Chapter 8 Homework 2: Principles of Data Reduction: Problems and Solutions
Exercise 8.2 (Casella and Berger 6.2) Let \(X_1,\cdots,X_n\) be independent random variables with densities \[\begin{equation} f_{X_i}(x|\theta)=\left\{\begin{aligned} & e^{i\theta-x} & \quad x\geq i\theta \\ & 0 & \quad x<i\theta \end{aligned} \right. \tag{8.2} \end{equation}\]
Prove that \(T=\min_i(X_i/i)\) is a sufficient statistic for \(\theta\).Exercise 8.5 (Casella and Berger 6.9) For each of the following distributions let \(X_1,\cdots,X_n\) be a random sample. Find a minimal sufficient statistic for \(\theta\).
Normal: \(f(x|\theta)=\frac{1}{\sqrt{2\pi}}e^{-(x-\theta)^2/2}\), \(-\infty<x<\infty\), \(-\infty<\theta<\infty\).
Location exponential: \(f(x|\theta)=e^{-(x-\theta)}\), \(\theta<x<\infty\), \(-\infty<\theta<\infty\).
Logistic: \(f(x|\theta)=\frac{e^{-(x-\theta)}}{(1+e^{-(x-\theta)})^2}\), \(-\infty<x<\infty\), \(-\infty<\theta<\infty\).
Cauchy: \(f(x|\theta)=\frac{1}{\pi[1+(x-\theta)^2]}\), \(-\infty<x<\infty\), \(-\infty<\theta<\infty\).
- Double exponential: \(f(x|\theta)=\frac{1}{2}e^{-|x-\theta|}\), \(-\infty<x<\infty\), \(-\infty<\theta<\infty\).
Proof. (a) In this case, sample mean is the minimal sufficient statistic for \(\theta\). Because \[\begin{equation} \begin{split} \frac{f(\mathbf{x}|\theta)}{f(\mathbf{y}|\theta)}&= \frac{exp\{-\frac{1}{2}[\sum_{i=1}^n(x_i-\bar{x})^2+n(\bar{x}-\theta)^2]\}} {exp\{-\frac{1}{2}[\sum_{i=1}^n(y_i-\bar{y})^2+n(\bar{y}-\theta)^2]\}}\\ &=exp\{-\frac{1}{2}[(n-1)(S_x^2+S_y^2)+n(\bar{x}^2-\bar{y}^2)+2n\theta(\bar{x}-\bar{y})]\} \end{split} \tag{8.7} \end{equation}\] The ratio will be a constant w.r.t. \(\theta\) only when \(\bar{x}=\bar{y}\).
In this case, \(\min_i\mathbf{X}=min\{X_1,\cdots,X_n\}\) is the minimal sufficient statistic. Becasue \[\begin{equation} \begin{split} \frac{f(\mathbf{x}|\theta)}{f(\mathbf{y}|\theta)}&= \frac{exp\{n(\bar{x}-\theta)\}\prod_{i=1}^nI_{(\theta,\infty)(x_i)}} {exp\{n(\bar{y}-\theta)\}\prod_{i=1}^nI_{(\theta,\infty)(x_i)}} &=exp\{-n(\bar{x}-\bar{y})\}\frac{I_{(\theta,\infty)}(\min_i(x_i))}{I_{(\theta,\infty)}(\min_i(y_i))} \end{split} \tag{8.8} \end{equation}\] which is a constant w.r.t. \(\theta\) only when \(\min_i(x_i)=\min_i(y_i)\).
In this case, the order statistics are the minimal sufficient statistic. Because \[\begin{equation} \begin{split} \frac{f(\mathbf{x}|\theta)}{f(\mathbf{y}|\theta)}&= \frac{\prod_{i=1}^ne^{-x_i}}{\prod_{i=1}^ne^{-y_i}} (\frac{\prod_{i=1}^n(1+e^{-y_i-\theta})}{\prod_{i=1}^n(1+e^{-x_i-\theta})})^2\\ &=\frac{\prod_{i=1}^ne^{-x_i}}{\prod_{i=1}^ne^{-y_i}} (\prod_{i=1}^n\frac{e^{-\theta}+e^{-y_i}}{e^{-\theta}+e^{-x_i}})^2 \end{split} \tag{8.9} \end{equation}\] which is a constant w.r.t. \(\theta\) only when order statistics for sample points \(\mathbf{x}\) and \(\mathbf{y}\) are equal.
