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\layout Title

Statistics A exam solutions (2003-2004)
\layout Author

David Rossell i Ribera
\layout Standard

I may have committed a number of errors in writing these solutions, but
 they should be fine for the most part.Use at your own risk!
\layout Section*

Statistics May 2003
\layout Section*

Problem 1
\layout Standard

a) 
\begin_inset Formula $g(p)$
\end_inset 

 is U-estimable iff 
\begin_inset Formula $\sum_{x=0}^{n}h(x)\left(\begin{array}{c}
n\\
x\end{array}\right)p^{x}(1-p)^{n-x}=g(p)$
\end_inset 

.
 The left hand side is a sum of polynomial of degree 
\begin_inset Formula $\leq n$
\end_inset 

, and hence the right hand side has to be a polynomial of degree 
\begin_inset Formula $\leq n$
\end_inset 

.
\layout Standard

b) From section a) we know that 
\begin_inset Formula $h(x)$
\end_inset 

 has to be such that the sumation equals 
\begin_inset Formula $p(1-p)^{2}=p-2p^{2}+p^{3}$
\end_inset 

.
 Hence it suffices to choose 
\begin_inset Formula $h(x)$
\end_inset 

 such that the coefficient affecting 
\begin_inset Formula $p$
\end_inset 

 in the left hand side is 
\begin_inset Formula $1$
\end_inset 

, for 
\begin_inset Formula $p^{2}$
\end_inset 

 it's 
\begin_inset Formula $-2$
\end_inset 

 and for 
\begin_inset Formula $p^{3}$
\end_inset 

 it's 
\begin_inset Formula $1$
\end_inset 

.
 To do this we need to get through some messy algebra to expand the left
 hand side expression.
 
\layout Standard

prova prova
\layout Standard

Hint: 
\begin_inset Formula $(1-p)^{n-x}=(1+(-p))^{n-x}=\sum_{k=0}^{n-x}\left(\begin{array}{c}
n-x\\
k\end{array}\right)1^{k}(-p)^{n-x-k}$
\end_inset 

 
\layout Standard

so 
\begin_inset Formula $\sum_{x=0}^{n}h(x)\left(\begin{array}{c}
n\\
x\end{array}\right)p^{x}(1-p)^{n-x}=$
\end_inset 


\begin_inset Formula $\sum_{x=0}^{n}\sum_{k=0}^{n-x}h(x)\left(\begin{array}{c}
n\\
x\end{array}\right)\left(\begin{array}{c}
n-x\\
k\end{array}\right)(-1)^{n-k}p^{n-k}$
\end_inset 


\layout Standard

c) 
\begin_inset Formula $g(x)=sin^{-1}(\sqrt{x})$
\end_inset 

 gives 
\begin_inset Formula $g'(x)=\frac{1}{2\sqrt{x}cos(sin^{-1}(x))}$
\end_inset 

, and thus 
\begin_inset Formula $\left(g'(x)\right)^{2}=\frac{1}{4xcos^{2}(sin^{-1}(x))}$
\end_inset 


\begin_inset Formula $=\frac{1}{4x\left(1-sin^{2}(sin^{-1}(x))\right)}=\frac{1}{4x\left(1-(\sqrt{x})^{2}\right)}$
\end_inset 


\begin_inset Formula $=\frac{1}{4x(1-x)}$
\end_inset 


\layout Standard

The delta method gives that 
\begin_inset Formula $\sqrt{n}\left(sin^{-1}\left(\sqrt{x/n}\right)-sin^{-1}\left(\sqrt{p}\right)\right)\rightarrow^{D}N\left(0,\frac{1}{4}\right)$
\end_inset 

.
\layout Section*

Problem 2
\layout Standard

a) Note that 
\begin_inset Formula $\int f_{\theta}(x)d\mu(x)=1\Rightarrow\int h(x)e^{\theta'T}d\mu(x)=e^{A(\theta)}$
\end_inset 

, which is convex.
 Hence 
\begin_inset Formula $e^{A(\alpha\theta_{1}+(1-\alpha)\theta_{2})}\leq\alpha e^{A(\theta_{1})}+(1-\alpha)e^{A(\theta_{2})}<\infty$
\end_inset 

.
\layout Standard

b) First, let's derivate w.r.t.
 
