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chapter7_exercise.lyx
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#LyX 2.3 created this file. For more info see http://www.lyx.org/
\lyxformat 544
\begin_document
\begin_header
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\end_header
\begin_body
\begin_layout Part*
Chapter 7.
Estimation
\end_layout
\begin_layout Section*
7.1 Statistical Inference
\end_layout
\begin_layout Paragraph
1.
\series medium
Identify the components of the statistical model (as defined in Definition
7.1.1) in example 7.1.3.
\end_layout
\begin_layout Paragraph
Answer:
\series medium
Following Definition 7.1.1, Statistical model consists of :
\end_layout
\begin_layout Enumerate
Random variables of interest (both observable and only hypothetically observable
)
\end_layout
\begin_layout Enumerate
A Specification of a joint distribution or a family of possible joint dsitributi
ons for the observable random variables
\end_layout
\begin_layout Enumerate
The identification of any parameters of those distributions that are assumed
unknown and possibly hypothetically observable
\end_layout
\begin_layout Enumerate
A specification for a joint distribution for the unknown parameters
\end_layout
\begin_layout Paragraph
In example 7.1.3:
\end_layout
\begin_layout Enumerate
Random variables of interest:
\end_layout
\begin_deeper
\begin_layout Enumerate
Observed variable:
\begin_inset Formula $X_{1},X_{2},...$
\end_inset
as i.i.d.
Bernoulli random variables with parameter
\begin_inset Formula $p$
\end_inset
,
\end_layout
\begin_layout Enumerate
\begin_inset Formula $p$
\end_inset
is hypothetically observable
\end_layout
\end_deeper
\begin_layout Enumerate
Distribution of
\begin_inset Formula $p$
\end_inset
is uniform distribution
\end_layout
\begin_layout Enumerate
Distribution of conditionaly
\begin_inset Formula $X_{1},X_{2},..$
\end_inset
if known
\begin_inset Formula $p$
\end_inset
is Bernoulli
\end_layout
\begin_layout Paragraph
2.
\series medium
Identify two statistical inferences mentioned in Example
\begin_inset Formula $7.1.3$
\end_inset
.
\end_layout
\begin_layout Enumerate
Assumed that
\begin_inset Formula $P$
\end_inset
had the uniform distribution on the interval [0,1], and we found the conditiona
l distribution of
\begin_inset Formula $P$
\end_inset
given the observed results of the study.
We also computed the conditional mean of
\begin_inset Formula $P$
\end_inset
given the study results as well as the M.S.E for trying to predict
\begin_inset Formula $P$
\end_inset
both before and after observing the results of the study.s
\end_layout
\begin_layout Paragraph
3.
\series medium
In Examples 7.1.4 and 5.7.8, identify the components of the statistical model
as defined in Definition 7.1.1.
\end_layout
\begin_layout Enumerate
\begin_inset Formula $Z_{k},k=1,2,...$
\end_inset
.
are observable variables and
\begin_inset Formula $\beta$
\end_inset
is hypothetically observable.
\end_layout
\begin_layout Chapter*
7.2 Prior and Posterior Distribution
\end_layout
\begin_layout Problem*
\series bold
1.
\series default
Consider again the situation described in Example 7.2.8.
This time, suppose that the experimenter believes that the prior distribution
of
\begin_inset Formula $\theta$
\end_inset
is the gamma distribution with parameters
\begin_inset Formula $1$
\end_inset
and
\begin_inset Formula $5000$
\end_inset
.
What would this experimenter compute as the value of
\begin_inset Formula $Pr(X_{6}>3000|\boldsymbol{x})$
\end_inset
.
\end_layout
\begin_layout Solution*
We remind that lifetimes of fluerescent lamps are independent exponential
random variables with parameter
\begin_inset Formula $\theta$
\end_inset
, so
\begin_inset Formula $X_{i}$
\end_inset
and
\begin_inset Formula $X_{j}$
\end_inset
is independent in condition given
\begin_inset Formula $\theta$
\end_inset
, or,
\begin_inset Formula $P(X_{i}|\theta,X_{j})=P(X_{i}|\theta)$
\end_inset
.
But
\begin_inset Formula $P(X_{n}|X_{1},...,X_{n-1})$
\end_inset
means that
\begin_inset Formula $X_{n}$
\end_inset
still depends on
\begin_inset Formula $X_{1},...,X_{n-1}$
\end_inset
if not given
\begin_inset Formula $\theta$
\end_inset
.
\end_layout
\begin_layout Standard
We want to find
\begin_inset Formula $P(x_{6}|x_{1},...,x_{5})$
\end_inset
.