In this case, the order statistics are the minimal sufficient statistic. Because \[\begin{equation} \frac{f(\mathbf{x}|\theta)}{f(\mathbf{y}|\theta)}=\frac{\prod_{i=1}^n(1+(y_i-\theta)^2)}{\prod_{i=1}^n(1+(x_i-\theta)^2)} \tag{8.10} \end{equation}\] Hence, suppose two sample points \(\mathbf{x}\) and \(\mathbf{y}\) only differ at \(x_j\) and \(y_j\), then from (8.10) we have \[\begin{equation} \frac{f(\mathbf{x}|\theta)}{f(\mathbf{y}|\theta)}=\frac{1+(y_j-\theta)^2}{1+(x_j-\theta)^2} \tag{8.11} \end{equation}\] which is not a constant w.r.t. \(\theta\). Hence in order to have it as a constant, we have to have the trivial case, i.e. order statistics, as the minimal sufficient statistics.
- In this case, the order statistics are the minimal sufficient statistic. Because \[\begin{equation} \begin{split} \frac{f(\mathbf{x}|\theta)}{f(\mathbf{y}|\theta)}&=exp(\sum_{i=1}^n|y_i-\theta|-\sum_{i=1}^n|x_i-\theta|)\\ &=exp(\sum_{j\in\{j:y_{(j)}<\theta\}}(\theta-y_{(j)})+\sum_{j\in\{j:y_{(j)}\geq\theta\}}(y_{(j)}-\theta)\\ &-\sum_{j\in\{j:x_{(j)}<\theta\}}(\theta-x_{(j)})-\sum_{j\in\{j:x_{(j)}\geq\theta\}}(x_{(j)}-\theta)) \end{split} \tag{8.12} \end{equation}\] From (8.12) we can see if there is at least one order statistic in sample points \(\mathbf{x}\) and \(\mathbf{y}\) that are differ, (8.12) will be depended on \(\theta\). Thus, order statistics are the minimal sufficient statistic.
Exercise 8.6 (Casella and Berger 6.12) A natural ancillary statistic in most problem is the sample size. For example, let N be a random variable taking values \(1,2,\cdots\) with known probabilities \(p_1,p_2,\cdots\), where \(\sum_{i=1}^{\infty}p_i=1\). Having observed \(N=n\), perform \(n\) Bernoulli trials with success probability \(\theta\), getting \(X\) successes.
Prove that the pair \((X,N)\) is minimal sufficient and N is ancillary for \(\theta\).
- Prove that the estimator \(\frac{X}{N}\) is unbiased for \(\theta\) and has variance \(\theta(1-\theta)E(1/N)\).
Proof. (a) For minimal sufficiency, consider two pairs of statistics \((x,n_1)\) and \((y,n_2)\) corresponding to two sample points \(\mathbf{x}\) and \(\mathbf{y}\). Then \[\begin{equation} \begin{split} \frac{p(x,n_1|\theta)}{p(y,n_2|\theta)}&=\frac{{n_1 \choose x}\theta^x(1-\theta)^{n_1-x}p_{n_1}}{{n_2 \choose y}\theta^y(1-\theta)^{n_2-y}p_{n_2}}\\ &\propto\theta^{x-y}(1\theta)^{(n_1-n_2)-(x-y)} \end{split} \tag{8.13} \end{equation}\] This is a constant w.r.t. \(\theta\) only when \(x=y\) and \(n_1=n_2\). Thus, the minimal sufficiency is proved.