\begin_inset Formula $\theta_{i}$
\end_inset 

.
 
\begin_inset Formula $\int h(x)e^{\theta'T}d\mu(x)=e^{A(\theta)}\Rightarrow\int h(x)e^{\theta'T}T_{i}(x)d\mu(x)=e^{A(\theta)}\frac{d}{d\theta_{i}}A(\theta)$
\end_inset 


\begin_inset Formula $\Rightarrow\frac{d}{d\theta_{i}}A(\theta)=\int h(x)e^{\theta'T-A(\theta)}T_{i}(x)d\mu(x)=E(T_{i}(X))$
\end_inset 

.
\layout Standard

Now let's derivate again w.r.t.
 
\begin_inset Formula $\theta_{j}$
\end_inset 

.
 
\begin_inset Formula $\frac{d^{2}}{d\theta_{i}d\theta_{j}}A(\theta)=\frac{d}{d\theta_{j}}\int h(x)e^{\theta'T-A(\theta)}T_{i}(x)d\mu(x)$
\end_inset 


\begin_inset Formula $=\int h(x)e^{\theta'T-A(\theta)}(T_{j}(x)-\frac{d}{d\theta_{j}}A(\theta))T_{i}(x)d\mu(x)$
\end_inset 


\begin_inset Formula $=E(T_{i}(X)T_{j}(X))-E(T_{i}(X))E(T_{j}(X))=COV(T_{i}(X),T_{j}(X))$
\end_inset 

.
\layout Standard

c) The likelihood is 
\begin_inset Formula $f_{p}(x)=\frac{n!}{\prod x_{i}!}exp\left\{ \sum_{i=0}^{r}x_{i}logp_{i}\right\} $
\end_inset 


\begin_inset Formula $=\frac{n!}{\prod x_{i}!}exp\left\{ nlogp_{0}+\sum_{i=0}^{r}x_{i}log\frac{p_{i}}{p_{0}}\right\} $
\end_inset 

.
 Now let 
\begin_inset Formula $\eta_{i}=log\frac{p_{i}}{p_{0}}\Rightarrow p_{0}=(1+\sum_{i=1}^{r}e^{\eta_{i}})$
\end_inset 

, so we get
\layout Standard


\begin_inset Formula $f_{\eta}(x)=\frac{n!}{\prod x_{i}!}exp\left\{ \sum x_{i}\eta_{i}-nlog(1+\sum e^{\eta_{i}})\right\} $
\end_inset 

.
 Hence 
\begin_inset Formula $A(\eta)=nlog(1+\sum e^{\eta_{i}})$
\end_inset 


\begin_inset Formula $\Rightarrow\frac{d}{d\eta_{i}}A(\eta)=\frac{ne^{\eta_{i}}}{1+\sum e^{\eta_{i}}}=np_{i}$
\end_inset 

.
\layout Standard

And so on...
\layout Section*

Problem 3
\layout Standard

a) Easy
\layout Standard

b) 
\begin_inset Formula $E($
\end_inset 


\begin_inset Formula $\hat{\sigma}^{2})=E\left(\frac{n-1}{n}S^{2}\right)=\frac{1}{n}E((n-1)S^{2})=\frac{1}{n}\sigma^{2}(n-1)$
\end_inset 

, so 
\begin_inset Formula $S^{2}=\frac{n}{n-1}\hat{\sigma}^{2}$
\end_inset 

 is unbiased and a function of the complete & sufficient statistic 
\begin_inset Formula $\Rightarrow$
\end_inset 

 UMVUE by Lehmann-Scheffe
\layout Standard

c) 
\begin_inset Formula $f(\mu,\sigma^{2})\propto(2\tau)^{n}e^{-\tau\sum(x_{i}-\mu)^{2}}\frac{\alpha^{g}}{\Gamma(g)}\tau^{g-1}e^{-\alpha\tau}\propto...$
\end_inset 


