Chapter8_exercise.lyx

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\begin_body

\begin_layout Chapter*
8.2
\end_layout

\begin_layout Exercise*
13.
 Prove 
\begin_inset Formula $\hat{\sigma^{2}}=\frac{1}{n}\sum_{i=1}^{n}(X_{i}-\mu)^{2}$
\end_inset

 is gamma distribution with parameters 
\begin_inset Formula $n/2$
\end_inset

 and 
\begin_inset Formula $n/(2\sigma^{2})$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Solution*
\begin_inset Formula $Y=\hat{\sigma^{2}}=\frac{\sigma^{2}}{n}\sum_{i=1}^{n}\frac{(X_{i}-\mu)^{2}}{\sigma^{2}}=\frac{\sigma^{2}}{n}Z$
\end_inset

, we can see 
\begin_inset Formula $Z$
\end_inset

 is 
\begin_inset Formula $\chi_{n}^{2}$
\end_inset

 distribution.
\end_layout

\begin_layout Solution*
We want to consider distribution of 
\begin_inset Formula $Y$
\end_inset

: 
\begin_inset Formula $g(Y)$
\end_inset

.
\end_layout

\begin_layout Solution*
Following property of distribution function of variable 
\series bold
Theorem 3.8.4 
\series default
: p.d.f.
 of 
\begin_inset Formula $Y=r(Z)$
\end_inset

 is 
\begin_inset Formula $g(Y)=f(s(Y))|ds(y)/dy|$
\end_inset

.
 (
\begin_inset Formula $s(r(z))=z$
\end_inset

)
\end_layout

\begin_layout Solution*
\begin_inset Formula $f(Z)=\gamma(\alpha=\frac{n}{2},\beta=\frac{1}{2})=\frac{\beta^{\alpha}}{\Gamma(\alpha)}x^{\alpha-1}e^{-\beta x}=\frac{1}{2^{n/2}\Gamma(n/2)}x^{(n/2)-1}e^{-x/2}$
\end_inset


\end_layout

\begin_layout Solution*
\begin_inset Formula $g(Y)=f(\frac{n}{\sigma^{2}}Y)\frac{n}{\sigma^{2}}=\frac{n}{\sigma^{2}}\frac{1}{2^{n/2}\Gamma(n/2)}\frac{nx}{\sigma^{2}}^{(n/2)-1}e^{-(nx/\sigma^{2})/2}=(\frac{n}{2\sigma^{2}})^{n/2}\frac{1}{\Gamma(n/2)}x^{n/2-1}e^{-nx/(2\sigma^{2})}\sim\gamma(\frac{n}{2},\frac{n}{2\sigma^{2}})$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Standard
We have concluded: If 
\begin_inset Formula $Z$
\end_inset

 has 
\begin_inset Formula $\chi_{n}^{2}$
\end_inset

 distribution, so 
\begin_inset Formula $\frac{\sigma^{2}}{n}Z$
\end_inset

 has distribution: 
\begin_inset Formula $\gamma(\frac{n}{2},\frac{n}{2\sigma^{2}})$
\end_inset


\end_layout

\begin_layout Chapter*
8.3 Joint Distribution of the Sample Mean and Sample Variance
\end_layout

\begin_layout Exercise*
1.
 
\begin_inset Formula $X_{1},...,X_{n}$
\end_inset

 form a random sample from the normal distribution with mean 
\begin_inset Formula $\mu$
\end_inset

 and variance 
\begin_inset Formula $\sigma^{2}$
\end_inset

.
 Show that 
\begin_inset Formula $\hat{\sigma^{2}}$
\end_inset

 has gamma distribution with parameters 
\begin_inset Formula $(n-1)/2$
\end_inset

 and 
\begin_inset Formula $n/(2\sigma^{2})$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Solution*
We had 
\begin_inset Formula $Z=n\hat{\sigma^{2}}/\sigma^{2}$
\end_inset

 has a chi-square distribution with 
\begin_inset Formula $n-1$
\end_inset

 degrees of freedom.
 
\begin_inset Formula $Y=\hat{\sigma^{2}}=\frac{\sigma^{2}}{n}Z$
\end_inset


\end_layout

\begin_layout Solution*
\begin_inset Formula $f(z)=\chi_{n-1}^{2}=\frac{1}{2^{(n-1)/2}\Gamma[(n-1/2]}x^{[(n-1)/2]-1}e^{-x/2}$
\end_inset


\end_layout

\begin_layout Solution*
\begin_inset Formula $g(Y)=g(\frac{n}{\sigma^{2}}Y)\frac{n}{\sigma^{2}}=\frac{n}{\sigma^{2}}\frac{1}{2^{(n-1)/2}\Gamma[(n-1/2]}(\frac{n}{\sigma^{2}}x)^{[(n-1)/2]-1}e^{-nx/(2\sigma^{2})}$
\end_inset


\end_layout

\begin_layout Solution*
\begin_inset Formula $=(\frac{n}{2\sigma^{2}})^{(n-1)/2}\frac{1}{\Gamma[(n-1)/2]}e^{-nx/(2\sigma^{2})}$
\end_inset


\end_layout

\begin_layout Solution*
\begin_inset Formula $=\Gamma(\frac{n-1}{2},\frac{n}{2\sigma^{2}})$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
2.
 We compare 
\begin_inset Formula $X^{-1}$
\end_inset

 and 
\begin_inset Formula $X^{T}$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
4.
 
\begin_inset Formula $X_{1},X_{2},X_{3}$
\end_inset

 are i.i.d.
 and that each has standard normal disbution.
 Also, suppose that 
\end_layout

\begin_layout Exercise*
\begin_inset Formula 
\[
Y_{1}=0.8X_{1}+0.6X_{2}
\]

\end_inset


\end_layout

\begin_layout Exercise*
\begin_inset Formula 
\[
Y_{2}=\sqrt{2}(0.3X_{1}-0.4X_{2}-0.5X_{3})
\]

\end_inset


\end_layout

\begin_layout Exercise*
\begin_inset Formula 
\[
Y_{3}=\sqrt{2}(0.3X_{1}-0.4X_{2}+0.5X_{3})
\]

\end_inset


\end_layout

\begin_layout Exercise*
Find the joint distribution of 
\begin_inset Formula $Y_{1},Y_{2},Y_{3}$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Solution*
Apply 
\series bold
Theorem 3.9.5 Multivariate Transformation.
 