For ancillarity, consider the distribution of \(N\), we have \[\begin{equation} f_N(n)=\sum_{x=0}^n{n \choose x}\theta^x(1-\theta)^{n-x}p_{n}=p_n \tag{8.14} \end{equation}\] Since the pdf of \(N\) does not depend on \(\theta\), it is an ancillary statistic for \(\theta\).
- Directly calculate the expectation we have \[\begin{equation} \begin{split} E(\frac{X}{N})&=\sum_{n=1}^{\infty}\sum_{x=0}^{n}\frac{x}{n}{n \choose x}\theta^x(1-\theta)^{n-x}p_{n}\\ &=\sum_{n=1}^{\infty}\frac{p_n}{n}\cdot \frac{n}{\theta}\\ &=\theta\sum_{n=1}^{\infty} p_n=\theta \end{split} \tag{8.15} \end{equation}\] Hence, this estimator is unbiased. For variance, we have \[\begin{equation} \begin{split} E((\frac{X}{N})^2)&=\sum_{n=1}^{\infty}\sum_{x=0}^{n}(\frac{x}{n})^2{n \choose x}\theta^x(1-\theta)^{n-x}p_{n}\\ &=\sum_{n=1}^{\infty}\frac{p_n}{n^2}\cdot (n\theta(1-\theta)+(n\theta)^2)\\ &=\theta(1-\theta)\sum_{n=1}^{\infty}\frac{p_n}{n}+\theta^2=\theta(1-\theta)E(\frac{1}{N})+\theta^2 \end{split} \tag{8.16} \end{equation}\] Hence, \[\begin{equation} Var(\frac{X}{N})=\theta(1-\theta)E(\frac{1}{N})+\theta^2-\theta^2=\theta(1-\theta)E(\frac{1}{N}) \tag{8.17} \end{equation}\] as we desired.
Exercise 8.7 (Casella and Berger 6.15) Let \(X_1,\cdots,X_n\) be i.i.d. \(N(\theta,a\theta^2)\) where a is a known constant and \(\theta>0\).
Show that the parameter space does not contain a two-dimensional open set.
Show that the statistic \(T=(\bar{X},S^2)\) is a sufficient statistic for \(\theta\), but the family of distributions is not complete.
Proof. (a) Since when mean parameter \(\theta\) is fixed, the variance parameter are also fixed as \(a\theta^2\). Hence, the parameter space is a quadratic curve on the plane truncated at \(\theta=0\). It does not contain any two-dimensional open set.
- Consider the joint pdf of \(X_1,\cdots,X_n\), we have \[\begin{equation} \begin{split} f(x_1,\cdots,x_n)&=(2\pi)^{-n/2}(a\theta^2)^{-n/2}exp(-\frac{1}{2a\theta^2}\sum_{i=1}^n(x_i-\theta)^2)\\ &=(2\pi)^{-n/2}(a\theta^2)^{-n/2}exp(-\frac{1}{2a\theta^2}[(n-1)S^2+n(\bar{x}-\theta)^2]) \end{split} \tag{8.18} \end{equation}\] By Factorization theorem, \(T=(\bar{X},S^2)\) is a sufficient statistic for \(\theta\).
Since \(E(S^2)=a\theta^2\) and \(E(\bar{X}^2)=\frac{a\theta^2}{n}+\theta^2\). Hence, \(E(\frac{an}{n+a}\bar{X}^2-S^2)=0\). Hence, if we choose \(g(\mathbf{T})=\frac{an}{n+a}\bar{X}^2-S^2\), then \(Eg(\mathbf{T})=0\) but does necessarily have \(g(\mathbf{T})=0\) almost surely. Thus, it is not complete.
Proof. Consider the joint pdf of \(X_1,\cdots,X_n\), we have \[\begin{equation} f(x_1,\cdots,x_n)=(\frac{\theta}{1-\theta})^n(1-\theta)^{\sum_{i=1}^nx_i} \tag{8.19} \end{equation}\] By Factorization Theorem, take \(h(x)=1\) we have \(\sum_{i=1}^nx_i\) is a sufficient statistic.