\begin_inset Formula $\propto\tau^{n+g-1}e^{-\tau(\alpha+\sum x_{i}^{2})}e^{-n\tau(\mu-\overline{X})^{2}}$
\end_inset 

, so 
\begin_inset Formula $f(\mu|\tau,x)$
\end_inset 

 is 
\begin_inset Formula $N(\bar{X},n\tau)$
\end_inset 

 and 
\begin_inset Formula $f(\tau|x)$
\end_inset 

 is 
\begin_inset Formula $Gamma(n+g,\alpha+\sum x_{i}^{2})$
\end_inset 

.
 The Bayes estimator is the posterior mean, 
\begin_inset Formula $\frac{n+g}{\alpha+\sum x_{i}^{2}}$
\end_inset 

.
\layout Standard

d) Under Squared Error Loss we need to consider the Bias & Variance of each
 estimator
\layout Standard

For 
\begin_inset Formula $\hat{\sigma}^{2}$
\end_inset 

.
 Bias is 
\begin_inset Formula $\sigma^{2}/n$
\end_inset 

 and variance 
\begin_inset Formula $\frac{\sigma^{4}2(n-1)}{n^{2}}$
\end_inset 

, so 
\begin_inset Formula $MSE=\sigma^{4}\left(\frac{2n-1}{n^{2}}\right)$
\end_inset 

.
\layout Standard

For 
\begin_inset Formula $S^{2}$
\end_inset 

.
 Bias is 
\begin_inset Formula $0$
\end_inset 

 and variance is 
\begin_inset Formula $\frac{2\sigma^{4}}{n-1}$
\end_inset 

, so 
\begin_inset Formula $MSE=\frac{2\sigma^{4}}{n-1}$
\end_inset 

.
\layout Standard

Since 
\begin_inset Formula $\frac{2n-1}{n^{2}}<\frac{2}{n-1}=\frac{2n}{n^{2}-n}$
\end_inset 

, 
\begin_inset Formula $S^{2}$
\end_inset 

 is inadmissible under Squared Error Loss.
\layout Section*

Problem 4
\layout Standard

a) Neymann-Pearson lemma.
\layout Standard

b) Let 
\begin_inset Formula $\theta_{1}<\theta_{2}$
\end_inset 

.
 The Likelihood Ratio takes the form 
\begin_inset Formula $\frac{\theta_{2}^{x}(1-\theta_{2})^{n-x}}{\theta_{1}^{x}(1-\theta_{1})^{n-x}}$
\end_inset 

, which is increasing with 
\begin_inset Formula $x$
\end_inset 

.
 Hence a MP test will reject for large values of 
\begin_inset Formula $x$
\end_inset 

.
\layout Standard


\begin_inset Formula $\phi(x)=\left\{ \begin{array}{cc}
1 & ,x>c\\
\beta & ,x=c\\
0 & ,x<c\end{array}\right.$
\end_inset 

, where 
\begin_inset Formula $P_{1/2}(X>c)+\beta P_{1/2}(X=c)$
\end_inset 

=
\begin_inset Formula $\alpha$
\end_inset 

.
\layout Standard

Playing with different values we get 
\begin_inset Formula $c=3$
\end_inset 

 and 
\begin_inset Formula $\beta=0.14$
\end_inset 

.
\layout Standard

c) By Karlin-Rubin the test in section b) is UMP.
\layout Standard

d) Denote 
\begin_inset Formula $\phi_{1}(X)$
\end_inset 

 the test defined in section b), which is the essentially unique UMP in
 the region 
\begin_inset Formula $\theta\in(\frac{1}{2},1]$
\end_inset 

.
 Hence the test 
\begin_inset Formula $\phi_{1}(X)$
\end_inset 

 is the only possible UMP, since no other test beats its in the region 
\begin_inset Formula $\theta\in(\frac{1}{2},1]$
\end_inset 

