复旦大学-432统计学-2019年

一、(15分) 甲袋中有n−1n-1个白球1个黑球, 乙袋中有nn个白球, 每次从两袋中各取一球进行交换, 求交换NN次后黑球还在甲袋的概率.

Solution:
设 pkp_{k} 为交换 kk 次后黑球还在甲袋的概率, 则根据全概率公式,

pk+1=pk⋅n−1n+(1−pk)⋅1n=(1−2n)pk+1n,p_{k+1}=p_{k} \cdot \frac{n-1}{n}+\left(1-p_{k}\right) \cdot \frac{1}{n}=\left(1-\frac{2}{n}\right) p_{k}+\frac{1}{n},

故有

pN−12=(1−2n)(pN−1−12)=⋯=(1−2n)N(p0−12),p_{N}-\frac{1}{2}=\left(1-\frac{2}{n}\right)\left(p_{N-1}-\frac{1}{2}\right)=\cdots=\left(1-\frac{2}{n}\right)^{N}\left(p_{0}-\frac{1}{2}\right),

即

pN=12+12(1−2n)N. p_{N}=\frac{1}{2}+\frac{1}{2}\left(1-\frac{2}{n}\right)^{N} \text {. }

二、(15分) f(x,y)=Ae−(2x+3y)I[x>0,y>0],f(x, y)=A e^{-(2 x+3 y)} I[x>0, y>0], 求
(1)(3分) AA;
(2)(3分) P(X<2,Y<1)P(X<2, Y<1);
(3)(3分) XX的边际密度;
(4)(3分) P(X<3∣Y<1)P(X<3 \mid Y<1);
(5)(3分) f(x∣y)f(x \mid y).

Solution:
(1) 由概率的正则性, 1=∫R2f(x,y)dxdy=A∫0+∞e−2xdx∫0+∞e−3ydy=A6⇒A=61=\int_{R^{2}} f(x, y) d x d y=A \int_{0}^{+\infty} e^{-2 x} d x \int_{0}^{+\infty} e^{-3 y} d y=\frac{A}{6} \Rightarrow A=6.
(2) P(X<2,Y<1)=∫02e−2xdx∫01e−3ydy=6(1−e−4)(1−e−3)P(X<2, Y<1)= \int_{0}^{2} e^{-2 x} d x \int_{0}^{1} e^{-3 y} d y=6\left(1-e^{-4}\right)\left(1-e^{-3}\right).
(3) fX(x)=∫0+∞6e−2x−3ydy=2e−2x,x>0f_{X}(x)=\int_{0}^{+\infty} 6 e^{-2 x-3 y} d y=2 e^{-2 x}, x>0.
(4) 由于 fX(x)=2e−2x,fY(y)=3e−3yf_{X}(x)=2 e^{-2 x}, f_{Y}(y)=3 e^{-3 y}, 故 X,YX, Y 相互独立, 因此

P(X<3∣Y<1)=P(X<3)=1−e−6.P(X<3 \mid Y<1)=P(X<3)=1-e^{-6} .

(5) f(x∣y)=f(x,y)fY(y)=2e−2x,x>0,y>0f(x \mid y)=\frac{f(x, y)}{f_{Y}(y)}=2 e^{-2 x}, x>0, y>0.

三、(10分) X1,X2,X_{1}, X_{2},i.i.d ∼N(μ,σ2),\sim N\left(\mu, \sigma^{2}\right), 求 Emax⁡{X1,X2}E \max \left\{X_{1}, X_{2}\right\}.

Solution:
令 Yi=Xi−μσ,i=1,2Y_{i}=\frac{X_{i}-\mu}{\sigma}, i=1,2, 故 max⁡{Y1,Y2}=max⁡{X1,X2}−μσ\max \left\{Y_{1}, Y_{2}\right\}=\frac{\max \left\{X_{1}, X_{2}\right\}-\mu}{\sigma}.
而

Emax⁡{Y1,Y2}=EY1I[Y1≥Y2]+EY2I[Y1<Y2]=2EY1I[Y1≥Y2],E \max \left\{Y_{1}, Y_{2}\right\}=E Y_{1} I_{\left[Y_{1} \geq Y_{2}\right]}+E Y_{2} I_{\left[Y_{1}<Y_{2}\right]}=2 E Y_{1} I_{\left[Y_{1} \geq Y_{2}\right]},