Intuitively, the sum of independent geometricly distributed random variables has a negative binominal distribution. \[\begin{equation} P(\sum_{i=1}^nX_i=m)={m-1 \choose n-1}\theta^n(1-\theta)^{m-n} \tag{8.20} \end{equation}\]
For completeness, using definition, consider the expectation of \(g(\mathbf{T})\), we have \[\begin{equation} Eg(\mathbf{T})=\sum_{t=n}^{\infty}g(t){t-1 \choose n-1}(\frac{\theta}{1-\theta})^n(1-\theta)^t \tag{8.21} \end{equation}\] which equals \(0\) only when \(g(t)=0\) becasue \(0<\theta<1\) and \({t-1 \choose n-1}(\frac{\theta}{1-\theta})^n(1-\theta)^t>0,\forall t\in\mathbb{N}\) and \(t\geq n\). Hence, \(\sum_{i=1}^nx_i\) is a complete statistic.Exercise 8.9 (Casella and Berger 6.22) Let \(X_1,\cdots,X_n\) be a random sample from a population with pdf \[\begin{equation} f(x|\theta)=\theta x^{\theta-1},0<x<1,\theta>0 \tag{8.22} \end{equation}\]
Is \(\sum_{i=1}^n X_i\) sufficient for \(\theta\)?
Find a complete sufficient statistic for \(\theta\).
Proof. (a) Consider the joint pdf of \(X_1,\cdots,X_n\), we have \[\begin{equation} f(x_1,\cdots,x_n)=\theta^n(\prod_{i=1}^nx_i)^{\theta-1} \tag{8.23} \end{equation}\] Thus, \(\prod_{i=1}^nx_i\) is a sufficient statistic, not \(\sum_{i=1}^n X_i\).
- Notice that \(f(x|\theta)=\theta e^{(\theta-1)\log(x)}\) belongs to exponential family and the support contains open set in \(\mathbb{R}\). Thus, by Theorem 6.2, \(\sum_{i=1}^n\log(X_i)=\log(\prod_{i=1}^nX_i)\) is a complete statistic. \(\log\) is an one-to-one transformation, hence \(\prod_{i=1}^nx_i\) is a complete statistic.
Exercise 8.10 (Casella and Berger 6.26) Use Minimal Sufficient Statistics Theorem (Theorem 8.1) to estabilish that given sample \(X_1,\cdots,X_n\), the following statistics are minimal sufficient.
Distribution: \(N(\theta,1)\), Statistic: \(\bar{X}\).
Distribution: \(Gamma(\alpha,\beta)\) with \(\alpha\) known, Statistic: \(\sum_{i=1}^nX_i\).
Distribution: \(Unif(0,\theta)\), Statistic: \(\max_iX_i\).
Distribution: \(Cauchy(\theta,1)\), Statistic: \(X_{(1)},\cdots,X_{(n)}\).
- Distribution: \(logistic(\mu,\beta)\), Statistic: \(X_{(1)},\cdots,X_{(n)}\).
Theorem 8.1 (Minimal Sufficient Statistics) Suppose that the family of densities \(\{f_0(\mathbf{x}),\cdots,f_k(\mathbf{x})\}\) all have common support. Then
The statistic \(T(\mathbf{X})=(\frac{f_1(\mathbf{X})}{f_0(\mathbf{X})},\frac{f_2(\mathbf{X})}{f_0(\mathbf{X})},\cdots,\frac{f_k(\mathbf{X})}{f_0(\mathbf{X})})\) is minimal sufficient for the family \(\{f_0(\mathbf{x}),\cdots,f_k(\mathbf{x})\}\).
If \(\mathcal{F}\) is a family of densities with common support, and
\(f_i(\mathbf{x})\in\mathcal{F},i=0,1,\cdots,k\),
\(T(\mathbf{x})\) is sufficient for \(\mathcal{F}\),
Exercise 8.11 (Casella and Berger 6.30) Let \(X_1,\cdots,X_n\) be a random sample from the pdf \(f(x|\mu)=e^{-(x-\mu)}\), where \(-\infty<\mu<x<\infty\).