\end_layout

\begin_layout Solution*
\begin_inset Formula 
\[
g(y_{1},...,y_{n})=\begin{cases}
f(s_{1},...,s_{n})|J|\\
0 & \text{otherwise}
\end{cases}
\]

\end_inset


\end_layout

\begin_layout Solution*
\begin_inset Formula 
\[
|J|\text{ is jacobian matrix}
\]

\end_inset


\end_layout

\begin_layout Solution*
\begin_inset Formula $x_{1}=s_{1}(y_{1},...,y_{n})$
\end_inset


\end_layout

\begin_layout Solution*
\begin_inset Formula $x_{2}=s_{2}(y_{1},...,y_{n})$
\end_inset


\end_layout

\begin_layout Solution*
\begin_inset Formula $...........................$
\end_inset


\end_layout

\begin_layout Solution*
\begin_inset Formula $x_{n}=s_{n}(y_{1},...,y_{n})$
\end_inset


\end_layout

\begin_layout Solution*
\begin_inset Formula $f(x_{1},x_{2},x_{3})=f(x_{1})f(x_{2})f(x_{3})$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
5.
 Suppose that random variables 
\begin_inset Formula $X_{1}$
\end_inset

 and 
\begin_inset Formula $X_{2}$
\end_inset

 are independent, each has the normal distribution with mean 
\begin_inset Formula $\mu$
\end_inset

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

.
 Prove 
\begin_inset Formula $X_{1}+X_{2}$
\end_inset

 and 
\begin_inset Formula $X_{1}-X_{2}$
\end_inset

 are independent.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Solution*
\begin_inset Formula $Cov[(X_{1}+X_{2}),(X_{1}-X_{2})]=E[(X_{1}+X_{2})(X_{1}-X_{2})]-E[(X_{1}+X_{2})]E[(X_{1}-X_{2})]$
\end_inset


\end_layout

\begin_layout Solution*
\begin_inset Formula $=E[X_{1}^{2}-X_{2}^{2}]-(E[X_{1}]+E[X_{2}])(E[X_{1}]-E[X_{2}])$
\end_inset


\end_layout

\begin_layout Solution*
\begin_inset Formula $=E[X_{1}^{2}]-E[X_{2}^{2}]-E[X_{1}]^{2}+E[X_{2}]^{2}$
\end_inset


\end_layout

\begin_layout Solution*
\begin_inset Formula $=\sigma_{1}^{2}-\sigma_{2}^{2}=0$
\end_inset

 
\end_layout

\begin_layout Solution*
So 
\begin_inset Formula $Cov()=0$
\end_inset

, so this 2 variables is independent.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
6.
 Suppose that 
\begin_inset Formula $X_{1},...,X_{n}$
\end_inset

 form a random sample from the normal distribution with mean 
\begin_inset Formula $\mu$
\end_inset

 and variance 
\begin_inset Formula $\sigma^{2}$
\end_inset

.
 Assuming that the sample size 
\begin_inset Formula $n$
\end_inset

 is 
\begin_inset Formula $16$
\end_inset

, determine the values of the following properties:
\end_layout

\begin_layout Exercise*
a.
 
\begin_inset Formula $Pr[\frac{1}{2}\sigma^{2}\le\frac{1}{n}\sum_{i=1}^{n}(X_{i}-\mu)^{2}\leq2\sigma^{2}]$
\end_inset


\end_layout

\begin_layout Exercise*
b.
 
\begin_inset Formula $Pr[\frac{1}{2}\sigma^{2}\le\frac{1}{n}\sum_{i=1}^{n}(X_{i}-\bar{X}_{n})^{2}\leq2\sigma^{2}]$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Solution*
a.
 
\begin_inset Formula $X=\frac{1}{n}\sum_{i=1}^{n}(X_{i}-\mu)^{2}$
\end_inset

, 
\begin_inset Formula $\frac{n}{\sigma^{2}}X$
\end_inset

 has 
\begin_inset Formula $\chi_{n}^{2}$
\end_inset

, 
\begin_inset Formula $n=16$
\end_inset

, so we need to find 
\begin_inset Formula $Pr[\frac{1}{2}*16\le16X\leq2*16]$
\end_inset


\end_layout

\begin_layout Solution*
b.
 
\begin_inset Formula $Pr[\frac{1}{2}*15\le15X\leq2*15]$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
7.
 Determined smallest 
\begin_inset Formula $n$
\end_inset


\end_layout

\begin_layout Exercise*
a.
 
\begin_inset Formula $Pr(\frac{\hat{\sigma}_{2}}{\sigma_{2}}\leq1.5)>0.95$
\end_inset


\end_layout

\begin_layout Exercise*
Try for 
\begin_inset Formula $n=1...$
\end_inset

 to find smallest 
\begin_inset Formula $n$
\end_inset

 satisfied 
\begin_inset Formula $P(\chi_{n}^{2}\le1.5n)\ge0.95$
\end_inset

 based on Chi-squared Distribution table.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
9.
 
\begin_inset Formula $\hat{\mu}$
\end_inset

 and 
\begin_inset Formula $\hat{\sigma}$
\end_inset

 is independent so not relevent.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Chapter*
8.4 The 
\begin_inset Formula $t$
\end_inset

 Distributions
\end_layout

\begin_layout Exercise*
1.
 
\begin_inset Formula $X$
\end_inset

 has 
\begin_inset Formula $t$
\end_inset

 distributions with 
\begin_inset Formula $m$
\end_inset

 degrees of freedom 
\begin_inset Formula $(m>2)$
\end_inset

.
 Show that 
\begin_inset Formula $Var(X)=m/(m-2)$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
2.
 Same example 8.4.3.
 Answer: 
\begin_inset Formula $\frac{-1.746}{4}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
3.
 Determine 
\begin_inset Formula $c$
\end_inset

 to 
\begin_inset Formula $\frac{c(X_{1}+X_{2})}{(X_{3}^{2}+X_{4}^{2}+X_{5}^{2})^{1/2}}$
\end_inset

 have 
\begin_inset Formula $t$
\end_inset

 distribution.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Solution*
\begin_inset Formula $t$
\end_inset

 distribution have : 
\begin_inset Formula $\frac{Z}{(\frac{Y}{m})^{2}}$
\end_inset

.
 with 
\begin_inset Formula $Z$
\end_inset

 has standard normal distribution, 
\begin_inset Formula $Y$
\end_inset

 has 
\begin_inset Formula $\chi_{m}^{2}$
\end_inset

.
 
\end_layout

\begin_layout Solution*
\begin_inset Formula $Z=\frac{X_{1}+X_{2}}{\sqrt{2}}$
\end_inset

 has standard normal distribution, 
\begin_inset Formula $Y=X_{3}^{2}+X_{4}^{2}+X_{5}^{2}$
\end_inset

 is 
\begin_inset Formula $\chi_{3}^{2}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
6.
 
\begin_inset Formula $n=20$
\end_inset

.
 
\begin_inset Formula $\frac{\sqrt{20}(\bar{X}_{n}-\mu)}{\sigma^{'}}$
\end_inset

 is 
\begin_inset Formula $t$
\end_inset

 distribution 
\begin_inset Formula $19$
\end_inset

 freedom degree.
\end_layout

\begin_layout Exercise*
\begin_inset Formula $Pr(\frac{\sqrt{20}(\bar{X}_{n}-\mu)}{\sigma^{'}}\leq1.729)=0.95$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
7.
 From Theorem 5.7.4, 
\begin_inset Formula $\lim_{x\rightarrow\infty}\Gamma(m)=(2\pi)^{1/2}m^{m-1/2}e^{-m}$
\end_inset

, similar for 
\begin_inset Formula $\Gamma(m+1)$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
8.
 We need to find max of 
\begin_inset Formula $Pr(-c<X<c)-Pr(-c<Y<c)=2\Phi(X)-1-(2\Phi(Y)-1)=2*(F(X)-F(Y))$
\end_inset

 or 
\begin_inset Formula $\max(F(X)-F(Y))$
\end_inset

, so we find maximum point of 
\begin_inset Formula $g=F(X)-F(Y)$
\end_inset

, 
\begin_inset Formula $g^{'}=f(x)-f(y)=0$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Chapter*
8.5 Confidence Intervals 
\end_layout

\begin_layout Exercise*
5.
 