\end_layout
\begin_layout Paragraph*
\begin_inset Formula $P(x_{6}|x_{1},...,x_{5})=\int_{\Omega}P(x_{6}|\theta)P(\theta|x_{1},...,x_{5})d\theta$
\end_inset
(5)
\end_layout
\begin_layout Standard
\begin_inset Formula $P(x_{6}|\theta)=\theta e^{-\theta x_{6}}$
\end_inset
\series bold
(6)
\series default
because
\begin_inset Formula $x_{i}$
\end_inset
is exponential distribution with parameter
\begin_inset Formula $\theta$
\end_inset
.
\end_layout
\begin_layout Paragraph*
Now, we find
\begin_inset Formula $P(\theta|x_{1},...,x_{5})$
\end_inset
,
\begin_inset Formula $P(\theta|x_{1},...,x_{5})=\frac{P(x_{1},...,x_{5}|\theta)\xi(\theta)}{g(x_{1},...,x_{5})}$
\end_inset
\series bold
(1)
\series default
with
\begin_inset Formula $g(x_{1},...,x_{5})$
\end_inset
is marginal joint distribution of
\begin_inset Formula $x_{1},...,x_{5}$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Formula $P(x_{1},...,x_{5}|\theta)=\theta^{5}e^{-\theta(x_{1}+...+x_{5})}=\theta^{5}e^{-\theta y}$
\end_inset
\series bold
(2)
\series default
with
\begin_inset Formula $y=x_{1}+...+x_{5}$
\end_inset
.
\end_layout
\begin_layout Paragraph*
\begin_inset Formula $\xi(\theta)$
\end_inset
is prior distribution of
\begin_inset Formula $\theta$
\end_inset
is the gamma distribution with parameters
\begin_inset Formula $1$
\end_inset
and
\begin_inset Formula $5000$
\end_inset
.
\begin_inset Formula $\xi(\theta)=\frac{\beta^{\alpha}}{\Gamma(\alpha)}x^{\alpha-1}e^{-\beta x}$
\end_inset
with
\begin_inset Formula $x>0$
\end_inset
.
So,
\begin_inset Formula $\xi(\theta)=5000e^{-5000\theta}$
\end_inset
\series bold
(3)
\series default
for
\begin_inset Formula $\theta>0$
\end_inset
.
\end_layout
\begin_layout Paragraph
\begin_inset Formula $g(x_{1},...,x_{5})=\int_{\Omega}P(x_{1},...,x_{5}|\theta)\xi(\theta)d\theta=\int_{0}^{\infty}\theta^{5}e^{-(5000+y)\theta}5000d\theta$
\end_inset
\end_layout
\begin_layout Standard
Following
\series bold
Theorem 5.7.3
\series default
for each
\begin_inset Formula $\alpha>0$
\end_inset
and each
\begin_inset Formula $\beta>0$
\end_inset
,
\begin_inset Formula $\int_{0}^{\infty}x^{\alpha-1}exp(\beta x)dx=\frac{\Gamma(\alpha)}{\beta^{\alpha}}$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Formula $g(x_{1},...,x_{5})=5000\frac{\Gamma(6)}{(5000+y)^{6}}$
\end_inset
\series bold
(4)
\end_layout
\begin_layout Paragraph
From (1),(2) and (3), (4):
\begin_inset Formula $P(\theta|x_{1}...,x_{5})=\frac{\theta^{5}e^{-\theta y}5000e^{-5000\theta}}{5000\frac{5!}{(5000+y)^{6}}}=\theta^{5}e^{-(5000+y)\theta}\frac{(5000+y)^{6}}{5!}$
\end_inset
\series bold
(7)
\end_layout
\begin_layout Standard
From (5), (6) and (7) :
\begin_inset Formula $P(x_{6}|x_{1},...,x_{5})=\int_{0}^{\infty}\theta e^{-\theta x_{6}}\theta^{5}e^{-(5000+y)\theta}\frac{(5000+y)^{6}}{5!}d\theta=\int_{0}^{\infty}\theta^{6}e^{-(5000+y+x_{6})\theta}\frac{(5000+y)^{6}}{5!}d\theta=\frac{(5000+y)^{6}}{5!}\frac{\Gamma(7)}{-(5000+y+x_{6})^{7}}$
\end_inset
\end_layout
\begin_layout Standard
We have
\begin_inset Formula $y=16178=x_{1}+..+x_{5}$
\end_inset
.
So
\begin_inset Formula $P(x_{6}>3000|x_{1}...x_{5})=\frac{(5000+16178)^{6}6!}{5!}\int_{3000}^{\infty}\frac{-1}{(5000+16178+x_{6})^{7}}dx_{6}$
\end_inset
.