Show that \(X_{(1)}=\min_{i}X_i\) is a complete sufficient statistic.
- Use Basu Theorem to show that \(X_{(1)}\) and \(S^2\) are independent.
Proof. Consider the joint pdf of \(X_1,\cdots,X_n\), \[\begin{equation} \begin{split} f(x_1,\cdots,x_n)&=\prod_{i=1}^nexp(-(x_i-\mu))I_{x_i>\mu}(x_i)\\ &=exp(-(\sum_{i=1}^nx_i-n\mu))I_{\min_ix_i>\mu}(\mathbf{x})\\ &=exp(-\sum_{i=1}^nx_i)exp(n\mu)I_{\min_ix_i>\mu}(\mathbf{x}) \end{split} \tag{8.24} \end{equation}\] Thus, by Factorization Theorem, \(X_{(1)}=\min_{i}X_i\) is a sufficient statistic.
As for completeness, consider the expectation of some function \(g(\mathbf{T})\), we have \[\begin{equation} Eg(\mathbf{T})=\int_{\mu}^{\infty}ne^{-n(y-\mu)}dy \tag{8.25} \end{equation}\] If \(Eg(\mathbf{T})=0\) then \(\int_{\mu}^{\infty}e^{-ny}dy=0\) for all \(\mu\). Taking derivatives w.r.t. \(\mu\), we have \[\begin{equation} -g(\mu)e^{-n\mu}=0 \tag{8.26} \end{equation}\] for all \(\mu\). Hence \(g(\mu)=0\) for all \(\mu\). \(X_{(1)}\) is also complete as we desire.
- By Basu Theorem, we only need to show that \(S^2\) is an ancillary statistic. Becasue \(f(x|\mu)\) is a location family. We can write \(X_i=Z_i+\mu\), where \(Z_i\) are samples from \(f(x|0)\). Also \(S^2=\frac{1}{n-1}\sum_{i=1}^n(Z_i-\bar{Z})^2\), whose distribution does not depend on \(\mu\). Hence, \(S^2\) is ancillary.
Exercise 8.14 (Casella and Berger 6.36) One advantage of using a minimal sufficient statistic is that unbiased estimators will have smaller variance, as the following exercise will show. Suppose \(T_1\) is sufficient and \(T_2\) is minimal sufficient, \(U\) is an unbiased estimator of \(\theta\), and define \(U_1=E(U|T_1)\) and \(U_2=E(U|T_2)\).
Show that \(U_2=E(U_1|T_2)\).
Now use the conditional variance formula to show that \(Var(U_2)\leq Var(U_1)\).
Proof. (a) Only consider discrete case here. Let us first proof the following lemma: if \(\mathbf{x}\) is a vector of ramdom variables with joint pmf \(f(\mathbf{x})\), then the joint distribution \(f(\mathbf{x},g(\mathbf{x}))\) is the same as \(f(\mathbf{x})\). This is straight forward for discrete random variables.