E[Y1I[Y1≥Y2]]=12π∫−∞+∞∫x+∞ye−x2+y22dxdy=12π∫π45π4sin⁡θdθ∫0+∞r2e−r22drE[ Y_{1} I_{\left[Y_{1} \geq Y_{2}\right]}]=\frac{1}{2 \pi} \int_{-\infty}^{+\infty} \int_{x}^{+\infty} y e^{-\frac{x^{2}+y^{2}}{2}} d x d y=\frac{1}{2 \pi} \int_{\frac{\pi}{4}}^{\frac{5 \pi}{4}} \sin \theta d \theta \int_{0}^{+\infty} r^{2} e^{-\frac{r^{2}}{2}} d r

其中 ∫0+∞r2e−r22dr=2∫0+∞r22e−r22d(r22)=2Γ(32)=π2\int_{0}^{+\infty} r^{2} e^{-\frac{r^{2}}{2}} d r=\sqrt{2} \int_{0}^{+\infty} \sqrt{\frac{r^{2}}{2}} e^{-\frac{r^{2}}{2}} d\left(\frac{r^{2}}{2}\right)=\sqrt{2} \Gamma\left(\frac{3}{2}\right)=\sqrt{\frac{\pi}{2}}, 故 Emax⁡{Y1,Y2}=1π,Emax⁡{X1,X2}=μ+σπE \max \left\{Y_{1}, Y_{2}\right\}=\frac{1}{\sqrt{\pi}}, E \max \left\{X_{1}, X_{2}\right\}=\mu+\frac{\sigma}{\sqrt{\pi}}.

四、(20分) EX=0,Var⁡(X)=σ2,E X=0, \operatorname{Var}(X)=\sigma^{2}, 证明对任意 ε>0,\varepsilon>0, 有
(1)(10分) P(∣X∣>ε)≤σ2ε2P(|X|>\varepsilon) \leq \frac{\sigma^{2}}{\varepsilon^{2}};
(2)(10分) P(X>ε)≤σ2σ2+ε2P(X>\varepsilon) \leq \frac{\sigma^{2}}{\sigma^{2}+\varepsilon^{2}}.

Solution:
(1) 记 I[∣X∣>ε]={1,∣X∣>ε,0,∣X∣≤ε.I_{[|X|>\varepsilon]}=\left\{\begin{array}{ll}1, & |X|>\varepsilon, \\ 0, & |X| \leq \varepsilon .\end{array}\right. 它是集合 {ω:∣X(ω)∣>ε}\{\omega:|X(\omega)|>\varepsilon\} 的示性函数, 可以看出:

I[∣X∣>ε]≤∣X∣2ε2I_{[|X|>\varepsilon]} \leq \frac{|X|^{2}}{\varepsilon^{2}}

故P(∣X∣>ε)=EI[∣X∣>ε]≤E∣X∣2ε2=σ2ε2P(|X|>\varepsilon)=E I_{[|X|>\varepsilon]} \leq \frac{E|X|^{2}}{\varepsilon^{2}}=\frac{\sigma^{2}}{\varepsilon^{2}}.
(2) 由马尔可夫不等式, 有

P(X>ε)=P(X+a>ε+a)≤E(X+a)2(ε+a)2=σ2+a2(ε+a)2,P(X>\varepsilon)=P(X+a>\varepsilon+a) \leq \frac{E(X+a)^{2}}{(\varepsilon+a)^{2}}=\frac{\sigma^{2}+a^{2}}{(\varepsilon+a)^{2}},

取 a=σ2εa=\frac{\sigma^{2}}{\varepsilon}, 则恰有

σ2+a2(ε+a)2=σ2+σ4ε2(ε+σ2ε)2=σ2σ2+ε2,\frac{\sigma^{2}+a^{2}}{(\varepsilon+a)^{2}}=\frac{\sigma^{2}+\frac{\sigma^{4}}{\varepsilon^{2}}}{\left(\varepsilon+\frac{\sigma^{2}}{\varepsilon}\right)^{2}}=\frac{\sigma^{2}}{\sigma^{2}+\varepsilon^{2}},

因此 P(X>ε)≤σ2σ2+ε2P(X>\varepsilon) \leq \frac{\sigma^{2}}{\sigma^{2}+\varepsilon^{2}}.