\begin_inset Formula $\int_{3000}^{\infty}\frac{-1}{(5000+36178+x_{6})^{7}}dx_{6}=\int_{3000}^{\infty}-(x_{6}+41178)^{-7}dx_{6}=-\frac{1}{-6}(x_{6}+41178)^{-6}|_{3000}^{\infty}=\frac{1}{-6x(3000+21178)^{6}}$
\end_inset
\end_layout
\begin_layout Paragraph*
So
\begin_inset Formula $P(x_{6}>3000|x_{1}...x_{5})=\frac{(5000+16178)^{6}6!}{5!}*\frac{1}{-6x(3000+21178)^{6}}=0.4163$
\end_inset
\end_layout
\begin_layout Problem*
\series bold
10.
\series default
\begin_inset Formula $X$
\end_inset
is to taken from the uniform distribution on the interval
\begin_inset Formula $[\theta-\frac{1}{2},\theta+\frac{1}{2}]$
\end_inset
, the value of
\begin_inset Formula $\theta$
\end_inset
is unknown, and the prior distribution of
\begin_inset Formula $\theta$
\end_inset
is the uniform distribution on the interval
\begin_inset Formula $[10,20]$
\end_inset
.
If observed value
\begin_inset Formula $X$
\end_inset
is
\begin_inset Formula $12$
\end_inset
, what is the poterior distribution of
\begin_inset Formula $\theta$
\end_inset
.
\end_layout
\begin_layout Solution*
\begin_inset Formula $P(\theta|X)=\frac{P(X|\theta)\xi(\theta)}{\int_{10}^{20}P(X|\theta)P(\theta)d\theta}$
\end_inset
.
\begin_inset Formula $P(\theta)=1/10$
\end_inset
,
\begin_inset Formula $P(X|\theta)=1$
\end_inset
.
\begin_inset Formula $X=12$
\end_inset
so
\begin_inset Formula $\theta$
\end_inset
is choosed in
\begin_inset Formula $[11.5,12.5]$
\end_inset
, because if choosing
\begin_inset Formula $\theta$
\end_inset
outside this interval , we can choose
\begin_inset Formula $X=12$
\end_inset
, or
\begin_inset Formula $X=12$
\end_inset
cannot be in
\begin_inset Formula $(\theta-12,\theta+1/2)$
\end_inset
.
So
\begin_inset Formula $\int_{10}^{20}P(X|\theta)\xi(\theta)d\theta=\int_{11.5}^{12.5}\frac{1}{10}d\theta=1/10$
\end_inset
.
So
\begin_inset Formula $P(\theta|X=12)=1,$
\end_inset
choose
\begin_inset Formula $\theta$
\end_inset
in
\begin_inset Formula $[11.5,12.5]$
\end_inset
.
So
\begin_inset Formula $P(\theta|X=12)$
\end_inset
is uniform distribution in
\begin_inset Formula $[11.5,12.5]$
\end_inset
.
\end_layout
\begin_layout Chapter*
7.3 Conjugate Prior Distributions
\end_layout
\begin_layout Exercise*
\series bold
16.
\end_layout
\begin_layout Exercise*
Based on Exercise 15, conjugate family of prior distribution for samples
from a normal distribution with a known value of the mean
\begin_inset Formula $\mu$
\end_inset
and unknown value of the standard deviation
\begin_inset Formula $\sigma$
\end_inset
is distribution:
\end_layout
\begin_layout Exercise*
\begin_inset Formula
\[
\xi(\theta)=\begin{cases}
\frac{\beta^{\alpha+1/2}}{\Gamma(\alpha+1/2)}x^{-2(\alpha-1/2)}e^{-2\beta/x^{2}} & x>0\\
0 & otherwise
\end{cases}
\]
\end_inset
\end_layout
\begin_layout Exercise*
Note: we already based on p.d.f of inverse gamma distribution is spend for
prior distribution normal with unknown
\series bold
variance.
\series default
With unknown
\series bold
standard deviation
\series default
we transform some computation in this distribution to present
\begin_inset Formula $x^{2}$
\end_inset
in distribution (particular, in term
\begin_inset Formula $e^{-2\beta/x^{2}}$
\end_inset
) instead of
\begin_inset Formula $e^{-\beta/x}$
\end_inset
, this function cannot compute post distribution with unknown
\series bold
standard deviation.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Chapter*
7.4 Bayes estimator
\end_layout
\begin_layout Exercise*
\series bold
1.
\series default
In a clinical trial, let probability of successful outcome
\begin_inset Formula $\theta$
\end_inset
have a prior distribution that is the uniform distribution on the interval
\begin_inset Formula $[0,1]$
\end_inset
- beta distribution 1, 1.
Suppose that the first patient has a successful outcome.
Find the Bayes estimates of
\begin_inset Formula $\theta$
\end_inset
that would be obtained for both the squared error and absolute error loss
functions.