Noticing this lemma, assume samples \(\mathbf{X}=(X_1,\cdots,X_n)\) has joint pdf \(f(\mathbf{x}|\theta)\) and pdfs of statistic \(T_1\) and \(T_2\) are \(g_1(\cdot|\theta)\) and \(g_2(\cdot|\theta)\), respectively. Then the joint distribution of any statistic and \(\mathbf{X}\) is still \(f(\mathbf{x}|\theta)\) and the joint distribution of \(T_1\) and \(T_2\) is \(g_1(\cdot|\theta)\) since \(T_2\) is minimal sufficient. Suppose \(T_2=r(T_1)\). Therefore, firstly we have \[\begin{equation} E(U|T_1=t)=E(U(\mathbf{x})|T_1(\mathbf{x})=t)=\sum_{\mathbf{x}\in\mathcal{X}}U(\mathbf{x})f_1(\mathbf{x}|T_1=t) \tag{8.30} \end{equation}\] Here \(f_1(\mathbf{x}|T_1)\) denote the conditional pdf of \(\mathbf{x}\) given \(T_1\). Define \(A(t):=\{\mathbf{x}\in\mathcal{X}:T(\mathbf{x})=t\}\). Then (8.30) can be written as \[\begin{equation} E(U|T_1=t_1)=\sum_{\mathbf{x}\in A(t_1)}U(\mathbf{x})\frac{f(\mathbf{x}|\theta)}{g_1(t_1|\theta)} \tag{8.31} \end{equation}\] And similarily, we have \[\begin{equation} E(U|T_2=t_2)=\sum_{\mathbf{x}\in B(t_2)}U(\mathbf{x})\frac{f(\mathbf{x}|\theta)}{g_2(t_2|\theta)} \tag{8.32} \end{equation}\] Furthermore, consider \(E(U_1|T_2)\) we have \[\begin{equation} \begin{split} E(U_1|T_2=t_2)&=\sum_{t_1\in C(t_2)}\sum_{\mathbf{x}\in A(t_1)}U(\mathbf{x})\frac{f(\mathbf{x}|\theta)}{g_1(t_1|\theta)}\frac{g_1(t_1|\theta)}{g_2(t_2|\theta)} &=\sum_{t_1\in C(t_2)}\sum_{\mathbf{x}\in A(t_1)}U(\mathbf{x})\frac{f(\mathbf{x}|\theta)}{g_2(t_2|\theta)} \end{split} \tag{8.33} \end{equation}\] with \(C(t_2):\{t(\mathbf{x}):r(t(\mathbf{x}))=t_2\}\). Hence, we only left to show that \(\bigcup_{t1\in C(t_2)}A(t_1)=B(t_2)\) then we are done.
Consider sample point \(\mathbf{x}\in\mathcal{X}\), for any \(\mathbf{x}\in\bigcup_{t1\in C(t_2)}A(t_1)\), then \(T_1(\mathbf{x})=t_1\) and \(r(t_1)=t_2\), hence \(T_2(\mathbf{x})=r(T_1(\mathbf{x}))=t_2\). we have \(\mathbf{x}\in B(t_2)\). Therefore, \(\bigcup_{t1\in C(t_2)}A(t_1)\subseteq B(t_2)\). On the other hand, for any \(\mathbf{x}\in B(t_2)\), \(T_2(\mathbf{x})=t_2\) which implies \(r(T_1(\mathbf{x}))=t_2\). Denote \(T_1(\mathbf{x})=t_1^*\). Then we have \(t_1^*\in C(t_2)\) and \(\mathbf{x}\in A(t_1^*)\). Hence we have proved \(\mathbf{x}\in \bigcup_{t_1^*\in C(t_2)}A(t_1^*)=\bigcup_{t1\in C(t_2)}A(t_1)\). Hence \(B(t_2)\subseteq\bigcup_{t1\in C(t_2)}A(t_1)\) and finally \(\bigcup_{t1\in C(t_2)}A(t_1)= B(t_2)\).
Thus, (8.33) and (8.32) gives same result and we have what we want.
- Since \(U_2=E(U_1|T_2)\), using law of total variance we have \[\begin{equation} Var(U_1)=Var(E(U_1|T_2))+E(Var(U_1|T_2))\geq Var(E(U_1|T_2))=Var(U_2) \tag{8.34} \end{equation}\]
Exercise 8.15 (Casella and Berger 6.37) Joshi and Nabar examine properties of linear estimators for the parameter in the so-called “Problem of the Nile”, where \((X,Y)\) has the joint density \[\begin{equation} f(x,y|\theta)=exp\{-(\theta x+y/\theta)\},\quad x>0,y>0 \tag{8.35} \end{equation}\]
For an i.i.d. sample of size n, show that the Fisher information is \(I(\theta)=2n/\theta^2\).
