布函数是
F ( y ) = { 0 , y < 0 , 1 − ( 1 − y ) 3 , 0 ≤ y < 1 , 1 , y ≥ 1. F(y)=\left\{\begin{array}{lc}0, & y<0, \\ 1-(1-y)^{3}, & 0 \leq y<1, \\ 1, & y \geq 1 .\end{array}\right. F(y)=⎩⎨⎧0,1−(1−y)3,1,y<0,0≤y<1,y≥1. 这恰好是 Beta ( 1 , 3 ) \operatorname{Beta}(1,3) Beta(1,3)的分布函数. E Y = ∫ 0 1 y d F ( y ) = 1 ⋅ F ( 1 ) − ∫ 0 1 [ 1 − ( 1 − y ) 3 ] d y = 1 4 E Y=\int_{0}^{1} y d F(y)=1 \cdot F(1)-\int_{0}^{1}\left[1-(1-y)^{3}\right] d y=\frac{1}{4} EY=∫01ydF(y)=1⋅F(1)−∫01[1−(1−y)3]dy=41
八、(30分) 有来自 X ∼ f ( x ) = { θ , 0 ≤ x < 1 1 − θ , 1 ≤ x < 2 , 0 , otherwise X \sim f(x)=\left\{\begin{array}{cl}\theta, & 0 \leq x<1 \\ 1-\theta, & 1 \leq x<2, \\ 0, & \text { otherwise }\end{array}\right. X∼f(x)=⎩⎨⎧θ,1−θ,0,0≤x<11≤x<2, otherwise 的 n n n 个简单随机样本, 求
(1)(15分) θ \theta θ 的矩估计;
(2)(15分) θ \theta θ 的最大似然估计, 已知 m = ∑ i = 1 n I [ X i < 1 ] m=\sum_{i=1}^{n} I\left[X_{i}<1\right] m=∑i=1nI[Xi<1].
Solution: (1) E X = ∫ 0 1 x θ d x + ∫ 1 2 x ( 1 − θ ) d x = 3 2 − θ E X=\int_{0}^{1} x \theta d x+\int_{1}^{2} x(1-\theta) d x=\frac{3}{2}-\theta EX=∫01xθdx+∫12x(1−θ)dx=23−θ, 因此 θ ^ 1 = 3 2 − X ˉ . \hat{\theta}_{1}=\frac{3}{2}-\bar{X}. θ^1=23−Xˉ.(2) 似然函数 L ( x 1 , … , x n ; θ ) = θ m ( 1 − θ ) n − m , L\left(x_{1}, \ldots, x_{n} ; \theta\right)=\theta^{m}(1-\theta)^{n-m}, L(x1,…,xn;θ)=θm(1−θ)n−m,
ln L = m ln θ + ( n − m ) ln ( 1 − θ ) \ln L=m \ln \theta+(n-m) \ln(1-\theta) lnL=mlnθ+(n−m)ln(1−θ), 对 θ \theta θ 求偏导, 得 ∂ ln L ∂ θ = m θ − n − m 1 − θ = l e t 0 , \frac{\partial \ln L}{\partial \theta}=\frac{m}{\theta}-\frac{n-m}{1-\theta} \stackrel{l e t}{=} 0, ∂θ∂lnL=θm−1−θn−m=let0,解得 θ ^ 2 = m n \hat{\theta}_{2}=\frac{m}{n} θ^2=nm.
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