\end_layout
\begin_layout Solution*
Following Theorem 7.3.4, the posterior distribution of
\begin_inset Formula $\theta$
\end_inset
is the gamma distribution with parameter
\begin_inset Formula $\alpha=1+1=2$
\end_inset
,
\begin_inset Formula $\beta=1$
\end_inset
\end_layout
\begin_layout Solution*
So with loss function is the squared error, the
\begin_inset Formula $\delta^{*}(x)=\mu=\frac{\alpha}{\alpha+\beta}=2/3$
\end_inset
.
Otherwise, with loss function is absolute error loss function, then
\begin_inset Formula $\delta^{*}(x)=P^{-1}(0.5)\approx0.7$
\end_inset
\end_layout
\begin_layout Exercise*
\series bold
12.
\series default
\begin_inset Formula $\xi_{A}=2\theta$
\end_inset
is Gamma distribution with parameter 2, 1 ;
\begin_inset Formula $\xi_{B}=4\theta^{3}$
\end_inset
is Gamma distribution with parameter 3, 1.
So apply theorem from 7.3 we can find posterior distribtion.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Problem*
\series bold
14
\series default
\begin_inset Formula $\hat{\psi}=E[x^{2}]\geq\mu^{2}\text{ (because variance > 0)}=\hat{\theta}^{2}$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Chapter*
7.6 Properties of Maximum likelihood estimators.
\end_layout
\begin_layout Problem*
1.
Suppose that
\begin_inset Formula $X_{1},...,X_{n}$
\end_inset
form a random sample from a distribution with the p.d.f.
given in Exercise 10 of Sec.
7.5.
Find M.L.E of
\begin_inset Formula $e^{-1/\theta}$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
\begin_inset Formula $X_{1},...,X_{n}$
\end_inset
form a random sample from a distribution
\begin_inset Formula $f(x|\theta)=\frac{1}{2}e^{-|x-\theta|}$
\end_inset
, we choose
\begin_inset Formula $\theta$
\end_inset
with M.L.E is median of
\begin_inset Formula $X_{1},...,X_{n}$
\end_inset
: sort(
\begin_inset Formula $X_{1},...,X_{n}$
\end_inset
) and choose median of this sorted array.
\end_layout
\begin_layout Solution*
M.L.E of
\begin_inset Formula $g(\theta)=e^{-1/\theta}$
\end_inset
(one-to-one function) is
\begin_inset Formula $e^{-1/\text{median(}X_{1},...,X_{n}\text{)}}$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Exercise*
2.
Suppose that
\begin_inset Formula $X_{1},..,X_{n}$
\end_inset
form a random sample from a Poisson distribution for which the mean is
unknown.
Determine the M.L.E.
of the standard deviation of the distribution.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
\begin_inset Formula $\mu=\lambda$
\end_inset
,
\begin_inset Formula $\sigma=\sqrt{\lambda}$
\end_inset
.
With M.L.E
\begin_inset Formula $\hat{\lambda}=\frac{1}{n}\sum_{i=1}^{n}X_{i}$
\end_inset
, so
\begin_inset Formula $\hat{\sigma}=\sqrt{\frac{1}{n}\sum_{i=1}^{n}X_{i}}$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Exercise*
3.
Suppose that
\begin_inset Formula $X_{1},...,X_{n}$
\end_inset
form a random sample from an exponential distribution for which the value
of the parameter
\begin_inset Formula $\beta$
\end_inset
is unknown.
Determine the M.L.E of the median of the distribution
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Solution*
First, we represent median of exponetial distribution following
\begin_inset Formula $\beta$
\end_inset
:
\begin_inset Formula $\int_{0}^{n}\beta e^{-\beta x}dx=0.5$
\end_inset
,
\begin_inset Formula $n$
\end_inset
is median number of exponential distribution.
\begin_inset Formula $n=g(\beta)=\frac{ln2}{\beta}$
\end_inset
.
Using M.L.E for finding
\begin_inset Formula $\beta$
\end_inset
:
\begin_inset Formula $\hat{\beta}=\frac{n}{\sum_{i=1}^{n}X_{i}}$
\end_inset
.
So
\begin_inset Formula $\hat{n}=g(\hat{\beta})=ln2.\bar{X}_{n}$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Exercise*
Lifetime of a certain type of lamp has an exponential distribution:
\begin_inset Formula $\beta$
\end_inset
is unknown.
A random sample of
\begin_inset Formula $n$
\end_inset
lamps of this type are tested for a period of
\begin_inset Formula $T$
\end_inset
hours and the number
\begin_inset Formula $X$
\end_inset
of lamps that fail during this period is observed, but the times at which
the failures occured are not noted.
Determine the M.L.E of
\begin_inset Formula $\beta$
\end_inset
based on the observed value of
\begin_inset Formula $X$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset