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Chapter12_exercise.lyx
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Chapter12_exercise.lyx
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#LyX 2.3 created this file. For more info see http://www.lyx.org/
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\begin_document
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\end_header
\begin_body
\begin_layout Chapter*
12.1
\end_layout
\begin_layout Exercise*
5.
c)
\begin_inset Formula $P(X,Y)=P(X)P(Y)$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Chapter*
12.2 Why is Simulation Useful ?
\end_layout
\begin_layout Exercise*
2.
In example, following Eq.
(12.2.6),
\begin_inset Formula $v=\frac{\hat{\sigma}^{2}}{\epsilon^{2}(1-\gamma)}$
\end_inset
, with
\begin_inset Formula $\epsilon=0.01,\gamma=0.95$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Exercise*
6.b.
Determine an appropriate:
\end_layout
\begin_layout Exercise*
\begin_inset Formula $X\sim Beta(3.5,3.7),Y\sim Beta(1.8,4.2)$
\end_inset
.
Find distribution of
\begin_inset Formula $\frac{X}{X+Y}$
\end_inset
\end_layout
\begin_layout Exercise*
Remind:
\begin_inset Formula $Beta(\alpha,\beta)=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}$
\end_inset
\end_layout
\begin_layout Exercise*
From Theorem 5.8.4, we can have
\begin_inset Formula $\frac{X}{X+Y}$
\end_inset
will have beta distribution with parameter:
\begin_inset Formula $\alpha$
\end_inset
,
\begin_inset Formula $\beta$
\end_inset
.
\end_layout
\begin_layout Exercise*
and
\begin_inset Formula $\sigma^{2}=\frac{\alpha\beta}{(\alpha+\beta)^{2}(\alpha+\beta+1)}$
\end_inset
(from Theorem 5.3.8)
\end_layout
\begin_layout Exercise*
So we can find appropriate number
\begin_inset Formula $v=\frac{\sigma^{2}}{\epsilon^{2}(1-\gamma)}$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Exercise*
7.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Exercise*
9
\end_layout
\begin_layout Standard
Reminder:
\end_layout
\begin_layout Standard
Following Eq.
(3.6.12):
\end_layout
\begin_layout Standard
If we have p.d.f of
\begin_inset Formula $Y$
\end_inset
is
\begin_inset Formula $f_{2}(y)$
\end_inset
and
\begin_inset Formula $g_{1}(x|y)$
\end_inset
is conditional p.d.f of
\begin_inset Formula $X$
\end_inset
given
\begin_inset Formula $Y=y$
\end_inset
.
Then p.d.f of
\begin_inset Formula $X$
\end_inset
is
\begin_inset Formula $f_{1}(x)=\int_{-\infty}^{\infty}g_{1}(x|y)f_{2}(y)dy$
\end_inset
.
\end_layout
\begin_layout Standard
Eq (5.7.10):
\end_layout
\begin_layout Standard
\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
In Example 12.2.12
\end_layout
\begin_layout Standard
We have
\begin_inset Formula $f(x|\mu)=\mu e^{-\mu x};f(\mu)=\frac{\beta^{\alpha}}{\Gamma(\alpha)}x^{\alpha-1}e^{-\beta x},\beta=1,\alpha=3$
\end_inset
.
\end_layout
\begin_layout Standard
So based on Eq (3.6.12):
\begin_inset Formula $f(x)=\int_{-\infty}^{\infty}\mu e^{-\mu x}\frac{\beta^{\alpha}}{\Gamma(\alpha)}\mu^{\alpha-1}e^{-\beta\mu}d\mu$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $=\frac{1}{\Gamma(3)}\int_{0}^{\infty}\mu e^{-\mu x}\mu^{2}e^{-\mu}d(\mu)$
\end_inset
.
\begin_inset Formula $=\frac{1}{\Gamma(3)}\int_{0}^{\infty}\mu^{3}e^{-\mu(\alpha+1)}d\text{\ensuremath{\mu}}$
\end_inset
based on Eq.
(5.7.10):
\end_layout
\begin_layout Standard
\begin_inset Formula $f(x)=\frac{1}{\Gamma(3)}\frac{\Gamma(4)}{(x+1)^{4}}=\frac{3}{(x+1)^{4}}$
\end_inset
.
