UA MATH566 统计理论 QE练习题1

UA MATH566 统计理论 QE练习题1

• 第四题
• 第五题
• 第六题

2014年1月理论题目4-6。

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第四题

Part (a)
Joint likelihood function of random sample is
$L(\theta) = \prod_{i=1}^n \left( \theta^{-c} cX_i^{c-1}e^{-(X_i/\theta)^c} \right) = \theta^{-nc}c^n (\prod_{i=1}^n X_i)^{c-1}e^{-\frac{1}{\theta^c}\sum_{i=1}^n X_i^c} \\ l(\theta) = \log L(\theta) = -nc \log \theta + n \log c + (c-1)\sum_{i=1}^n \log X_i - \frac{1}{\theta^c} \sum_{i=1}^n X_i^c$L(θ)=i=1∏n​(θ−ccXic−1​e−(Xi​/θ)c)=θ−nccn(i=1∏n​Xi​)c−1e−θc1​∑i=1n​Xic​l(θ)=logL(θ)=−nclogθ+nlogc+(c−1)i=1∑n​logXi​−θc1​i=1∑n​Xic​

Take derivative of log-likelihood and let it be zero,
$l'(\theta) = -\frac{nc}{\theta} + \frac{c}{\theta^{c+1}} \sum_{i=1}^n X_i^c = 0$l′(θ)=−θnc​+θc+1c​i=1∑n​Xic​=0

Solve this equation and we can find MLE of $\theta$θ,
$\hat{\theta} = \left( \frac{1}{n}\sum_{i=1}^n X_i^c \right)^{1/c}$θ^=(n1​i=1∑n​Xic​)1/c

Part (b)
Compute the density function of $X_1^c$X1c​,
$P(X_1^c \le y) = P(X_1 \le y^{1/c}) = F_X(y^{1/c}) \\ f_{X_1^c}(y) = \frac{1}{c}y^{1/c-1} \theta^{-c}cy^{c-1/c}e^{-(y^{1/c}/\theta)^c} = \theta^{-c}e^{-\theta^{-c}y},y>0$P(X1c​≤y)=P(X1​≤y1/c)=FX​(y1/c)fX1c​​(y)=c1​y1/c−1θ−ccyc−1/ce−(y1/c/θ)c=θ−ce−θ−cy,y>0

Obviously $X_1^c \sim EXP(\theta^c)$X1c​∼EXP(θc)，equivalently, $\Gamma(1,\theta^{-c})$Γ(1,θ−c). By additivity of Gamma distribution, $\sum_{i=1}^n X_i^c \sim \Gamma(n,\theta^{-c})$∑i=1n​Xic​∼Γ(n,θ−c). By scale transformation, $\frac{1}{n}\sum_{i=1}^n X_i^c \sim \Gamma(n,n\theta^{-c})$n1​∑i=1n​Xic​∼Γ(n,nθ−c). Let $Y = \frac{1}{n}\sum_{i=1}^n X_i^c$Y=n1​∑i=1n​Xic​,
$EY^{1/c} = \int_{0}^{\infty} y^{1/c} \frac{(n\theta^{-c})^n y^{n-1}}{\Gamma(n)}e^{-n\theta^{-c}y}dy = \frac{(n\theta^{-c})^n }{\Gamma(n)} \int_{0}^{\infty} y^{1/c+n-1}e^{-n\theta^{-c}y}dy \\ = \frac{(n\theta^{-c})^{-1/c} }{\Gamma(n)} \int_{0}^{\infty} (n\theta^{-c}y)^{1/c+n-1}e^{-n\theta^{-c}y}dn\theta^{-c}y = \frac{(n\theta^{-c})^{-1/c} \Gamma(1/c+n)}{\Gamma(n)}\\= \theta\frac{n^{-1/c} \Gamma(1/c+n)}{\Gamma(n)} \\ \Rightarrow E \left( \frac{Y^{1/c}\Gamma(n)}{n^{1/c} \Gamma(1/c+n)} \right) = \theta$EY1/c=∫0∞​y1/cΓ(n)(nθ−c)nyn−1​e−nθ−cydy=Γ(n)(nθ−c)n​∫0∞​y1/c+n−1e−nθ−cydy=Γ(n)(nθ−c)−1/c​∫0∞​(nθ−cy)1/c+n−1e−nθ−cydnθ−cy=Γ(n)(nθ−c)−1/cΓ(1/c+n)​=θΓ(n)n−1/cΓ(1/c+n)​⇒E(n1/cΓ(1/c+n)Y1/cΓ(n)​)=θ

So we get an unbiased estimator of $\theta$θ. See joint likelihood $L(\theta)$L(θ), by Neyman-Fisher Theorem, we know $\sum_{i=1}^n X_i^c$∑i=1n​Xic​ is sufficient statistics. Plus,
$L(\theta) = c^n (\prod_{i=1}^n X_i)^{c-1}\exp\left( -\frac{1}{\theta^c}\sum_{i=1}^n X_i^c - nc \log \theta \right)$L(θ)=cn(i=1∏n​Xi​)c−1exp(−θc1​i=1∑n​Xic​−nclogθ)

indicating it belongs to exponential family. So $\sum_{i=1}^n X_i^c$∑i=1n​Xic​ is also complete. Since MLE is definitely a function of $\sum_{i=1}^n X_i^c$∑i=1n​Xic​, by Lehmann-Scheffe Theorem, $\frac{Y^{1/c}\Gamma(n)}{n^{1/c} \Gamma(1/c+n)}$n1/cΓ(1/c+n)Y1/cΓ(n)​ is a UMVUE of $\theta$θ.

Part ( c)
For arbitrary $\theta_1 \le \theta_0 \le \theta_2$θ1​≤θ0​≤θ2​, let’s compute likelihood ratio
$\frac{L(\theta_1)}{L(\theta_2)} = \frac{\theta_1^{-nc}c^n (\prod_{i=1}^n X_i)^{c-1}e^{-\frac{1}{\theta_1^c}\sum_{i=1}^n X_i^c}}{\theta_2^{-nc}c^n (\prod_{i=1}^n X_i)^{c-1}e^{-\frac{1}{\theta_2^c}\sum_{i=1}^n X_i^c}} \\ = \left( \frac{\theta_2}{\theta_1} \right)^{nc} \exp \left( (1/\theta_2^c - 1/\theta_1^c)\sum_{i=1}^n X_i^c \right)$L(θ2​)L(θ1​)​=θ2−nc​cn(∏i=1n​Xi​)c−1e−θ2c​1​∑i=1n​Xic​θ1−nc​cn(∏i=1n​Xi​)c−1e−θ1c​1​∑i=1n​Xic​​=(θ1​θ2​​)ncexp((1/θ2c​−1/θ1c​)i=1∑n​Xic​)

where $(1/\theta_2^c - 1/\theta_1^c)<0$(1/θ2c​−1/θ1c​)<0. To make this likelihood ratio smaller than some number, $\sum_{i=1}^n X_i^c$∑i=1n​Xic​ should be greater than $k_{\alpha}$kα​ such that
$P(\sum_{i=1}^n X_i^c > k_{\alpha}) = \alpha$P(i=1∑n​Xic​>kα​)=α

We have discussed above that $\sum_{i=1}^n X_i^c \sim \Gamma(n,\theta^{-c})$∑i=1n​Xic​∼Γ(n,θ−c), so $2\sum_{i=1}^n X_i^c/\theta_0^c \sim \chi_{2n}^2$2∑i=1n​Xic​/θ0c​∼χ2n2​. Let $k_{\alpha}$kα​ is the upper $\alpha$α-quantile of $\chi_{2n}^2$χ2n2​. And the rejection region is $\{X_1,X_2,\cdots,X_n:2\sum_{i=1}^n X_i^c/\theta_0^c > k_{\alpha}\}${X1​,X2​,⋯,Xn​:2∑i=1n​Xic​/θ0c​>kα​}.

第五题

Part (a)
Joint likelihood function of random sample is
$L(\theta) = \prod_{i=1}^n \theta^{-1}X_i^{(1-\theta)/\theta} = \theta^{-n}\exp \left( \frac{\theta-1}{2\theta} [-2\sum_{i=1}^n \log X_i]\right)$L(θ)=i=1∏n​θ−1Xi(1−θ)/θ​=θ−nexp(2θθ−1​[−2i=1∑n​logXi​])

By Neyman-Fisher Theorem, $T(X) =-2\sum_{i=1}^n \log X_i$T(X)=−2∑i=1n​logXi​ is sufficient statistics. For two different groups of random sample,
$\frac{L(\theta|\textbf{X})}{L(\theta|\textbf{Y})} = \frac{\theta^{-n}\exp \left( \frac{\theta-1}{2\theta} [-2\sum_{i=1}^n \log X_i]\right)}{\theta^{-n}\exp \left( \frac{\theta-1}{2\theta} [-2\sum_{i=1}^n \log Y_i]\right)} \\ = \exp \left( \frac{\theta-1}{2\theta} [2\sum_{i=1}^n \log Y_i-2\sum_{i=1}^n \log X_i]\right)$L(θ∣Y)L(θ∣X)​=θ−nexp(2θθ−1​[−2∑i=1n​logYi​])θ−nexp(2θθ−1​[−2∑i=1n​logXi​])​=exp(2θθ−1​[2i=1∑n​logYi​−2i=1∑n​logXi​])

To make this ratio independent of $\theta$θ, $T(X) = T(Y)$T(X)=T(Y) holds. So $T(X)$T(X) is minimal sufficient statistics.

Part(b)
Compute
$P(Y \le y) = P(-2\log X_1 \le y) = P(X_1 \ge e^{-y/2}) = 1 - F_X(e^{-y/2}) \\ f_Y(y) = \frac{1}{2}e^{-y/2}f_X(e^{-y/2}) = \frac{1}{2}e^{-y/2} \theta^{-1}e^{-y(1-\theta)/2\theta} = \frac{1}{2\theta}e^{-y/2\theta},y>0$P(Y≤y)=P(−2logX1​≤y)=P(X1​≥e−y/2)=1−FX​(e−y/2)fY​(y)=21​e−y/2fX​(e−y/2)=21​e−y/2θ−1e−y(1−θ)/2θ=2θ1​e−y/2θ,y>0

So $Y \sim EXP(2\theta)$Y∼EXP(2θ), equivalently $\Gamma(1,1/2\theta)$Γ(1,1/2θ).

Part ( c)
By additivity of Gamma distribution, $T(X) \sim \Gamma(n,1/2\theta)$T(X)∼Γ(n,1/2θ). By scale transformation, $T(X)/2\theta \sim \Gamma(n,1)$T(X)/2θ∼Γ(n,1). Let $L$L be 2.5%-quantile of $\Gamma(n,1)$Γ(n,1), $U$U be 97.5%-quantile of $\Gamma(n,1)$Γ(n,1)，and then the 95% confidential interval is
$L \le T(X) /2\theta\le U \Rightarrow \frac{T(X)}{2U}\le \theta \le \frac{T(X)}{2L}$L≤T(X)/2θ≤U⇒2UT(X)​≤θ≤2LT(X)​

Part (d)
The expected length of confidential interval is
$E[T(X)\left( \frac{1}{2L} - \frac{1}{2U} \right)] = \frac{n\theta(U-L)}{LU}$E[T(X)(2L1​−2U1​)]=LUnθ(U−L)​

(I can think of a different way but I need to attach the solution here.)

（参考UA MATH564 概率论VI 数理统计基础3 卡方分布的正态近似）

第六题

Part (a)
Joint likelihood of random sample is
$L(\beta) = \prod_{i=1}^n \frac{(x_i \beta)^{Y_i}e^{-x_i\beta}}{Y_i !} = \left( \prod_{i=1}^n \frac{x_i^{Y_i}}{Y_i !} \right) \beta^{n\bar{Y}}e^{-n\beta \bar{x}} \\ l(\beta) = \log \left( \prod_{i=1}^n \frac{x_i^{Y_i}}{Y_i !} \right) + n\bar{Y} \log \beta - n \bar{x} \beta$L(β)=i=1∏n​Yi​!(xi​β)Yi​e−xi​β​=(i=1∏n​Yi​!xiYi​​​)βnYˉe−nβxˉl(β)=log(i=1∏n​Yi​!xiYi​​​)+nYˉlogβ−nxˉβ

MLE of $\beta$β is given by $\argmax l(\beta)$argmaxl(β), which is
$\frac{n\bar{Y}}{\beta} - n\bar{x} = 0\Rightarrow \hat{\beta} = \frac{\bar{Y}}{\bar{x}}$βnYˉ​−nxˉ=0⇒β^​=xˉYˉ​

Part (b)
$E\hat{\beta} = \frac{1}{\bar{x}} E\bar{Y} = \frac{1}{n\bar{x}} \sum_{i=1}^n EY_i = \frac{\beta \sum_{i=1}^n x_i }{n\bar{x}} = \beta \\ Var(\hat{\beta}) = \frac{1}{n^2\bar{x}^2} \sum_{i=1}^n VarY_i = \frac{\beta \sum_{i=1}^n x_i}{n^2\bar{x}^2} = \frac{\beta}{n\bar{x}}$Eβ^​=xˉ1​EYˉ=nxˉ1​i=1∑n​EYi​=nxˉβ∑i=1n​xi​​=βVar(β^​)=n2xˉ21​i=1∑n​VarYi​=n2xˉ2β∑i=1n​xi​​=nxˉβ​

Part ( c)
Posterior kernel of $\beta$β is
$\pi(\beta|\textbf{Y}) \propto L(\beta)\pi(\beta|w,b_0) \propto \beta^{n\bar{Y}}e^{-n\beta \bar{x}} \beta^{wb_0-1}e^{-w\beta} = \beta^{n\bar{Y}+wb_0-1}e^{-\beta(w+n\bar{x})}$π(β∣Y)∝L(β)π(β∣w,b0​)∝βnYˉe−nβxˉβwb0​−1e−wβ=βnYˉ+wb0​−1e−β(w+nxˉ)

This is the kernel of $\Gamma(n\bar{Y}+wb_0,\frac{1}{w+n\bar{x}})$Γ(nYˉ+wb0​,w+nxˉ1​). So posterior density of $\beta$β is
$\pi(\beta|\textbf{Y}) = \frac{(w+n\bar{x})^{n\bar{Y}+wb_0}}{\Gamma(n\bar{Y}+wb_0)}\beta^{n\bar{Y}+wb_0-1}e^{-\beta(w+n\bar{x})},\beta>0$π(β∣Y)=Γ(nYˉ+wb0​)(w+nxˉ)nYˉ+wb0​​βnYˉ+wb0​−1e−β(w+nxˉ),β>0

Part (d)
Posterior mean of $\beta$β is
$E[\beta|\textbf{Y}] = \frac{n\bar{Y}+wb_0}{w+n\bar{x}} = \frac{n\bar{x}\hat{\beta}}{w+n\bar{x}} +\frac{wb_0}{w+n\bar{x}}$E[β∣Y]=w+nxˉnYˉ+wb0​​=w+nxˉnxˉβ^​​+w+nxˉwb0​​

definitely weighted average of $\hat{\beta}$β^​ and $b_0$b0​. When $w \to 0$w→0, $E[\beta|\textbf{Y}] \to \hat{\beta}$E[β∣Y]→β^​.