\begin_inset Formula $X_{1},X_{2},...,X_{n}$
\end_inset

 form a random sample from the normal distribution with unknown mean 
\begin_inset Formula $\mu$
\end_inset

 and unknown variance 
\begin_inset Formula $\sigma^{2}$
\end_inset

.
 Describe method for constructing a confidence interval for 
\begin_inset Formula $\sigma^{2}$
\end_inset


\end_layout

\begin_layout Solution*
We have 
\begin_inset Formula $V(\theta,X)=\frac{\sum_{i=1}^{n}(X_{i}-\bar{X}_{n})^{2}}{\sigma^{2}}$
\end_inset

 is 
\begin_inset Formula $\chi_{n-1}^{2}$
\end_inset

 distribution, that have c.d.f is 
\begin_inset Formula $G$
\end_inset

.
 Assume that 
\begin_inset Formula $\gamma_{2}-\gamma_{1}=\gamma$
\end_inset

, and 
\begin_inset Formula $c_{1}=G^{-1}(\gamma_{1})$
\end_inset

 and 
\begin_inset Formula $c_{2}=G^{-1}(\gamma_{2})$
\end_inset

.
 So 
\begin_inset Formula $Pr[c_{1}\le\frac{\sum_{i=1}^{n}(X_{i}-\bar{X}_{n})^{2}}{\sigma^{2}}\le c_{2}]=\gamma$
\end_inset

 , then 
\begin_inset Formula $Pr[\frac{A}{c_{2}}\leq\sigma^{2}\le\frac{A}{c_{1}}]$
\end_inset

, with 
\begin_inset Formula $A=\sum_{i=1}^{n}(X_{i}-\bar{X}_{n})^{2}$
\end_inset

.
\end_layout

\begin_deeper
\begin_layout Exercise*
6.
 Similar to 
\series bold
Example 8.5.7.
 
\series default
We have 
\begin_inset Formula $\frac{1}{\mu}\sum_{i=1}^{n}X_{i}$
\end_inset

 is 
\begin_inset Formula $\Gamma(\alpha=3,\beta=1)$
\end_inset

 distribution, have c.d.
 is 
\begin_inset Formula $G$
\end_inset

, so Assume that 
\begin_inset Formula $\gamma_{2}-\gamma_{1}=\gamma$
\end_inset

 (like in Exercise 5).
 So we do like in Exercise 5.
\end_layout

\end_deeper
\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
7.
 Using Formulation in 
\series bold
Example 8.5.1
\series default
.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
9.
 c.
 
\begin_inset Formula $Y_{2}-1/2\le\theta\leq Y_{1}+1/2$
\end_inset

 is consistent with observed data.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Chapter*
8.6 Bayesian Analysis of Samples from a Normal Distribution
\end_layout

\begin_layout Exercise*
1.
 
\begin_inset Formula $X$
\end_inset

 has the normal distribution with mean 
\begin_inset Formula $\mu$
\end_inset

 and precision 
\begin_inset Formula $\tau$
\end_inset

.
 Show 
\begin_inset Formula $Y=aX+b$
\end_inset

 has normal distribution with mean 
\begin_inset Formula $a\mu+b$
\end_inset

 and precision 
\begin_inset Formula $\tau/a^{2}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Standard
Based on 
\series bold
Theorem 5.6.4, 
\series default
We have 
\begin_inset Formula $Y$
\end_inset

 will have normal distribution 
\begin_inset Formula $a\mu+b$
\end_inset

 and 
\begin_inset Formula $a^{2}\sigma^{2}$
\end_inset

, so precision is 
\begin_inset Formula $\frac{1}{a^{2}\sigma^{2}}=\frac{\tau}{a^{2}}$
\end_inset


\end_layout

\begin_layout Exercise*
2.
 
\begin_inset Formula $X_{1},...,X_{n}$
\end_inset

 form a random sample from the normal distribution with unknown mean 
\begin_inset Formula $\mu$
\end_inset

 (
\begin_inset Formula $-\infty<\mu<\infty$
\end_inset

 ) and known precision 
\begin_inset Formula $\tau$
\end_inset

.
 Suppose prior distribution of 
\begin_inset Formula $\mu$
\end_inset

 is normal distribution with mean 
\begin_inset Formula $\mu_{0}$
\end_inset

 and precision 
\begin_inset Formula $\lambda_{0}$
\end_inset

.
 
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Standard
Apply 
\series bold
Theorem 7.3.3 
\series default
Conjugate prior distribution for normal distribution with known variance
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
\mu_{1}=\frac{\sigma^{2}\mu_{0}+nv_{0}^{2}\bar{x_{n}}}{\sigma^{2}+nv_{0}^{2}}\text{ (7.3.1)}
\]

\end_inset

 and 
\begin_inset Formula 
\[
v_{1}^{2}=\frac{\sigma^{2}v_{0}^{2}}{\sigma^{2}+nv_{0}^{2}}\text{ (7.3.2)}
\]

\end_inset


\end_layout

\begin_layout Standard
we replace 
\begin_inset Formula $\sigma^{2}=1/\tau$
\end_inset

 and 
\begin_inset Formula $v_{0}^{2}=1/\lambda_{0}$
\end_inset

 to (7.3.1), we have 
\begin_inset Formula $\mu_{1}=\frac{\lambda_{0}\mu_{0}+n\tau\bar{x}_{n}}{\lambda_{0}+n\tau}$
\end_inset


\end_layout

\begin_layout Standard
Also, from (7.3.2), we have 
\begin_inset Formula $1/v_{1}^{2}=\lambda_{0}+n\tau$
\end_inset

.
\end_layout

\begin_layout Exercise*
3.
 Apply Exercise 24.c and Exercise 23 (Conjugate prior for Exponential distributio
n family) in 
\series bold
Chapter 7.3 
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
4.
 
\begin_inset Formula $V=(n-1)\sigma^{'2}\tau=(n-1)\sigma^{'2}/\sigma^{2}=\frac{\sum_{i=1}^{n}(X_{i}-\bar{X}_{n})^{2}}{\sigma^{2}}$
\end_inset

 from Chapter 8.3, we conclude 
\begin_inset Formula $V$
\end_inset

 has 
\begin_inset Formula $\chi_{n-1}^{2}$
\end_inset

 distribution.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
5.
 