So
\begin_inset Formula $F(x)=\int f(x)=\frac{-1}{(x+1)^{3}}=0.5\rightarrow x$
\end_inset
\end_layout
\begin_layout Exercise*
10.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Standard
a.
Median of Exponential distribution is
\begin_inset Formula $\frac{\ln2}{\lambda}$
\end_inset
.
\end_layout
\begin_layout Standard
b.
with
\begin_inset Formula $\lambda=1$
\end_inset
, M.S.E of sample median is
\begin_inset Formula $\theta=E[(Y-\frac{\ln2}{\lambda})]^{2}=E[(Y-ln2)]^{2}$
\end_inset
.
So In general, with
\begin_inset Formula $\lambda$
\end_inset
, MSE
\begin_inset Formula $=\frac{1}{\lambda^{2}}E[(Y-ln2)]^{2}=\frac{\theta}{\lambda^{2}}$
\end_inset
.
\end_layout
\begin_layout Standard
c.
Simulation method for estimating
\begin_inset Formula $\theta$
\end_inset
.
\end_layout
\begin_layout Standard
Generate
\begin_inset Formula $v$
\end_inset
of
\begin_inset Formula $n$
\end_inset
sample for
\begin_inset Formula $i=1,...,v$
\end_inset
find
\begin_inset Formula $Y^{i}$
\end_inset
is median sample of that
\begin_inset Formula $n$
\end_inset
sample.
Then We compute
\begin_inset Formula $Z_{i}=(Y^{i}-ln2/\lambda)^{2}$
\end_inset
.
Then compute
\begin_inset Formula $\bar{Z}=\frac{1}{n}\sum Z_{i}$
\end_inset
.
(From Example 12.2.6)
\end_layout
\begin_layout Exercise*
12.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Standard
Apply
\begin_inset Formula $Var(aX+bY+c)=a^{2}Var(X)+b^{2}Var(Y)+2abCov(X,Y)$
\end_inset
.
\end_layout
\begin_layout Standard
with
\begin_inset Formula $aX+bY+c$
\end_inset
in this situation equals
\begin_inset Formula $g(\bar{Y},\bar{W})=g(E(Y),E(W))+g_{1}(E(Y),E(W))(\bar{Y}-E(Y))+g_{2}(E(Y),E(W))(\bar{W}-E(W))$
\end_inset
.
\end_layout
\begin_layout Exercise*
13.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $Z=g(Y,W)=\bar{W}-\bar{Y}^{2}$
\end_inset
.
\begin_inset Formula $g_{1}=\frac{\partial g}{\partial Y}=-2Y$
\end_inset
.
\begin_inset Formula $g_{2}=\frac{\partial g}{\partial W}=1$
\end_inset
\end_layout
\begin_layout Standard
Apply exercise 12 but get
\begin_inset Formula $y_{0}=\bar{Y}$
\end_inset
,
\begin_inset Formula $w_{0}=\bar{W}$
\end_inset
\end_layout
\begin_layout Standard
So
\begin_inset Formula $g(y,w)=g_{1}(\bar{Y},\bar{W})^{2}\sigma_{YY}+2g_{1}(\bar{Y},\bar{W})g_{2}(\bar{Y},\bar{W})\sigma_{yw}+g_{2}(\bar{Y},\bar{W})\sigma_{ww}$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Formula $=4\bar{Y}^{2}\sigma_{YY}-4\bar{Y}\sigma_{YW}+\sigma_{WW}$
\end_inset
.
From Example 12.2.10,
\begin_inset Formula $\sigma_{YY}=Z\text{ }$
\end_inset
,
\begin_inset Formula $\sigma_{YW}=C$
\end_inset
,
\begin_inset Formula $\sigma_{WW}=V$
\end_inset
.
\end_layout
\begin_layout Standard
So we have result.
\end_layout
\begin_layout Paragraph*
Note:
\series medium
From Exercise 12, 13 we can conclude if we have to compute variance of function
of 2 variables we can use Taylor expansion as exercise 12 to separate joint
distribution to one distribution, then we can compute variance for each
term based on individual variance (maybe we have to compute extra covariance).
\end_layout
\begin_layout Exercise*
14.
Generating
\begin_inset Formula $n$
\end_inset
sample , then compute
\begin_inset Formula $\hat{\sigma}$
\end_inset
of
\begin_inset Formula $n$
\end_inset
elements.
Compute
\begin_inset Formula $\mu=\bar{X}$
\end_inset
, them compute
\begin_inset Formula $E((X-\mu)^{3})=\frac{1}{n}(X_{i}-\bar{X})^{3}.$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Chapter*
12.3 Simulating Specific Distribution
\end_layout
\begin_layout Exercise*
2.
\begin_inset Formula $U_{1},...,U_{n}$
\end_inset
is uniform distribution on the interval
\begin_inset Formula $[0,1]$
\end_inset
.
Convert to
\begin_inset Formula $n$
\end_inset
uniform distribution on the interval
\begin_inset Formula $[a,b]$
\end_inset
:
\begin_inset Formula $U_{x}a+b-a$
\end_inset
.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Exercise*
3.
In this exercise, the two most important part is generating sample from
two distributions:
\begin_inset Formula $g_{1}(x)=\frac{1}{2x^{1/2}}$
\end_inset
and
\begin_inset Formula $g_{2}(x)=\frac{1}{2(1-x)^{1/2}}$
\end_inset
.
\end_layout
\begin_layout Exercise*
First, we have to find c.d.f
\begin_inset Formula $G$
\end_inset
for each density function, then we find its
\begin_inset Formula $x=G^{-1}(y)=f(y)$
\end_inset
, finally
\begin_inset Formula $f(y)$
\end_inset
of
\begin_inset Formula $y$
\end_inset
having uniform disitribution on
\begin_inset Formula $[0,1]$
\end_inset
will be simulated.
\end_layout
\begin_layout Exercise*
For
\begin_inset Formula $g_{1}(x)=\frac{1}{2x^{1/2}}$
\end_inset
.
\begin_inset Formula $G(x)=x^{1/2}.$
\end_inset
\begin_inset Formula $G^{-1}(y)=y^{2}$
\end_inset
.
So if generating
\begin_inset Formula $y$
\end_inset
from uniform distribution on
\begin_inset Formula $[0,1]$
\end_inset
,
\begin_inset Formula $y^{2}$
\end_inset
will have distribution
\begin_inset Formula $g_{1}(x)$
\end_inset
.
\end_layout
\begin_layout Exercise*
Similarly,
\begin_inset Formula $1-y^{2}$
\end_inset
have
\begin_inset Formula $g_{2}(x)$
\end_inset
distirbution, with
\begin_inset Formula $y$
\end_inset
having uniform distribution on
\begin_inset Formula $[0,1]$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Exercise*
6.
a.
Laplace distribution with parameters are
\begin_inset Formula $1,1$
\end_inset
:
\begin_inset Formula $f(x|1,1)=\frac{1}{2}e^{-|x-1|}$
\end_inset
.
\end_layout
\begin_layout Exercise*
c.d.f
\begin_inset Formula $G(x)=\begin{cases}
\frac{1}{2}exp(x-1) & x\le1\\
1-\frac{1}{2}exp(-x+1) & x\ge1
\end{cases}$
\end_inset
.
\end_layout
\begin_layout Exercise*
\begin_inset Formula $G^{-1}(y)=\begin{cases}
1+log(2y) & log(2y)\le0\\
1-log(1-2y) & log(1-2y)\le0
\end{cases}$
\end_inset
\end_layout
\begin_layout Exercise*
if
\begin_inset Formula $y\in Uniform[0,1]$
\end_inset
them
\begin_inset Formula $G^{-1}(Y)$
\end_inset
will have Laplace distribution.
This is way to simulate Laplace: First we simulate
\begin_inset Formula $Y$
\end_inset
in uniform
\begin_inset Formula $[0,1]$
\end_inset
, if
\begin_inset Formula $log(2y)\le0$
\end_inset
them
\begin_inset Formula $1+log(2y)$
\end_inset
will be coresponding value sampling in Laplace distribution, otherwise,
if
\begin_inset Formula $log(1-2y)\le0$
\end_inset
, then
\begin_inset Formula $1-log(1-2y)$
\end_inset
will be sample value.
\end_layout
\begin_layout Exercise*
b.
To sample normal we find
\begin_inset Formula $k$
\end_inset
for
\begin_inset Formula $\frac{f(x)}{g(x)}\le k$
\end_inset
, with
\begin_inset Formula $f(x)$
\end_inset
is normal distribution and
\begin_inset Formula $g(x)$
\end_inset
is laplace distribution.
\begin_inset Formula $\frac{f(x)}{g(x)}\sim\frac{e^{-x^{2}/2}}{e^{-x}}=A$
\end_inset
, maxinmize
\begin_inset Formula $A$
\end_inset
with find derivation of A equals
\begin_inset Formula $0$
\end_inset
with find
\begin_inset Formula $x=1$
\end_inset
, then
\begin_inset Formula $k=e^{1/2}$
\end_inset
.
So we apply procedure similar to section after 12.3.3.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Exercise*
7.
F-formation have form:
\begin_inset Formula $\frac{X/n}{Y/m}$
\end_inset
with
\begin_inset Formula $X,Y$
\end_inset
have
\begin_inset Formula $\chi^{2}$
\end_inset
freedom is
\begin_inset Formula $n,m$
\end_inset
.
So we simulate
\begin_inset Formula $n$
\end_inset
normal distribution value:
\begin_inset Formula $X_{1},...,X_{n}$
\end_inset
.
Similarly,
\begin_inset Formula $Y_{1,},...,Y_{m}$
\end_inset
.
Then simulate for
\begin_inset Formula $F_{n,m}:\frac{(\sum_{i}X_{i})/n}{(\sum_{j}Y_{j})/m}$
\end_inset
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Exercise*
8.
c.
\begin_inset Formula $v=[\Phi^{-1}(\frac{1+\gamma}{2})\frac{\sigma}{\epsilon}]^{2}\sim[\Phi^{-1}(\frac{1+\gamma}{2})\frac{\hat{\sigma}}{\epsilon}]^{2}$
\end_inset
.
But we didn't know c.d.f
\begin_inset Formula $\Phi$
\end_inset
of
\begin_inset Formula $|X-Y|$
\end_inset
.
With
\begin_inset Formula $X,Y$
\end_inset
is two
\begin_inset Formula $t$
\end_inset
distribution.
\end_layout
\begin_layout Standard
\begin_inset Formula $\rightarrow$
\end_inset
We use simulation Monte-Carlo
\begin_inset Formula $Z=\frac{1}{n}\sum_{i}^{n}|X_{i}-Y_{i}|$
\end_inset
.
So
\begin_inset Formula $Z$
\end_inset
have approximately normal distribution with
\begin_inset Formula $\mu=E(|X-Y|)$
\end_inset
(Central limit theorem), then
\begin_inset Formula $\Phi$
\end_inset
is c.d.f.
of standard normal distribution.
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Exercise*
9.
We have to find
\begin_inset Formula $g(x)$
\end_inset
such that
\begin_inset Formula $kg(x)\ge f(x)$
\end_inset
??
\end_layout
\begin_layout Standard
\begin_inset Separator plain
\end_inset
\end_layout
\begin_layout Exercise*
10.
Using uniform distribution to simulate
\begin_inset Formula $n$
\end_inset
sample in discrete distribution for 4 elements
\begin_inset Formula $i=1,2,3,4$
\end_inset
.
Then, posterior distribution of
\begin_inset Formula $P_{i}$
\end_inset
is beta distribution with parameters :
\begin_inset Formula $\alpha_{0}+x_{i}$
\end_inset
,
\begin_inset Formula $\beta_{0}+n_{i}-x_{i}$
\end_inset
, as defined in Example 12.2.3
\begin_inset Formula $(n_{1}+n_{2}+n_{3}+n_{4}=n)$
\end_inset
.
From exercise we have
\begin_inset Formula $\alpha_{0}=1,\beta_{0}=1$
\end_inset
, so we have
\begin_inset Formula $4$
\end_inset
distribution.
Then use simulation for beta in Example.
12.3.3 to simulate
\begin_inset Formula $5000$
\end_inset
vector and then, compute
\begin_inset Formula $\hat{p}_{1\_max}=\frac{n_{1\_max}}{n}$
\end_inset
.
\begin_inset Formula $n_{1\_max}$
\end_inset
is number vector having
\begin_inset Formula $p_{1}$
\end_inset
max.
\end_layout
\begin_layout Exercise*
b.
Compute number of sample
\begin_inset Formula $n$
\end_inset
such that,
\begin_inset Formula $P(|Z-p_{1\_max}|\le0.01)\ge0.95$