.
 We'll show there's no UMP by seeing that 
\begin_inset Formula $\phi_{1}(X)$
\end_inset 

 is no UMP.
\layout Standard

Further the power 
\begin_inset Formula $\gamma_{1}(\theta)$
\end_inset 

 of this test increases with 
\begin_inset Formula $\theta$
\end_inset 

 (one of the corollaries of Karlin-Rubin gives this), and hence 
\begin_inset Formula $\gamma_{1}(\theta)<\alpha$
\end_inset 

 for 
\begin_inset Formula $\theta<1/2$
\end_inset 

.
 But the test 
\begin_inset Formula $\phi_{2}(X)=a$
\end_inset 

 has greater power and it's level 
\begin_inset Formula $\alpha$
\end_inset 

, and hence there's no UMP.
\layout Section*

Problem 5
\layout Standard

a) 
\begin_inset Formula $\Rightarrow)$
\end_inset 

 Let 
\begin_inset Formula $Y=|X|$
\end_inset 

.
 
\begin_inset Formula $E(Y)=\int Y^{p}dP=\int_{\{ Y\leq y\}}Y^{p}dP+\int_{\{ Y>y\}}Y^{p}dP$
\end_inset 


\begin_inset Formula $\geq\int_{\{ Y\leq y\}}Y^{p}dP+y^{p}P(Y\geq y)$
\end_inset 


\begin_inset Formula $\Rightarrow y^{p}P(Y\geq y)\leq E(Y)-\int_{\{ Y\leq y\}}Y^{p}dP$
\end_inset 


\begin_inset Formula $\rightarrow_{y\rightarrow\infty}0$
\end_inset 

, as desired.
\layout Standard


\begin_inset Formula $\Leftarrow)$
\end_inset 

 
\begin_inset Formula $E(|X|^{p-\epsilon})=\int(p-\epsilon)|x|^{p-\epsilon-1}P(|X|>x)dx=$
\end_inset 


\begin_inset Formula $\int_{0}^{M}(p-\epsilon)|x|^{p-\epsilon-1}P(|X|>x)dx+\int_{M}^{\infty}(p-\epsilon)|x|^{-(1+\epsilon)}|x|^{p}P(|X|>x)dx$
\end_inset 


\begin_inset Formula $<\{|x|^{p}P(|X|>x)<N$
\end_inset 

 for large enough 
\begin_inset Formula $N\}$
\end_inset 


\begin_inset Formula $<\int_{0}^{M}(p-\epsilon)|x|^{p-\epsilon-1}P(|X|>x)dx+N\int_{M}^{\infty}(p-\epsilon)|x|^{-(1+\epsilon)}dx<\infty$
\end_inset 

, since the left term is the integral of a bounded function in a bounded
 interval, and the right term is finite since 
\begin_inset Formula $|x|$
\end_inset 

 is raised to the negative of a number greater than one.
\layout Standard

b) 
\begin_inset Formula $\hat{\beta}_{n}=\bar{X}/3$
\end_inset 

.
 By Central Limit Theorem, 
\begin_inset Formula $\sqrt{n}(\bar{X}-3\beta)\rightarrow N(0,3\beta^{2})$
\end_inset 

 and by Delta Method 
\begin_inset Formula $\sqrt{n}(\bar{X}/3-\beta)\rightarrow N(0,\sigma^{2})$
\end_inset 

, where 
\begin_inset Formula $\sigma^{2}=3\beta^{2}\left(\frac{d}{d\beta}(\beta/3)\right)^{2}=\beta^{2}/3$
\end_inset 

.
\layout Standard

c) 
\begin_inset Formula $P\left(-z_{\alpha/2}<\frac{\sqrt{n}(\bar{X}-3\beta)}{\sqrt{\hat{\beta}_{n}/3}}<z_{\alpha/2}\right)\rightarrow1-\alpha$
\end_inset 

 by Slutsky and we obtain
\layout Standard


\begin_inset Formula $P\left(\bar{X}/3-z_{\alpha/2}\sqrt{\hat{\beta}_{n}/3n}<\beta<\bar{X}/3+z_{\alpha/2}\sqrt{\hat{\beta}_{n}/3n}\right)\rightarrow1-\alpha$
\end_inset 

.
\layout Section*

Problem 6
\layout Standard

a) MGF of Gamma is 
\begin_inset Formula $M_{X}(t)=(1-\beta t)^{\alpha}$
\end_inset 

 and hence 
\begin_inset Formula $M_{X_{1}+X_{2}}(t)=(1-\beta t)^{\alpha_{1}+\alpha_{2}}$
\end_inset 