\begin_inset Formula $\mu$
\end_inset

 and 
\begin_inset Formula $\tau$
\end_inset

 have the joint normal-gamma distribution, 
\begin_inset Formula $E(\mu)=\mu_{0};Var(\mu)=\frac{\beta_{0}}{\lambda_{0}(\alpha_{0}-1)}$
\end_inset


\end_layout

\begin_layout Exercise*
And, 
\begin_inset Formula $\tau$
\end_inset

 has gamma distribution, so 
\begin_inset Formula $E(\tau)=\frac{\alpha_{0}}{\beta_{0}}$
\end_inset

, 
\begin_inset Formula $Var(\tau)=\frac{\alpha_{0}}{\beta_{0}^{2}}$
\end_inset

, So 
\begin_inset Formula $\alpha_{0}=2,\beta_{0}=4$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
6.
 
\begin_inset Formula $\alpha_{0}=1$
\end_inset

 so it is cannot be that in 
\begin_inset Formula $Var(\mu)$
\end_inset

.
 
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
7.
 Following from prove of Theorem 8.6.3, if 
\begin_inset Formula $E(U)=0$
\end_inset

, if 
\begin_inset Formula $\alpha_{0}>1/2$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
8 
\begin_inset Formula $Pr(\text{\ensuremath{\mu>0}})=Pr((\frac{\lambda_{0}\alpha_{0}}{\beta_{0}})^{2}(\mu-\mu_{0})>-(\frac{\lambda_{0}\alpha_{0}}{\beta_{0}})^{2}\mu_{0})$
\end_inset

, While we have 
\begin_inset Formula $(\frac{\lambda_{0}\alpha_{0}}{\beta_{0}})^{2}(\mu-\mu_{0})$
\end_inset

 is 
\begin_inset Formula $t$
\end_inset

 distribution with 
\begin_inset Formula $2\alpha_{0}$
\end_inset

 degrees of freedom.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
9.
 
\begin_inset Formula $\bar{x}_{n}=182.17$
\end_inset

 and 
\begin_inset Formula $s_{n}^{2}=88678.5$
\end_inset

, 
\end_layout

\begin_layout Exercise*
\begin_inset Formula $\mu_{1}=183.95$
\end_inset

 
\begin_inset Formula $\lambda_{1}=20$
\end_inset

 
\begin_inset Formula $\alpha_{1}=11$
\end_inset

 
\begin_inset Formula $\beta_{1}=50925.37$
\end_inset


\end_layout

\begin_layout Exercise*
\begin_inset Formula $U=0.0657(\mu-183.95)$
\end_inset

 is 
\begin_inset Formula $t$
\end_inset

 distribution with 
\begin_inset Formula $2\alpha_{1}=22$
\end_inset

 degrees of freedom.
 The quantile 0.95 of 
\begin_inset Formula $t$
\end_inset

 distribtuion is 1.717
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Standard
a) Shortest possible interval fo posterior 
\begin_inset Formula $\mu$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Formula $Pr(-1.717<U<1.717)=0.9$
\end_inset

, 
\begin_inset Formula $Pr(157.8<\text{\ensuremath{\mu}}<210)$
\end_inset


\end_layout

\begin_layout Standard
b) Shortest possible confidence interval , confidence coefficient is 0.90.
\end_layout

\begin_layout Standard
\begin_inset Formula $\sigma^{'}=72.72$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $n=18$
\end_inset

, so following Eq.
 8.4.4.
 has 
\begin_inset Formula $t$
\end_inset

 distribution with 
\begin_inset Formula $17$
\end_inset

 degrees of freedom.
 0.95 quantile of it is 1.74
\end_layout

\begin_layout Standard
so Shortest possible confidence interval is 
\begin_inset Formula $(\bar{X_{n}}-1.74\frac{\sigma^{'}}{n^{1/2}},\bar{X_{n}}+1.74\frac{\sigma^{'}}{n^{1/2}})=(152.55,211.789)$
\end_inset

 , it close to 
\begin_inset Formula $a)$
\end_inset

.
\end_layout

\begin_layout Exercise*
10.
 
\begin_inset Formula $\alpha_{0}=4$
\end_inset

, 
\begin_inset Formula $\beta_{0}=2$
\end_inset

, 
\begin_inset Formula $\mu_{0}=0$
\end_inset

, so based on 
\begin_inset Formula $Pr(|\mu|<1.412)=0.5$
\end_inset

 and distribution of 
\begin_inset Formula $U=(\frac{\lambda_{0}\alpha_{0}}{\beta_{0}})^{1/2}(\mu-\mu_{0})$
\end_inset

 is 
\begin_inset Formula $t$
\end_inset

 distribution with freedom degree is 
\begin_inset Formula $2\alpha_{0}=4$
\end_inset

.
 So find 
\begin_inset Formula $T_{4}^{-1}(0.75)$
\end_inset

, and then find 
\begin_inset Formula $\lambda_{0}=[\frac{1}{1.412}T_{8}^{-1}(0.75)\frac{\beta_{0}}{\alpha_{0}}]^{2}=(\frac{1}{1.412}*0.706*0.5)^{2}=0.0625$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
11.
 
\begin_inset Formula $n=10$
\end_inset

 
\begin_inset Formula $\bar{x}_{n}=1$
\end_inset

 
\begin_inset Formula $s_{n}^{2}=8$
\end_inset

.
 Find shortest intervals such that posterior probability that 
\begin_inset Formula $\mu$
\end_inset

 lies in the interval 
\begin_inset Formula $0.95$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Standard
Following Eq.
 8.6.1 and Eq.
 8.6.2, we have 
\begin_inset Formula $\mu_{1}=0.994$
\end_inset

 
\begin_inset Formula $,\lambda_{1}=10.0625$
\end_inset

 ,
\begin_inset Formula $\alpha_{1}=9$
\end_inset

, 
\begin_inset Formula $\beta_{1}=6.031$
\end_inset


\end_layout

\begin_layout Standard
Shortest 0.95 interval is :
\end_layout

\begin_layout Standard
\begin_inset Formula 
\[
(\mu_{1}-T_{18}^{-1}(0.975)[\frac{\beta_{1}}{\alpha_{1}\lambda_{1}}]^{1/2},\mu_{1}+T_{18}^{-1}(0.975)[\frac{\beta_{1}}{\alpha_{1}\lambda_{1}}]^{1/2})=(0.452,1.536)
\]

\end_inset


\end_layout

\begin_layout Exercise*
14.
 Following Theorem 8.6.2.
 
\begin_inset Formula $(\frac{\lambda_{1}\alpha_{1}}{\beta_{1}})^{1/2}(\mu-\mu_{1})$
\end_inset

 is 
\begin_inset Formula $t$
\end_inset

 distribution with 
\begin_inset Formula $2\alpha_{1}=11$
\end_inset

 degree freedom.
 
\begin_inset Formula $T_{11}^{-1}(0.95)=1.796$
\end_inset


\end_layout

\begin_layout Exercise*
So 
\begin_inset Formula $-1.796<(\frac{\lambda_{1}\alpha_{1}}{\beta_{1}})^{1/2}(\mu-\mu_{1})<1.796$
\end_inset

, 
\begin_inset Formula $-1.796<(\frac{\lambda_{1}\alpha_{1}}{\beta_{1}})^{1/2}(\mu-\mu_{1})<1.796$
\end_inset

.
 So 
\begin_inset Formula $Pr(1.1<\mu<1.58)=0.9$
\end_inset

, It so close to Example 8.5.4 - 
\begin_inset Formula $(1.379,1.569)$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
15.
 