[注]: 为什么要取 a=σ2εa=\frac{\sigma^2}{\varepsilon}? 这里我们记 g(a)=σ2+a2(ε+a)2g(a) = \frac{\sigma^2 + a^2}{(\varepsilon+a)^2}, 这是aa 的函数, 求出它的最大值点即可, 发现恰好在 a=σ2εa=\frac{\sigma^2}{\varepsilon} 取到.

五、(10分) X1,X2X_{1}, X_{2}, i.i.d. ∼N(0,1),\sim N(0,1), 求 X1X2\frac{X_{1}}{X_{2}} 的分布.

Solution: 重复考察, 略去.

六、(20分) X1,X2,…,Xn,i.i.d∼N(μ,σ2),μX_{1}, X_{2}, \ldots, X_{n}, i . i . d \sim N\left(\mu, \sigma^{2}\right), \mu 已知, 证明:
(1)(10分) 1n∑i=1n(Xi−μ)2\frac{1}{n} \sum_{i=1}^{n}\left(X_{i}-\mu\right)^{2} 是 σ2\sigma^{2} 的有效估计;
(2)(10分) 1nπ2∑i=1n∣Xi−μ∣\frac{1}{n} \sqrt{\frac{\pi}{2}} \sum_{i=1}^{n}\left|X_{i}-\mu\right| 是 σ\sigma 的无偏估计, 但不有效.

Solution:
(1) 先计算 σ2\sigma^{2} 的 Fisher 信息量, 根据定义

I(σ2)=E[∂ln⁡f(X;σ2)∂σ2]2=14σ4E[(X−μσ)2−1]2,I\left(\sigma^{2}\right)=E\left[\frac{\partial \ln f\left(X ; \sigma^{2}\right)}{\partial \sigma^{2}}\right]^{2}=\frac{1}{4 \sigma^{4}} E\left[\left(\frac{X-\mu}{\sigma}\right)^{2}-1\right]^{2},

恰好 E[(X−μσ)2−1]2E\left[\left(\frac{X-\mu}{\sigma}\right)^{2}-1\right]^{2} 是 χ2\chi^{2} (1) 的方差, 故 I(σ2)=12σ4I\left(\sigma^{2}\right)=\frac{1}{2 \sigma^{4}}. 因此, σ2\sigma^{2} 的 C-R 下界为 1nI(σ2)=2nσ4.\frac{1}{n I\left(\sigma^{2}\right)}=\frac{2}{n} \sigma^{4} . 再计算 1n∑i=1n(Xi−μ)2\frac{1}{n} \sum_{i=1}^{n}\left(X_{i}-\mu\right)^{2} 的期望方差, 由于 1σ2∑i=1n(Xi−μ)2∼χ2(n)\frac{1}{\sigma^{2}} \sum_{i=1}^{n}\left(X_{i}-\mu\right)^{2} \sim \chi^{2}(n), 期望是 nn, 方差是 2n2 n, 因此

E[1n∑i=1n(Xi−μ)2]=σ2,Var⁡[1n∑i=1n(Xi−μ)2]=2nσ4,E\left[\frac{1}{n} \sum_{i=1}^{n}\left(X_{i}-\mu\right)^{2}\right]=\sigma^{2}, \quad \operatorname{Var}\left[\frac{1}{n} \sum_{i=1}^{n}\left(X_{i}-\mu\right)^{2}\right]=\frac{2}{n} \sigma^{4},

它是无偏估计, 方差又恰好达到 C-R 下界, 故是有效估计.
(2) 首先, 令 g(x)=xg(x)=\sqrt{x}, 则 σ\sigma 的 C−R\mathrm{C}-\mathrm{R} 下界为 [g′(σ2)]2nI(σ2)=12nσ2\frac{\left[g^{\prime}\left(\sigma^{2}\right)\right]^{2}}{n I\left(\sigma^{2}\right)}=\frac{1}{2 n} \sigma^{2}, 我们再去计算 1nπ2∑i=1n∣Xi−μ∣\frac{1}{n} \sqrt{\frac{\pi}{2}} \sum_{i=1}^{n}\left|X_{i}-\mu\right| 的期望方差:

1σE∣X1−μ∣=∫−∞+∞∣x∣12πe−x22dx=2π∫0+∞xe−x22dx=2π⇒E∣X1−μ∣=2πσ,\frac{1}{\sigma} E\left|X_{1}-\mu\right|=\int_{-\infty}^{+\infty}|x| \frac{1}{\sqrt{2 \pi}} e^{-\frac{x^{2}}{2}} d x=\sqrt{\frac{2}{\pi}} \int_{0}^{+\infty} x e^{-\frac{x^{2}}{2}} d x=\sqrt{\frac{2}{\pi}} \Rightarrow E\left|X_{1}-\mu\right|=\sqrt{\frac{2}{\pi}} \sigma,

E∣X1−μ∣2=σ2, 故 Var⁡(∣X1−μ∣)=σ2−2πσ2=(1−2π)σ2, E\left|X_{1}-\mu\right|^{2}=\sigma^{2} \text {, 故 } \operatorname{Var}\left(\left|X_{1}-\mu\right|\right)=\sigma^{2}-\frac{2}{\pi} \sigma^{2}=\left(1-\frac{2}{\pi}\right) \sigma^{2} \text {, }

故E[1nπ2∑i=1n∣Xi−μ∣]=σ,Var⁡[1nπ2∑i=1n∣Xi−μ∣]=(π2−1)nσ2E\left[\frac{1}{n} \sqrt{\frac{\pi}{2}} \sum_{i=1}^{n}\left|X_{i}-\mu\right|\right]=\sigma, \operatorname{Var}\left[\frac{1}{n} \sqrt{\frac{\pi}{2}} \sum_{i=1}^{n}\left|X_{i}-\mu\right|\right]=\frac{\left(\frac{\pi}{2}-1\right)}{n} \sigma^{2}, 它是无偏估计, 但 没有达到 C-R 下界, 不是有效估计.

七、(20分) 总体的分布函数F(x)F(x)连续单增, X(1),X(2),…,X(n)X_{(1)}, X_{(2)}, \ldots, X_{(n)} 是来自该总体的随机样本的次序统计量, Yi=F(X(i)),Y_{i}=F\left(X_{(i)}\right), 求
(1)(10分) EYi,Var⁡(Yi)E Y_{i}, \operatorname{Var}\left(Y_{i}\right);
(2)(10分) (Y1,Y2,…,Yn)T\left(Y_{1}, Y_{2}, \ldots, Y_{n}\right)^{T} 的协方差矩阵.

Solution:
(1) 由于 U1=F(X1),U2=F(X2),⋯ ,Un=F(Xn)U_{1}=F\left(X_{1}\right), U_{2}=F\left(X_{2}\right), \cdots, U_{n}=F\left(X_{n}\right) 独立同服从 [0,1][0,1] 上均匀分布, 故 Yi=U(i)Y_{i}=U_{(i)} 恰好就是均匀分布的次序统计量, 根据次序统计量的定义, 有:

fi(y)=n!(i−1)!(n−i)!fU(y)Pi−1{U≤y}Pn−i{U≥y}, 一一代入, 有 fi(y)=n!(i−1)!(n−i)!yi−1(1−y)n−i,0≤y≤1, 恰好是 Beta⁡(i,n−i+1).\begin{gathered} f_{i}(y)=\frac{n !}{(i-1) !(n-i) !} f_{U}(y) P^{i-1}\{U \leq y\} P^{n-i}\{U \geq y\}, \text { 一一代入, 有 } \\ f_{i}(y)=\frac{n !}{(i-1) !(n-i) !} y^{i-1}(1-y)^{n-i}, 0 \leq y \leq 1 \text {, 恰好是 } \operatorname{Beta}(i, n-i+1) . \end{gathered}

根据贝塔分布的性质 EYi=in+1,Var⁡(Yi)=i(n−i+1)(n+1)2(n+2)E Y_{i}=\frac{i}{n+1}, \operatorname{Var}\left(Y_{i}\right)=\frac{i(n-i+1)}{(n+1)^{2}(n+2)}.
(2) 我们考虑 (Yi,Yj),i<j\left(Y_{i}, Y_{j}\right), i<j 的联合分布, 根据次序统计量的定义, 有

fi,j(x,y)=n!(i−1)!(j−i−1)!(n−j)!fU(x)fU(y)Pi−1{U≤x}Pj−i−1{x≤U≤y}Pn−j{U≥y}f_{i, j}(x, y)=\frac{n !}{(i-1) !(j-i-1) !(n-j) !} f_{U}(x) f_{U}(y) P^{i-1}\{U \leq x\} P^{j-i-1}\{x \leq U \leq y\} P^{n-j}\{U \geq y\}

即

fi,j(x,y)=n!(i−1)!(j−i−1)!(n−j)!xi−1(y−x)j−i−1(1−y)n−j,0≤x≤y≤1.f_{i, j}(x, y)=\frac{n !}{(i-1) !(j-i-1) !(n-j) !} x^{i-1}(y-x)^{j-i-1}(1-y)^{n-j}, 0 \leq x \leq y \leq 1.

协方差问题归为计算积分 ∫01∫0yxy⋅xi−1(y−x)j−i−1(1−y)n−jdxdy\int_{0}^{1} \int_{0}^{y} x y \cdot x^{i-1}(y-x)^{j-i-1}(1-y)^{n-j} d x d y, 而如果我 们把 yy 视作 y=x+(y−x)y=x+(y-x), 那么这个积分可以化为两部分:

∫01∫0yxi+1(y−x)j−i−1(1−y)n−jdxdy=(i+1)!(j−i−1)!(n−j)!(n+2)!,∫01∫0yxi(y−x)j−i(1−y)n−jdxdy=i!(j−i)!(n−j)!(n+2)!,\begin{gathered} \int_{0}^{1} \int_{0}^{y} x^{i+1}(y-x)^{j-i-1}(1-y)^{n-j} d x d y=\frac{(i+1) !(j-i-1) !(n-j) !}{(n+2) !}, \\ \int_{0}^{1} \int_{0}^{y} x^{i}(y-x)^{j-i}(1-y)^{n-j} d x d y=\frac{i !(j-i) !(n-j) !}{(n+2) !}, \end{gathered}

(首先请看 fi,j(x,y)f_{i, j}(x, y) 的表达式, 会发现

(n+2)!(i+1)!(j−i−1)!(n−j)!xi+1(y−x)j−i−1(1−y)n−j\frac{(n+2) !}{(i+1) !(j-i-1) !(n-j) !} x^{i+1}(y-x)^{j-i-1}(1-y)^{n-j}

恰也是一个密度函数, 故通过概率的正则性就很容易得到上述两个积分值) 故有 E[YiYj]=i(j+1)(n+1)(n+2)E [Y_{i} Y_{j}]=\frac{i(j+1)}{(n+1)(n+2)}, 故 Cov⁡(Yi,Yj)=i(n+1−j)(n+1)2(n+2)\operatorname{Cov}\left(Y_{i}, Y_{j}\right)=\frac{i(n+1-j)}{(n+1)^{2}(n+2)}. 因此协方差矩阵是 Σ=(aij)n×n\Sigma=\left(a_{i j}\right)_{n \times n}, 其中 aij=i(n+1−j)(n+1)2(n+2),i<ja_{i j}=\frac{i(n+1-j)}{(n+1)^{2}(n+2)}, i<j.

八、(20分) 有来自总体U(θ,2θ)U(\theta,2\theta)的随机样本X1,⋯ ,XnX_1,\cdots,X_n, 求θ\theta的矩估计和MLE, 并验证无偏性和相合性.

Solution:
先求矩估计, 求期望得EX1=3θ2EX_1=\frac{3\theta}{2}, 由替换原理得θ^M=23Xˉ\hat{\theta}_M=\frac{2}{3}\bar{X}. 求期望容易看出它无偏, 由强大数律知它是强相合估计.
再求MLE, 写出似然函数是

L(θ)=1θnI{X(n)<2θ}I{X(1)>θ}=I{X(n)2<θ<X(1)}θn,L\left( \theta \right) =\frac{1}{\theta ^n}I_{\left\{ X_{\left( n \right)}<2\theta \right\}}I_{\left\{ X_{\left( 1 \right)}>\theta \right\}}=\frac{I_{\left\{ \frac{X_{\left( n \right)}}{2}<\theta <X_{\left( 1 \right)} \right\}}}{\theta ^n},

可以看出1θn\frac{1}{\theta^n}关于θ\theta单调递减, 故θ\theta取最小值时为MLE, 即θ^L=X(n)2\hat{\theta}_L=\frac{X_{(n)}}{2}. 利用变换Yi=Xi−θθ∼U(0,1)Y_i=\frac{X_i-\theta}{\theta}\sim U(0,1), 知Y(n)=X(n)−θθ∼Beta(n,1)Y_{(n)}=\frac{X_{(n)}-\theta}{\theta}\sim Beta(n,1), 故有

E(Y(n))=nn+1,E(θ^L)=12E(X(n))=12[θE(Y(n))+θ]=2n+12n+2θ.E\left( Y_{\left( n \right)} \right) =\frac{n}{n+1},\quad E\left( \hat{\theta}_L \right) =\frac{1}{2}E\left( X_{\left( n \right)} \right) =\frac{1}{2}\left[ \theta E\left( Y_{\left( n \right)} \right) +\theta \right] =\frac{2n+1}{2n+2}\theta .

所以θ^L\hat{\theta}_L不无偏, 但渐近无偏, 再看相合性, 有

P(∣θ^L−θ∣>ε1)=P(∣X(n)−2θ∣>ε2)=P(∣Y(n)−1∣>ε3)=P(Y(n)<1−ε3)=(1−ε3)n,\begin{aligned} P\left( \left| \hat{\theta}_L-\theta \right|>\varepsilon _1 \right) &=P\left( \left| X_{\left( n \right)}-2\theta \right|>\varepsilon _2 \right)\\ &=P\left( \left| Y_{\left( n \right)}-1 \right|>\varepsilon _3 \right)\\ &=P\left( Y_{\left( n \right)}<1-\varepsilon _3 \right)\\ &=\left( 1-\varepsilon _3 \right) ^n,\\ \end{aligned}

级数收敛, 故它是强相合估计.

九、(20分) 设X1,⋯ ,XnX_1,\cdots,X_n是来自N(μ,1)N(\mu,1)的随机样本, 考虑假设检验问题

H0:μ=1vsH1:μ=2H_0:\mu = 1 \quad \mathrm{vs} \quad H_1:\mu=2

给定拒绝域W={Xˉ>1.6}W=\{\bar{X}>1.6\}, 回答下述问题:
(1)(10分) n=10n=10, 求犯两类错误的概率α\alpha, β\beta;
(2)(10分) 要求第二类错误β≤0.01\beta\le0.01, 求样本量的取值范围.

Solution:
(1) 先算第一类错误, 原假设成真时Xˉ∼N(1,1n)\bar{X}\sim N(1,\frac{1}{n}), 故

α=Pμ=1(Xˉ>1.6)=Pμ=1(n(Xˉ−1)>0.6n)=1−Φ(0.6n)=1−Φ(0.610).\begin{aligned} \alpha &=P_{\mu =1}\left( \bar{X}>1.6 \right)\\ &=P_{\mu =1}\left( \sqrt{n}\left( \bar{X}-1 \right) >0.6\sqrt{n} \right)\\ &=1-\Phi \left( 0.6\sqrt{n} \right)\\ &=1-\Phi \left( 0.6\sqrt{10} \right) .\\ \end{aligned}

再算第二类错误, 备择假设成真时Xˉ∼N(2,1n)\bar{X}\sim N(2,\frac{1}{n}), 故

β=Pμ=2(Xˉ≤1.6)=Pμ=1(n(Xˉ−2)≤−0.4n)=Φ(−0.4n)=Φ(−0.410).\begin{aligned} \beta &=P_{\mu =2}\left( \bar{X}\le 1.6 \right)\\ &=P_{\mu =1}\left( \sqrt{n}\left( \bar{X}-2 \right) \le -0.4\sqrt{n} \right)\\ &=\Phi \left( -0.4\sqrt{n} \right)\\ &=\Phi \left( -0.4\sqrt{10} \right) .\\ \end{aligned}

(2) 根据第(1)问计算, 令Φ(−0.4n)≤0.01\Phi \left( -0.4\sqrt{n} \right) \le 0.01, 得

−0.4n≤−2.33⟹n≥2.3320.42⟹n≥34.-0.4\sqrt{n}\le -2.33\Longrightarrow n\ge \frac{2.33^2}{0.4^2}\Longrightarrow n\ge 34.