For the estimators \(T=\sqrt{\sum Y_i/\sum X_i}\) and \(U=\sqrt{\sum X_i\sum Y_i}\), show that
the information in \(T\) alnoe is \([2n/(2n+1)]I(\theta)\).
the information in \((T,U)\) is \(I(\theta)\).
\((T,U)\) is jointly sufficient but not complete.
Proof. (a) For one observation \((X,Y)\) we have \[\begin{equation} I(\theta)=-E(\frac{\partial^2}{\partial\theta^2}\log f(X,Y|\theta))=\theta^{-3}E(2Y) \tag{8.36} \end{equation}\] Since \(Y\sim Exp(\theta)\), \(E(Y)=\theta\). Hence, for one data point \(I(\theta)=2/\theta^2\). Then for \(n\) data, \(I(\theta)=2n/\theta^2\).
- The cdf of T is \[\begin{equation} \begin{split} P(T\leq t)&=P(\frac{\sum Y_i}{\sum X_i}\leq t^2)=P(\frac{2\sum Y_i/\theta}{2\sum X_i\theta}\leq t^2/\theta^2)\\ &=P(F_{2n,2n}\leq t^2/\theta^2) \end{split} \tag{8.37} \end{equation}\] The last equation holds becasue \(2Y_i/\theta\) and \(2X_i\theta\) are all independent \(Exp(1)\) which is also \(\chi_2^2\). Thus, the density of \(T\) is \[\begin{equation} f_T(t)=\frac{\Gamma(2n)}{\Gamma{n}^2}\frac{2}{t}(\frac{t^2}{t^2+\theta^2})^n(\frac{\theta^2}{t^2+\theta^2})^n \tag{8.38} \end{equation}\] and the second derivatives w.r.t. \(\theta\) of the log density is \(\frac{2n}{\theta^2}(1-\frac{2}{(t^2/\theta^2+1)^2})\) and hence the information in \(T\) is \(\frac{2n}{\theta^2}[1-2 E(\frac{1}{F_{2n,2n}^2+1})^2]\). The expected value is \[\begin{equation} E(\frac{1}{F_{2n,2n}^2+1})^2=\frac{\Gamma(2n)}{\Gamma{n}^2}\int^{\infty}_0\frac{1}{(1+\omega)^2} \frac{\omega^{n-1}}{(1+\omega)^{2n}}d\omega=\frac{n+1}{2(2n+1)} \tag{8.39} \end{equation}\] Thus, the information of \(T\) is \([2n/(2n+1)]I(\theta)\).
For part (ii), let \(W=\sum X_i\) and \(V=\sum Y_i\). \(W\) and \(V\) are independent, and \(W\sim Gamma(n,\frac{1}{\theta})\), \(V\sim Gamma(n,\theta)\). Use this we can find joint distribution of \((T,U)\) as \[\begin{equation} f(t,u|\theta)=\frac{2}{\Gamma(n)^2t}u^{2n-1}exp(-\frac{u\theta}{t}-\frac{ut}{\theta}),\quad u>0,t>0 \tag{8.40} \end{equation}\] Thus, the ionformation is then \[\begin{equation} -E(-\frac{2UT}{\theta^3})=E(\frac{2V}{\theta^3})=\frac{2n\theta}{\theta^3}=I(\theta) \tag{8.40} \end{equation}\] as we desired.
For part (iii), the joint pdf of samples is \[\begin{equation} f(\mathbf{x},\mathbf{y})=exp(-\theta(\sum x_i)-(\sum y_i)/\theta) \tag{8.41} \end{equation}\] Hence \((W,V)\) is sufficient, so is \((T,U)\) since they are a one-to-one function of \((W,V)\).
However, since \(EU^2=E(WV)=(n/\theta)(n\theta)=n^2\), we can define function \(g(t,u)=u^2-n^2\) then \(E(g(\mathbf{T}))=0\) for all \(\theta\) does not imply \(g(\mathbf{T})=0\) almost surely. Thus, \((T,U)\) is not complete.