\begin_inset Formula $n=11$
\end_inset

.
 Prior 
\begin_inset Formula $(\mu,\begin{tabular}{c}
 \tau\end{tabular})$
\end_inset

 from the normal-gamma family with 
\begin_inset Formula $\alpha_{0}=2$
\end_inset

, 
\begin_inset Formula $\beta_{0}=1$
\end_inset

, 
\begin_inset Formula $\mu_{0}=3.5$
\end_inset

, 
\begin_inset Formula $\lambda_{0}=2$
\end_inset

.
 The sample yields an average of 
\begin_inset Formula $\bar{x}_{n}=7.2$
\end_inset

 and 
\begin_inset Formula $s_{n}^{2}=20.3$
\end_inset

.
\end_layout

\begin_layout Exercise*
a.
 Find posterior hyperparameters
\end_layout

\begin_layout Exercise*
\begin_inset Formula $\mu_{1}=\frac{\lambda_{0}\mu_{0}+n\bar{x}_{n}}{\lambda_{0}+n}=6.631$
\end_inset

, 
\begin_inset Formula $\lambda_{1}=\lambda_{0}+n=13$
\end_inset

, 
\begin_inset Formula $a_{1}=a_{0}+\frac{n}{2}=7.5$
\end_inset

, 
\begin_inset Formula $\beta_{1}=\beta_{0}+\frac{1}{2}s_{n}^{2}+\frac{n\lambda_{0}(\bar{x}_{n}-\mu_{0})^{2}}{2(\lambda_{0}+n)}=22.73$
\end_inset


\end_layout

\begin_layout Exercise*
b.
 Find 
\begin_inset Formula $95\%$
\end_inset

 of posteriro distribution of 
\begin_inset Formula $\mu$
\end_inset

.
 
\end_layout

\begin_layout Exercise*
\begin_inset Formula $(\mu_{1}-T_{2\alpha_{1}}^{-1}(\frac{1+\lambda}{2})[\frac{\beta_{1}}{\alpha_{1}\lambda_{1}}]^{1/2},\mu_{1}+T_{2\alpha_{1}}^{-1}(\frac{1+\lambda}{2})[\frac{\beta_{1}}{\alpha_{1}\lambda_{1}}]^{1/2})=0.95$
\end_inset

; 
\begin_inset Formula $T_{2\alpha_{1}}^{-1}(\frac{1+\lambda}{2})=T_{15}^{-1}(\frac{1+0.95}{2})=2.131$
\end_inset


\end_layout

\begin_layout Exercise*
So 
\begin_inset Formula $(5.6,7.66)$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
17.
 Using usual improper prior to compute posterior distribution of the parameters.
\end_layout

\begin_layout Exercise*

\series bold
Usal imporper prior: 
\series default

\begin_inset Formula $\alpha_{0}=\beta_{0}=\lambda_{0}=0$
\end_inset

 and 
\begin_inset Formula $\alpha_{0}=-1/2$
\end_inset

.
\end_layout

\begin_layout Exercise*
Data of Example 8.6.2: 
\begin_inset Formula $n=10$
\end_inset

, 
\begin_inset Formula $\bar{x}_{n}=1.379$
\end_inset

 and 
\begin_inset Formula $s_{n}^{2}=0.9663$
\end_inset


\end_layout

\begin_layout Exercise*
\begin_inset Formula $\mu_{1}=\frac{\lambda_{0}\mu_{0}+n\bar{x}_{n}}{\lambda_{0}+n}=1.379,\lambda_{1}=\lambda_{0}+n=10,a_{1}=a_{0}+\frac{n}{2}=4.5,\beta_{1}=\beta_{0}+\frac{1}{2}s_{n}^{2}+\frac{n\lambda_{0}(\bar{x}_{n}-\mu_{0})^{2}}{2(\lambda_{0}+n)}=0.48315$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
19.
 Similar to Exercise: 15.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
20.
 
\begin_inset Formula $\alpha_{0}=\beta_{0}=\lambda_{0}=0$
\end_inset

 and 
\begin_inset Formula $\alpha_{0}=-1/2$
\end_inset

, 
\begin_inset Formula $\bar{x}_{n}=0.125$
\end_inset

 and 
\begin_inset Formula $s_{n}^{2}=2102.9$
\end_inset

.
 Similar : 15
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Chapter*
8.7 The unbiased Estimator.
\end_layout

\begin_layout Exercise*
1.
 
\begin_inset Formula $Var_{\theta}(X_{i})=\theta$
\end_inset

.
\end_layout

\begin_layout Exercise*
b.
 
\begin_inset Formula $M.L.E=MLE(\mu)=\bar{X}_{n}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
5.
 
\begin_inset Formula $X$
\end_inset

 has Poisson distribution with unknown mean 
\begin_inset Formula $\lambda$
\end_inset

.
 Find a statistic 
\begin_inset Formula $\delta(X)$
\end_inset

 that will be an unbiased estimator of 
\begin_inset Formula $e^{\lambda}$
\end_inset

.
 
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Standard
Like in the hint.
 
\end_layout

\begin_layout Standard
we have: 
\begin_inset Formula $\sum_{x=0}^{\infty}\frac{\delta(x)\lambda^{x}}{x!}=e^{2\lambda}$
\end_inset

.
 
\begin_inset Formula $g(\lambda)=e^{2\lambda}$
\end_inset

.
 Assume that we have 
\begin_inset Formula $g(x)$
\end_inset

 with 
\begin_inset Formula $x=2\lambda$
\end_inset

, then applying Taylor series:
\end_layout

\begin_layout Standard
\begin_inset Formula $e^{2\lambda}=\sum_{i=0}^{\infty}\frac{g^{(k)}(0)(2\lambda-0)^{k}}{k!}$
\end_inset

.
 With 
\begin_inset Formula $g^{k}(x)$
\end_inset

 is derivative degree 
\begin_inset Formula $k$
\end_inset

 of 
\begin_inset Formula $g(x)$
\end_inset

.
 
\begin_inset Formula $g(x)=e^{x}$
\end_inset

so 
\begin_inset Formula $g^{k}(0)=1$
\end_inset

.
 so 
\end_layout

\begin_layout Standard
\begin_inset Formula $e^{2\lambda}=\sum_{i=0}^{\infty}\frac{g^{(k)}(0)(2\lambda-0)^{k}}{k!}=\sum_{i=0}^{\infty}\frac{(2\lambda)^{k}}{k!}$
\end_inset

.
 