\begin_inset Formula $\Rightarrow X_{1}+X_{2}\sim Gamma(\alpha_{1}+\alpha_{2},\beta)$
\end_inset 


\layout Standard

b) Let 
\begin_inset Formula $\left\{ \begin{array}{cc}
Y_{1}=X_{1}/X_{2} & \Rightarrow X_{1}=Y_{1}Y_{2}\\
Y_{2}=X_{2} & \Rightarrow X_{2}=Y_{2}\end{array}\right.$
\end_inset 

.
 The Jacobian is 
\begin_inset Formula $|J|=Y_{2}$
\end_inset 

 and hence 
\begin_inset Formula $f_{Y_{1}Y_{2}}(y_{1},y_{2})=f_{X_{1}X_{2}}(y_{1}y_{2},y_{2})y_{2}$
\end_inset 

and
\begin_inset Formula $f_{Y_{1}}(y_{1})=\frac{\Gamma(\alpha_{1}+\alpha_{2}-1)}{\Gamma(\alpha_{1})\Gamma(\alpha_{2})\beta}\frac{y_{1}^{\alpha_{1}-1}}{(1+y_{1})^{\alpha_{1}+\alpha_{2}-1}}I_{R^{+}}(y_{1})$
\end_inset 


\layout Section*

Problem 7
\layout Standard

a) Usual definition.
\layout Standard

b) Proof of Basu's theorem.
\layout Standard

c) Fix 
\begin_inset Formula $\sigma^{2}\in R^{+}$
\end_inset 

.
 Since 
\begin_inset Formula $\frac{S^{2}}{\sigma^{2}}(n-1)\sim\chi_{n-1}^{2}$
\end_inset 

, 
\begin_inset Formula $S^{2}$
\end_inset 

is ancillary for 
\begin_inset Formula $\mu$
\end_inset 

.
 Also we know that 
\begin_inset Formula $\bar{X}$
\end_inset 

 is complete & sufficient for 
\begin_inset Formula $\mu$
\end_inset 

 (check that the exponential family is full rank).
 Then, by Basu's theorem since 
\begin_inset Formula $\bar{X}$
\end_inset 

 complete & sufficient for 
\begin_inset Formula $\mu$
\end_inset 

 and 
\begin_inset Formula $S^{2}$
\end_inset 

 ancillary for 
\begin_inset Formula $\sigma^{2}$
\end_inset 

 we have that 
\begin_inset Formula $\bar{X}$
\end_inset 

 and 
\begin_inset Formula $S^{2}$
\end_inset 

 are independent.
\layout Standard

Now, 
\begin_inset Formula $\sigma^{2}$
\end_inset 

 was arbitrary so the result holds for all 
\begin_inset Formula $\sigma^{2}$
\end_inset 

.
\layout Section*

Problem 8
\layout Standard

a) The Poisson belongs to the 1 parameter exponential family, so it has
 the MLR property.
 By Karlin Rubin the UMP test will reject for large values of the sufficient
 statistic, which is 
\begin_inset Formula $\sum X_{i}$
\end_inset 

.
 We know that 
\begin_inset Formula $\sum X_{i}\sim Poisson(n\lambda)$
\end_inset 

.
 Let 
\begin_inset Formula $k$
\end_inset 

 be the 
\begin_inset Formula $max\left\{ x:P_{\lambda_{0}}(\sum X_{i}<x)\leq\alpha\right\} $
\end_inset 

.
 Then the following test is UMP
\layout Standard


\begin_inset Formula $\phi(X)=\left\{ \begin{array}{cc}
1 & ,\sum X_{i}<k\\
\beta & ,\sum X_{i}=k\\
0 & ,\sum X_{i}>k\end{array}\right.$
\end_inset 

, where 
\begin_inset Formula $\beta=\frac{\alpha-P_{\lambda_{0}}(\sum X_{i}<k)}{P_{\lambda_{0}}(\sum X_{i}=x)}$
\end_inset 


\layout Standard

b) A conservative P-value would be 
\begin_inset Formula $P_{\lambda_{0}}(\sum X_{i}<\sum x_{i})$
\end_inset 

, where 
\begin_inset Formula $\sum x_{i}$
\end_inset 

 is the observed value of the suff stat.
 A P-value is a test statistic with rejection region 
\begin_inset Formula $[0,\alpha]$
\end_inset 

.
\layout Section*

Statistics January 2004
\layout Section*

Problem 1
\layout Standard

a) MGF is 
\begin_inset Formula $M_{X}(t)=(1-\beta t)^{\alpha}$
\end_inset 

, so 
\begin_inset Formula $M_{X_{1}+X_{2}}(t)=(1-\beta t)^{\alpha_{1}+\alpha_{2}}$
\end_inset 


\begin_inset Formula $\Rightarrow X_{1}+X_{2}\sim Gamma(\alpha_{1}+\alpha_{2},\beta)$
\end_inset 


\layout Standard

b) Let 
\begin_inset Formula $\left\{ \begin{array}{cc}
W_{1}=\frac{X_{1}}{X_{1}+X_{2}} & \Rightarrow X_{1}=W_{1}W_{2}\\
W_{2}=X_{1}+X_{2} & \Rightarrow X_{2}=W_{2}(1-W_{1})\end{array}\right.$
\end_inset 

, so the Jacobian is 
\begin_inset Formula $|J|=w_{2}$
\end_inset 

 and 
\begin_inset Formula $f_{w_{1}w_{2}}(w_{1},w_{2})=f_{x_{1}x_{2}}(w_{1}w_{2},w_{2}(1-w_{1}))w_{2}$
\end_inset 

.
 By integrating wrt 
\begin_inset Formula $w_{2}$
\end_inset 

 we find that 
\begin_inset Formula $f_{w_{1}}(w_{1})=\frac{\Gamma(\alpha_{1}+\alpha_{2})}{\Gamma(\alpha_{1})\Gamma(\alpha_{2})}w_{1}^{\alpha_{1}-1}(1-w_{1})^{\alpha_{2}-1}I_{(0,1)}(w_{1})$
\end_inset 

.
\layout Standard

c) 
\begin_inset Formula $P(R=r)=P(\frac{1}{Z}=r)=2^{-1/r}$
\end_inset 

 for 
\begin_inset Formula $r=1,\frac{1}{2},\frac{1}{3}...$
\end_inset 


\layout Section*

Problem 2
\layout Standard

a) Use Cauchy-Schwartz inequality together with some regularity conditions.
\layout Standard

b) 
\begin_inset Formula $f_{\theta}(x)\propto\theta^{x}(1-\theta)^{n-x}$
\end_inset 

 so 
\begin_inset Formula $\frac{d}{d\theta}logf_{\theta}(x)=0\Rightarrow\hat{\theta}=x/n$
\end_inset 

.
 Then 
\begin_inset Formula $E\left(\frac{x}{\theta}-\frac{n-x}{1-\theta}\right)^{2}=\frac{n}{\theta(1-\theta)}$
\end_inset 

 and hence 
\begin_inset Formula $V_{\theta}(T(X))\geq\frac{\theta(1-\theta)}{n}$
\end_inset 

 and 
\begin_inset Formula $V_{\theta}(\bar{X})=\frac{V(X)}{n^{2}}=\frac{\theta(1-\theta)}{n}$
\end_inset 

 attains the Cramer-Rao LB.
\layout Section*

Problem 3
\layout Standard

a) 
\begin_inset Formula $\bar{X}=\frac{1}{n}\sum X_{i}$
\end_inset 

 is a linear combination of independent & normal rv's and hence it's normally
 distributed with mean 
\begin_inset Formula $\mu$
\end_inset 

 and variance 
\begin_inset Formula $1/n$
\end_inset 

.
 Equivalently 
\begin_inset Formula $\frac{\bar{X}-\mu}{1/\sqrt{n}}\sim N(0,1)$
\end_inset 

, and the rejection region is given by finding 
\begin_inset Formula $z_{1},z_{2}$
\end_inset 

 such that 
\begin_inset Formula $P_{H_{0}}(z_{1}<Z<z_{2})=1-\alpha$
\end_inset 

, e.g.
 
\begin_inset Formula $z_{1}=-z_{\alpha/2}$
\end_inset 

 and 
\begin_inset Formula $z_{2}=z_{\alpha/2}$
\end_inset 

.
\layout Standard

b) 
\begin_inset Formula $\gamma(\mu)=1-P_{H_{0}}(-z_{\alpha/2}<Z<z_{\alpha/2})$
\end_inset 

.
 For general 
\begin_inset Formula $\mu$
\end_inset 

, 
\begin_inset Formula $\bar{X}\sim N(\mu,1/n)\Rightarrow Z=\sqrt{n}(\bar{X}-10)\sim N(\sqrt{n}(\mu-10),1)$
\end_inset 

, and hence we can use the quantiles of this latter normal distribution
 to get the desired result.
\layout Standard

c) Not UMP because the UMP for 
\begin_inset Formula $\mu>10$
\end_inset 

 rejects for large values of 
\begin_inset Formula $X$
\end_inset 

 (by Karlin-Rubin) and for 
\begin_inset Formula $\mu<10$
\end_inset 

 rejects for small values of 
\begin_inset Formula $X$
\end_inset 

, i.e.
 the UMP test for 
\begin_inset Formula $\mu>10$
\end_inset 

 is different than the UMP test for 
\begin_inset Formula $\mu<10$
\end_inset 

 and since they're both unique there can be no UMP test.
\layout Standard

d) LR test rejects for large values of 
\begin_inset Formula $\frac{e^{-\frac{1}{2}\sum(x_{i}-\bar{X})^{2}}}{e^{-\frac{1}{2}\sum(x_{i}-10)^{2}}}$
\end_inset 

 or equivalently for large 
\begin_inset Formula $\sum(x_{i}-10)^{2}-\sum(x_{i}-\bar{X})^{2}$
\end_inset 

.
 With some algebra we can see that this is equivalent to rejecting for large
 
\begin_inset Formula $|Z|$
\end_inset 

.
 Rejecting for 
\begin_inset Formula $|Z|>z_{\alpha/2}$
\end_inset 

gives a level 
\begin_inset Formula $\alpha$
\end_inset 

 test.
\layout Section*

Problem 4
\layout Standard

Times series, not necessary.
\layout Section*

Problem 5
\layout Standard

a) Factorization theorem.
\layout Standard

b) By Rao-Blackwell any unbiased estimator can be improved by finding its
 exectation conditional on the suff stat, and so it follows that the best
 unbiased estimator has to be a function of the suff .
 Note that the UMVU may not be unique if we don't have completeness.
\layout Standard

Now, noting that the likelihood will be of the form 
\begin_inset Formula $log\left(g(T(x),\theta\right)+log(h(x))$
\end_inset 

, we can drop the second term since it doesn't depend on 
\begin_inset Formula $\theta$
\end_inset 

 and maximize 
\begin_inset Formula $log\left(g(T(x),\theta)\right)$
\end_inset 

 which will depend on the data only through 
\begin_inset Formula $T(x)$
\end_inset 

, the sufficient stat.
\layout Standard

The Bayes estimator is of the form 
\begin_inset Formula $E(\theta|X)=\int\theta f(\theta|X)d\theta$
\end_inset 

 and 
\begin_inset Formula $f(\theta|X)=\frac{g(T(x),\theta)h(x)\pi(\theta)}{\int g(T(x),\theta)h(x)\pi(\theta)d\theta}$
\end_inset 


\begin_inset Formula $=\frac{g(T(x),\theta)\pi(\theta)}{\int g(T(x),\theta)\pi(\theta)d\theta}$
\end_inset 

 which depends on the data only through 
\begin_inset Formula $T(x)$
\end_inset 

.
\layout Standard

c) 
\begin_inset Formula $lim_{n\rightarrow\infty}E_{\theta}(\delta_{n}(x))=\theta$
\end_inset 

, for all 
\begin_inset Formula $\theta\in\Theta$
\end_inset 

.
\layout Standard

d) (i) Likelihood is 
\begin_inset Formula $f_{\mu}(x)=(2\pi)^{-n/2}e^{-\frac{1}{2}\sum(x_{i}-\mu_{i})^{2}}$
\end_inset 

, and taking log and then derivative and setting equal to zero gives 
\begin_inset Formula $x_{i}=\mu_{i}$
\end_inset 

.
 Hence 
\begin_inset Formula $(\hat{\mu}_{1}...\hat{\mu}_{n})=(x_{1}...x_{n})$
\end_inset 

 and by the invariance property of MLEs we get 
\begin_inset Formula $\hat{\theta}_{n}=\frac{1}{n}\sum\hat{\mu}_{i}^{2}=\frac{1}{n}\sum x_{i}^{2}$
\end_inset 

.
\layout Standard

(ii) We can find 
\begin_inset Formula $E(X_{j}^{2})=\mu_{j}^{2}+1$
\end_inset 

 by using the 2nd derivative of the characteristic function of 
\begin_inset Formula $X_{j}$
\end_inset 

.
 Hence 
\begin_inset Formula $E(\hat{\theta}_{n})=1+\frac{1}{n}\sum\mu_{j}^{2}$
\end_inset 

 and thus 
\begin_inset Formula $\theta_{n}^{*}$
\end_inset 

 is unbiased.
 Now, we know that 
\begin_inset Formula $(x_{1}...x_{n})$
\end_inset 

 is a sufficient statistic and it can be seen that the exponential family
 is full rank, so the sufficient statistic is also complete.
 Hence Lehman-Scheffe applies and 
\begin_inset Formula $\theta_{n}^{*}$
\end_inset 

 is UMVU.
 Also, 
\begin_inset Formula $V(\theta_{n}^{*})=V(\hat{\theta}_{n}-1)=V(\hat{\theta}_{n})$
\end_inset 

 and 
\begin_inset Formula $Bias(\theta_{n}^{*})<Bias(\hat{\theta}_{n})$
\end_inset 

, so 
\begin_inset Formula $\hat{\theta}_{n}$
\end_inset 

 is inadmissible under squared error loss.
\layout Standard

(iii) If we can show that 
\begin_inset Formula $\theta_{n}^{*}\rightarrow^{P}\theta$
\end_inset 

, then 
\begin_inset Formula $\hat{\theta}_{n}\rightarrow\theta+1$
\end_inset 

 by the Continuous Mapping Principle.
 I'm not sure how to do it.
\layout Section*

Problem 6
\layout Standard

Easy but long!
\layout Section*

Problem 7
\layout Standard

a) The original model is 
\begin_inset Formula $y_{i}\sim N(\beta_{0}+\beta_{1}x_{i},\sigma^{2}\rho_{i})$
\end_inset 

 independent, so 
\begin_inset Formula $E(y_{i}/\rho_{i})=\beta_{0}/\rho_{i}+\beta_{1}x_{i}\rho_{i}+E(\epsilon_{i}/\rho_{i})$
\end_inset 


\begin_inset Formula $\Rightarrow E(\delta_{i})=E(\epsilon_{i}/\rho_{i})=0$
\end_inset 

.
 Hence 
\begin_inset Formula $\delta_{i}$
\end_inset 

 are independent with zero means and variance 
\begin_inset Formula $\sigma^{2}/\rho_{i}$
\end_inset 

.
\layout Standard

b) The usual least squares estimators give 
\begin_inset Formula $\hat{\beta}_{1}=\frac{\sum(v_{i}-\bar{v})(z_{i}-\bar{z})}{\sum(v_{i}-\bar{v})^{2}}$
\end_inset 

 where 
\begin_inset Formula $\bar{v}$
\end_inset 

 and 
\begin_inset Formula $\bar{z}$
\end_inset 

 are the sample means, and 
\begin_inset Formula $\hat{\beta}_{0}=\bar{z}-\hat{\beta}_{1}\bar{v}$
\end_inset 

.
 With some algebra we can see that they're not unbiased.
\layout Standard

c) and d) Long!
\layout Section*

Problem 8
\layout Standard

Biostatistics.
\the_end