\end_layout

\begin_layout Standard
So 
\begin_inset Formula $\delta(x)=2^{X}$
\end_inset

 is unbiased estimator of 
\begin_inset Formula $e^{\lambda}$
\end_inset


\end_layout

\begin_layout Exercise*
6, From example.
 8.7.6, 
\begin_inset Formula $T_{c}=c\sum_{i=1}^{n}(X_{i}-\bar{X}_{n})^{2}$
\end_inset

, transform it to 
\begin_inset Formula $\chi^{2}$
\end_inset

 with dividing for 
\begin_inset Formula $\sigma$
\end_inset

, then following Eq.(8.7.7): 
\begin_inset Formula $Var(T_{c})=2(n-1)c^{2}\sigma^{4}$
\end_inset

, 
\begin_inset Formula $\frac{Var(T_{n})}{Var(T_{n-1})}=\frac{n-1}{n}$
\end_inset

, So 
\begin_inset Formula $Var(T_{n})<Var(T_{n-1})$
\end_inset

, then we can conclude: MSE of 
\begin_inset Formula $\hat{\sigma_{0}^{2}}$
\end_inset

 is smaller thant the M.S.E of 
\begin_inset Formula $\hat{\sigma_{1}^{2}}.$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
7.
 
\begin_inset Formula $\delta(X_{1},...,X_{n})$
\end_inset

 with 
\begin_inset Formula $X_{1},...,X_{n}$
\end_inset

 form n Bernoulli trials for which the parameter 
\begin_inset Formula $p$
\end_inset

 is unknown.
\end_layout

\begin_layout Exercise*
So 
\begin_inset Formula $\delta(X_{1},...,X_{n})$
\end_inset

 is list of value 
\begin_inset Formula $\{a:p(a)\}$
\end_inset

 and 
\begin_inset Formula $p(a)=p^{k}(1-p)^{n-k}$
\end_inset

, so it is polynomial in 
\begin_inset Formula $p$
\end_inset

 whose degree does not exceed 
\begin_inset Formula $n$
\end_inset

 .
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
8.
 
\begin_inset Formula $X$
\end_inset

 has the geometric distribution with unknown parameter 
\begin_inset Formula $p$
\end_inset

 (
\begin_inset Formula $0<p<1$
\end_inset

).
 
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
9.
 
\begin_inset Formula $X$
\end_inset

 has the possion distribution with unknown mean 
\begin_inset Formula $\lambda$
\end_inset

.
 The only unbiased estimator of 
\begin_inset Formula $e^{-2\lambda}$
\end_inset

 is the estimator 
\begin_inset Formula $\delta(X)$
\end_inset

 such that 
\begin_inset Formula $\delta(X)=1$
\end_inset

 if 
\begin_inset Formula $X$
\end_inset

 is an even integer and 
\begin_inset Formula $\delta(X)=-1$
\end_inset

 if X is odd interger.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Standard
Like in Exercise 5, taylor series of 
\begin_inset Formula $e^{-\lambda}=\sum_{x=1}^{n}\frac{(-1)^{x}(\lambda)^{x}}{x!}$
\end_inset

, so '
\begin_inset Formula $\delta(X)=1$
\end_inset

 if 
\begin_inset Formula $X$
\end_inset

 is an even integer and 
\begin_inset Formula $\delta(X)=-1$
\end_inset

 if X is odd interger' is unbiased estimator of 
\begin_inset Formula $e^{-2\lambda}$
\end_inset

.
 Actually, we have equation: 
\begin_inset Formula $\sum_{x=0}^{\infty}\frac{\delta(x)\lambda^{x}}{x!}=\sum_{x=1}^{n}\frac{(-1)^{x}(\lambda)^{x}}{x!}$
\end_inset

, so we have only one function.
\end_layout

\begin_layout Exercise*
10.This is Geometric distribution.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $E[\frac{k-1}{N-1}]=\frac{k}{N}$
\end_inset

 when 
\begin_inset Formula $n\rightarrow\infty$
\end_inset

, and 
\begin_inset Formula $\frac{k}{N}=p$
\end_inset

.
\end_layout

\begin_layout Exercise*
11.
 
\begin_inset Formula $\mu_{1}=\mu_{2}=\theta$
\end_inset

.
 
\begin_inset Formula $\hat{\theta}=\alpha\bar{X}_{m}+(1-\alpha)\hat{Y}_{n}$
\end_inset


\end_layout

\begin_layout Exercise*
a, 
\begin_inset Formula $\forall\alpha,m,n$
\end_inset

 we have 
\begin_inset Formula $E[\hat{\theta}]=\alpha E[\bar{X}_{m}]+(1-\alpha)E[\bar{Y}_{n}]=\theta$
\end_inset

.
\end_layout

\begin_layout Exercise*
b.
\begin_inset Formula $Var(\hat{\theta})=Var(\alpha\bar{X}_{m}+(1-\alpha)\bar{Y}_{n})=\alpha^{2}\frac{\sigma_{1}^{2}}{m}+(1-\alpha)^{2}\frac{\alpha_{2}^{2}}{n}$
\end_inset

.
 We want to find 
\begin_inset Formula $\min[Var(\hat{\theta})]=\min[4n\alpha^{2}+(1-\alpha)^{2}m]=min[(4n+m)\sigma^{2}-2\sigma m+m].$
\end_inset

 We can see 
\begin_inset Formula $f(\sigma)=(4n+m)\sigma^{2}-2\sigma m+m$
\end_inset

 is quaradic convec fucntion so it achive minimum at 
\begin_inset Formula $f^{'}(\sigma)=0$
\end_inset

, then 
\begin_inset Formula $\sigma=\frac{m}{4n+m}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
12.
 b.
 Min[
\begin_inset Formula $\sum_{i=1}^{k}p_{i}\frac{\sigma_{i}^{2}}{n_{i}}$
\end_inset

] and 
\begin_inset Formula $\sum_{i=1}^{k}n_{i}-n=0$
\end_inset

.
 Using 
\series bold
Larange multiper
\series default
: 
\begin_inset Formula $Min(A):\sum_{i=1}^{k}p_{i}(\frac{\sigma_{i}^{2}}{n_{i}})+\lambda\sum_{i=1}^{k}n_{i}=\sum_{i=1}^{k}(p_{i}\frac{\sigma_{i}^{2}}{n_{i}}+\lambda n_{i})$
\end_inset


\end_layout

\begin_layout Exercise*
We consider derivative of A on 
\begin_inset Formula $\lambda$
\end_inset

, 
\begin_inset Formula $\frac{dA}{d\lambda}=\sum n_{i}-n=0$
\end_inset

, and with 
\begin_inset Formula $n_{i}$
\end_inset

, 
\begin_inset Formula $-\frac{p_{i}\sigma_{i}^{2}}{n_{i}^{2}}+\lambda=0\rightarrow n_{i}=\frac{(p_{i}\sigma_{i}^{2})^{1/2}}{\lambda^{1/2}}$
\end_inset

, 
\begin_inset Formula $\sum n_{i}=n=\frac{\sum_{i}(p_{i}\sigma_{i}^{2})^{1/2}}{\lambda^{1/2}}$
\end_inset

, so 
\begin_inset Formula $\lambda^{1/2}=\sum_{i}(p_{i}\sigma_{i}^{2})^{1/2}/n$
\end_inset

.
 So 
\begin_inset Formula $n_{i}=n*\frac{(p_{i}\sigma_{i}^{2})^{1/2}}{\sum_{i}(p_{i}\sigma_{i}^{2})^{1/2}}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
13.
 Making use of Sec 4.7.
 From theorem 4.7.1.
 Law of Total probability for expectations.
 Let 
\begin_inset Formula $X$
\end_inset

 and 
\begin_inset Formula $Y$
\end_inset

 be random variables such that 
\begin_inset Formula $Y$
\end_inset

 has finite mean.
 Then 
\end_layout

\begin_layout Exercise*
\begin_inset Formula 
\[
E[E(Y|X)]=E(Y)
\]

\end_inset


\end_layout

\begin_layout Exercise*
so 
\begin_inset Formula $E[E[\delta(X)|T]]=E[\delta(X)]=\theta$
\end_inset

, so 
\begin_inset Formula $E[\delta(X)|T]$
\end_inset

, is also an unbiased estimator of 
\begin_inset Formula $\theta$
\end_inset


\end_layout

\begin_layout Exercise*
b.
 
\begin_inset Formula $Var_{\theta}(\delta_{0})\le Var_{\theta}(\delta)$
\end_inset

.
 From 
\series bold
Theorem 4.7.4: 
\begin_inset Formula $Var(Y)=E[Var(Y|X)]+Var[E(Y|X)]$
\end_inset


\series default
, so 
\begin_inset Formula $Var(\delta(X))=E[Var(\delta(X)|T)]+Var[E(\delta(X)|T)]$
\end_inset

.
 So ..
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
14.
 
\begin_inset Formula $X_{1},...,X_{n}$
\end_inset

 form a random sample from the uniform distribution on the interval 
\begin_inset Formula $[0,\theta]$
\end_inset

.
 
\begin_inset Formula $Y_{n}=max(X_{1},...,X_{n})$
\end_inset

.
 Show that 
\begin_inset Formula $[(n+1)/n]Y_{n}$
\end_inset

 is an unbiased estimator of 
\begin_inset Formula $\theta$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $F(Y_{n})=Pr(Y_{n}<y)=Pr(X_{1}<y)....Pr(X_{n}<y)=(\frac{y}{\theta})^{n}$
\end_inset

.
 So 
\begin_inset Formula $f(Y_{n})=F^{'}(Y_{n})=n\frac{y^{n-1}}{\theta^{n}}$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Formula $E[Y_{n}]=\int_{0}^{\theta}f(y)ydy=\int_{o}^{\theta}n\frac{y^{n}}{\theta^{n}}dy=\frac{n}{n+1}\frac{y^{n+1}}{\theta^{n}}|_{0}^{\theta}=\frac{n}{n+1}\theta$
\end_inset

.
\end_layout

\begin_layout Standard
So 
\begin_inset Formula $E[\frac{n+1}{n}Y_{n}]=\frac{n+1}{n}E[Y_{n}]=\frac{n+1}{n}\frac{n}{n+1}\theta=\theta$
\end_inset

, so 
\begin_inset Formula $\frac{n+1}{n}Y_{n}$
\end_inset

 is an unbiased estimator of 
\begin_inset Formula $\theta$
\end_inset

.
\end_layout

\begin_layout Exercise*
15.
 
\begin_inset Formula $X:\{1,2,3,4,5\}$
\end_inset

.
 
\begin_inset Formula $f(1|\theta)=\theta^{3}$
\end_inset

, 
\begin_inset Formula $f(2|\theta)=\theta^{2}(1-\theta)$
\end_inset

, 
\begin_inset Formula $f(3|\theta)=2\theta(1-\theta)$
\end_inset

, 
\begin_inset Formula $f(4|\theta)=\theta(1-\theta)^{2}$
\end_inset

, 
\begin_inset Formula $f(5|\theta)=(1-\theta)^{3}$
\end_inset


\end_layout

\begin_layout Exercise*
a, 
\begin_inset Formula $\theta^{3}+\theta^{2}-\theta^{3}+2\theta-2\theta^{2}+\theta^{3}-2\theta^{2}+\theta-\theta^{3}+3\theta^{2}-3\theta+1=1$
\end_inset


\end_layout

\begin_layout Exercise*
b.We have:  
\begin_inset Formula $E_{\theta}[\delta_{c}]=\theta^{3}+(2-2c)(\theta^{2}-\theta^{3})+c(2\theta-2\theta^{2})+(1-2c)(\theta^{3}-2\theta^{2}+\theta)=\theta^{3}+2\theta^{2}-2\theta^{3}-2c\theta^{2}+2c\theta^{3}+2\theta c-2c\theta^{2}+\theta^{3}-2\theta^{2}+\theta-2c\theta^{3}+4c\theta^{2}-2c\theta$
\end_inset

=
\begin_inset Formula $\theta$
\end_inset

.
 
\end_layout

\begin_layout Exercise*
So 
\begin_inset Formula $\delta_{c}(X)$
\end_inset

 is unbiased estimator of 
\begin_inset Formula $\theta$
\end_inset

 for each constant 
\begin_inset Formula $c$
\end_inset

.
\end_layout

\begin_layout Exercise*
c.
 We want to find 
\begin_inset Formula $Min(Var(\delta_{c}(X)))$
\end_inset

.
 
\begin_inset Formula $Var(\delta_{c}(X))=\theta^{3}+(2-2c-\theta)^{2}(\theta^{2}-\theta^{3})+(c-\theta)^{2}(2\theta-2\theta^{2})+(1-2c-\theta)^{2}(\theta^{3}-2\theta^{2}+\theta)$
\end_inset

 or 
\begin_inset Formula $=E[(\delta_{c}(X))^{2}]-[E[\delta_{c}(C)]^{2}]=E[(\delta_{c}(X))^{2}]-\theta^{2}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
16.
 In exercise 3:
\end_layout

\begin_layout Exercise*
From theorem 8.7.1: 
\begin_inset Formula $\hat{\sigma}_{1}=\frac{1}{n-1}\sum_{i=1}^{n}(X_{i}-\bar{X}_{n})^{2}$
\end_inset

 is an unbiased of variacne 
\end_layout

\begin_layout Exercise*
and From exercise 2: 
\begin_inset Formula $(1/n)\sum_{i=1}^{n}X_{i}^{2}$
\end_inset

 is unbiased estimator of 
\begin_inset Formula $E[X^{2}]$
\end_inset

.
\end_layout

\begin_layout Exercise*
So 
\begin_inset Formula $(1/n)\sum_{i=1}^{n}X_{i}^{2}-\frac{1}{n-1}\sum_{i=1}^{n}(X_{i}-\bar{X}_{n})^{2}$
\end_inset

 is unbiased estimator of 
\begin_inset Formula $[E(X)]^{2}$
\end_inset

.
\end_layout

\begin_layout Exercise*
\begin_inset Formula $X_{1}=2$
\end_inset

 , 
\begin_inset Formula $X_{2}=-1$
\end_inset

 so value of unbiased estimaotor =
\begin_inset Formula $\frac{5}{2}-\frac{1}{1}(1/4+9/4)=0$
\end_inset

.
 FLAW, so value estimator of 
\begin_inset Formula $[E(X)]^{2}=0????$
\end_inset

.
 Meanwhile value's estimator of 
\begin_inset Formula $E(X)=1/2$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Chapter*
8.8 Fisher Information
\end_layout

\begin_layout Exercise*
9.
 
\begin_inset Formula $X$
\end_inset

 taken from the normal distribution with mean 
\begin_inset Formula $0$
\end_inset

 and unknown standard deviation 
\begin_inset Formula $\sigma>0$
\end_inset

.
 An unbiased estimator of 
\begin_inset Formula $\sigma$
\end_inset

 : 
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
10.
 Suppose that 
\begin_inset Formula $X_{1},...,X_{n}$
\end_inset

 form a random sample from the normal distribution with mean 
\begin_inset Formula $0$
\end_inset

 and unknown 
\begin_inset Formula $\sigma>0$
\end_inset

.
 Find the lower bound specified by the information inequality for the variance
 of any inbiased estimator of 
\begin_inset Formula $\log\sigma$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Solution*
Lower bound 
\begin_inset Formula $=\frac{[m^{'}(\theta)]^{2}}{nI(\theta)}$
\end_inset

.
 
\begin_inset Formula $m^{'}(\theta)=\frac{1}{\sigma}$
\end_inset

, 
\begin_inset Formula $I(\theta)=E[\lambda^{''}(X|\theta)]=\frac{1}{\sigma^{2}}$
\end_inset

.
 So lower bound is 
\begin_inset Formula $\frac{1}{n}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
11.
 From exercise 23 from Sec 7.3.
 
\begin_inset Formula $f(x|\theta)=a(\theta)b(x)exp[c(\theta)d(x)]$
\end_inset

.
 Show that estimator 
\begin_inset Formula $T=\sum_{i=1}^{n}d(X_{i})$
\end_inset

 is efficient estimator.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Standard
It means 
\begin_inset Formula $T$
\end_inset

 hold equation 
\begin_inset Formula $(8.8.15)$
\end_inset

: 
\begin_inset Formula $T=u(\theta)\lambda_{n}^{'}(X|\theta)+v(\theta)$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Formula $\lambda_{n}(X|\theta)=\log(f_{n}(x|\theta))$
\end_inset

.
 
\begin_inset Formula $f_{n}(x|\theta)=[a(\theta)]^{n}\prod_{i=1}^{n}b(x_{i})exp(c(\theta)\sum_{i=1}^{n}d(x_{i}))$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\lambda_{n}(X|\theta)=\log(f_{n}(x|\theta))=n\log(a(\theta))+log(\prod_{i=1}^{n}b(x_{i}))+c(\theta)\sum_{i=1}^{n}d(x_{i})$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Formula $\lambda_{n}^{'}(X|\theta)=\frac{n}{a(\theta)}a^{'}(\theta)+c^{'}(\theta)\sum_{i=1}^{n}d(x_{i})$
\end_inset

.
 So 
\begin_inset Formula $\sum_{i=1}^{n}d(x_{i})=\frac{1}{c^{'}(\theta)}\lambda_{n}^{'}(X|\theta)-\frac{n}{a(\theta)c^{'}(\theta)}a^{'}(\theta)$
\end_inset

.
 So 
\begin_inset Formula $u(\theta)=\frac{1}{c^{'}(\theta)}$
\end_inset

 and 
\begin_inset Formula $v(\theta)=-\frac{n}{a(\theta)c^{'}(\theta)}a^{'}(\theta)$
\end_inset


\end_layout

\begin_layout Exercise*
12.
 Suppose that 
\begin_inset Formula $X_{1},...,X_{n}$
\end_inset

 form a random sample from a normal distribution for mean is known and the
 variance is unknown.
 Construct an efficient estimator .
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Standard
Efficient estimator if 
\begin_inset Formula $T=u(\theta)\lambda_{n}^{'}(X|\theta)+v(\theta)$
\end_inset

.
 
\begin_inset Formula $\lambda_{n}(X|\theta)=\log(f_{n}(x|\theta))$
\end_inset

.
 
\end_layout

\begin_layout Standard
\begin_inset Formula $f(x|\theta)=\frac{1}{\sqrt{2\pi}\sigma}exp(\frac{-(x-\mu)^{2}}{\sigma^{2}}),$
\end_inset

so 
\begin_inset Formula $f_{n}(x|\theta)=(2\pi)^{-n/2}\sigma^{-n}exp(\frac{-\sum_{i=1}^{n}(x-\mu)^{2}}{\sigma^{2}})$
\end_inset

.
 
\begin_inset Formula $\log(f_{n}(x|\theta))=\frac{-n}{2}log(2\pi)-n\log(\sigma)-\sum_{i=1}^{n}(x-\mu)^{2}/\sigma^{2}$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset Formula $\lambda_{n}^{'}(x|\theta)=\frac{-n}{\sigma}+2\sum_{i=1}^{n}(x-\mu)^{2}/\sigma^{3}$
\end_inset

.
 We choose 
\begin_inset Formula $u(\theta)=\frac{\sigma}{2}$
\end_inset

 and 
\begin_inset Formula $v(\theta)=\frac{n}{2}$
\end_inset

.
 So 
\begin_inset Formula $T=\sum_{i=1}^{n}(x-\mu)^{2}/\sigma^{2}$
\end_inset

.
 
\begin_inset Formula $T$
\end_inset

 is 
\begin_inset Formula $\chi_{n}^{2}$
\end_inset

, 
\begin_inset Formula $\alpha=n/2,\beta=1/2$
\end_inset

.
 So 
\begin_inset Formula $E[T]=\alpha/\beta=n$
\end_inset

, 
\begin_inset Formula $Var(T)=\alpha/\beta^{2}=2n$
\end_inset


\end_layout

\begin_layout Exercise*
14.
 Following Theorem 8.8.5, Variance is 
\begin_inset Formula $1/[nI(\theta)]^{1/2}$
\end_inset

.
 So we need find 
\begin_inset Formula $I(\theta)$
\end_inset

 of Gamma distribution with 
\begin_inset Formula $\alpha$
\end_inset

 unknown.
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
15.
 From Bayesian Point of View in Page 524.
 Posterior distribution of 
\begin_inset Formula $\theta$
\end_inset

 is normal distribution with mean is M.L.E 
\begin_inset Formula $\hat{\theta}_{n}$
\end_inset

 and variance 
\begin_inset Formula $1/[nI(\hat{\theta}_{n})]$
\end_inset

.
 
\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Standard
My result: Posterior: mean 
\begin_inset Formula $\bar{x}_{n}$
\end_inset

; variance: 
\begin_inset Formula $\frac{\sigma^{2}}{2n}$
\end_inset

.
 Notice: 
\begin_inset Formula $1/[nI(\hat{\theta}_{n})]$
\end_inset

 not 
\begin_inset Formula $1/[nI_{n}(\hat{\theta}_{n})]$
\end_inset


\end_layout

\begin_layout Exercise*
16.
 Easy to find 
\begin_inset Formula $I(p)=\frac{1}{(1-p)p}$
\end_inset


\end_layout

\begin_layout Standard
\begin_inset Separator plain
\end_inset


\end_layout

\begin_layout Exercise*
19.
 Easy to find 
\begin_inset Formula $\lambda^{''}(X|\theta)=\frac{-n}{\theta^{2}}$
\end_inset


\end_layout

\end_body
\end_document