Homework1 for STAT6202
Autor:
Yang Liu
Last Updated:
hace 6 años
License:
LaTeX Project Public License 1.3c
Resumen:
STAT6202 Homework template
\begin
Discover why over 20 million people worldwide trust Overleaf with their work.
STAT6202 Homework template
\begin
Discover why over 20 million people worldwide trust Overleaf with their work.
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\begin{document}
\title{Homework 1 STAT 6202}%replace X with the appropriate number
\author{Yang Liu\\ %replace with your name
Instructor: Professor Tapan K. Nayak } %if necessary, replace with your course title
\maketitle
\hrule
\section*{1.} Let $X_1,\cdots,X_n$ be i.i.d. Bernoulli variables with success probability $\theta$, when $n>2$, and let $T=\sum_{i=1}^{n}X_i$. Derive the conditional distribution $X_1,\cdots,X_n$ given $T=t$.
\begin{proof}
Since $X_1,\cdots,X_n\stackrel{i.i.d.}{\sim}Bernoulli(\theta)$, and $T=\sum_{i=1}^nX_i\sim Binomial (n,\theta)$
\begin{align*}
P\left(X_1=x_1,\cdots,X_n=x_n\right)& =\prod_{i=1}^{n}\theta^{x_i}(1-\theta)^{1-x_i} \\
P\left(X_1=x_1,\cdots,X_n=x_n, T=\sum_{i=1}^nX_i=t\right)& =\theta^t(1-\theta)^{n-t}\\
P\left(\left.X_{1}=x_{1},\cdots,X_{n}=x_{n}\right|\sum_{i=1}^{n}X_{i}=t\right)& =\frac{\theta^t(1-\theta)^{n-t}}{\left(\begin{array}{c}
n\\
t
\end{array}\right)\theta^{t}\left(1-\theta\right)^{n-t}}\\
& = \frac{1}{\left(\begin{array}{c}
n\\
t
\end{array}\right)}
\end{align*}
\end{proof}
\section*{2.} Suppose $X_1$ and $X_2$ are iid $Poisson(\theta)$ random variables and let $T = X_1 + 2X_2$.
\begin{enumerate}[label=(\alph*)]
\item Find the conditional distribution of $(X_1, X_2)$ given $T = 7$.
\item For $\theta = 1$ and $\theta = 2$, respectively, calculate all probabilities in the above conditional
distribution and present the two conditional distributions numerically.
\end{enumerate}
\begin{proof}
\begin{enumerate}[label=(\alph*)]
\item Since $X_1,X_2\stackrel{i.i.d.}{\sim }Poisson (\theta)$, then we have
\begin{align*}
\left \lbrace X_1+2X_2=7 \right\rbrace= \left\lbrace (X_1=1,X_2=3), (X_1=3,X_2=2),(X_1=5,X_2=1),(X_1=7,X_2=0)\right\rbrace
\end{align*}
and $(X_1=1,X_2=3), (X_1=3,X_2=2),(X_1=5,X_2=1),(X_1=7,X_2=0)$ are mutually exclusive, then
\begin{align*}
P\left(T=7\right)& =P\left(X_1=1,X_2=3\right)+P \left(X_1=3,X_2=2\right)\\&\ \ +P\left(X_1=5,X_2=1\right)+P(X_1=7,X_2=0)\\
& = \frac{\theta}{1}e^{-\theta}\cdot \frac{\theta^3}{3!}e^{-\theta}+ \frac{\theta^3}{3!}e^{-\theta}\cdot \frac{\theta^2}{2!}e^{-\theta}+\frac{\theta^5}{5!}e^{-\theta}\cdot \frac{\theta^1}{1!}e^{-\theta}+ \frac{\theta^7}{7!}e^{-\theta}\cdot e^{-\theta}\\
& = \frac{\theta^4e^{-2\theta}}{6}\left(1+ \frac{\theta}{2}+\frac{\theta^2}{20}+\frac{\theta^3}{840}\right)
\end{align*}
Then the conditional distribution of $(X_1,X_2)$ given $T=7$ is
\begin{align*}
P\left(\left.X_1=1,X_2=3\right|T=7\right)& = \frac{P\left(X_1=1,X_2=3\right)}{P\left(T=7\right)}\\
& = \frac{ \frac{\theta^4e^{-2\theta}}{6}}{ \frac{\theta^4e^{-2\theta}}{6}\left(1+ \frac{\theta}{2}+\frac{\theta^2}{20}+\frac{\theta^3}{840}\right)}\\
& = \frac{840}{840+420\theta+42\theta^2+\theta^3}\\
P\left(\left.X_1=3,X_2=2\right|T=7\right)& =\frac{P\left(X_1=3,X_2=2\right)}{P\left(T=7\right)} \\& =\frac{ \frac{\theta^4e^{-2\theta}}{6}\cdot \frac{\theta}{2}
}{ \frac{\theta^4e^{-2\theta}}{6}\left(1+ \frac{\theta}{2}+\frac{\theta^2}{20}+\frac{\theta^3}{840}\right)}\\
& = \frac{420\theta}{840+420\theta+42\theta^2+\theta^3} \\P\left(\left.X_1=5,X_2=1\right|T=7\right)& =\frac{P\left(X_1=5,X_2=1\right)}{P\left(T=7\right)} \\
& =\frac{ \frac{\theta^4e^{-2\theta}}{6}\cdot \frac{\theta^2}{20}
}{ \frac{\theta^4e^{-2\theta}}{6}\left(1+ \frac{\theta}{2}+\frac{\theta^2}{20}+\frac{\theta^3}{840}\right)}\\
& = \frac{42\theta^2}{840+420\theta+42\theta^2+\theta^3}\\
P\left(\left.X_1=7,X_2=0\right|T=7\right)& =\frac{P\left(X_1=7,X_2=0\right)}{P\left(T=7\right)} \\
& =\frac{ \frac{\theta^4e^{-2\theta}}{6}\cdot \frac{\theta^3}{840}
}{ \frac{\theta^4e^{-2\theta}}{6}\left(1+ \frac{\theta}{2}+\frac{\theta^2}{20}+\frac{\theta^3}{840}\right)}\\
& = \frac{\theta^3}{840+420\theta+42\theta^2+\theta^3}
\end{align*}
\item
The conditional distribution of $(X_1,X_2)|T=7$ is given in table 1.
\begin{table}[H]
\centering
\caption{Conditional distribution of $(X_1,X_2)$ given $T=7$}
\begin{tabular}{c|cccc}
\hline
$P(X_1=x_1,X_2=x_2|T=7)$ & $(x_1=1,x_2=3)$ & $(x_1=3,x_2=2)$ & $(x_1=5,x_2=1)$& $(x_1=7,x_2=0)$ \\
\hline
$\theta=1$ & $\frac{840}{1303}$ & $\frac{420}{1303}$ & $\frac{42}{1303}$& $\frac{1}{1303}$ \\
\hline
$\theta=2$ & $\frac{840}{1856}$& $\frac{840}{1856}$ & $\frac{169}{1856}$ & $\frac{8}{1856}$ \\
\hline
\end{tabular}
\end{table}
\end{enumerate}
\end{proof}
\section*{3.} Let $X_1,\cdots,X_n$ be i.i.d. random variables with mean $\mu$ and variance $\sigma^2$. Let $\bar{X}$ denote the sample mean and $V=\sum_{i=1}^{n}\left(X_i-\bar{X}\right)^2$.
\begin{enumerate}[label=(\alph*)]
\item Derive the expected value of $\bar{X}$ and $V$.
\item Further suppose that $X_1,\cdots,X_n$ are normally distributed. Let $A_{n\times n} = ((a_{ij}))$
be an orthogonal matrix whose first row is $(\frac{1}{\sqrt{n}},\cdots,\frac{1}{\sqrt{n}})$ and let $Y = AX$, where
$Y = (Y_1,\cdots, Y_n)'$ and $X = (X_1,\cdots,X_n)$ are (column) vectors. (It is not necessary to
know $a_{ij}$ for $i = 2,\cdots, n$, $j = 1,\cdots, n$ for answering the following questions.)
\begin{enumerate}[label=(\roman*)]
\item Find $\sum_{j=1}^n a_{ij}$ for $i=1,\cdots,n$ and show that
$\sum_{i=1}^n Y_i^2 = \sum_{i=1}^nX_i^2$ (Use properties
of orthogonal matrices.)
\item Express $\bar{X}$ and $V$ in terms (or as functions) of $Y_1,\cdots,Y_n$.
\item Use (only) \textit{transformation of variables} approach to find the joint distribution of
$Y_1,\cdots,Y_n$. Are $Y_1,\cdots,Y_n$ independently distributed and what are their marginal distributions?
\item Prove that $\bar{X}$ and $V$ are independent given their marginal distributions.
\end{enumerate}
\end{enumerate}
\begin{proof}
\begin{enumerate}[label=(\alph*)]
\item Since $E[X_i]=\mu$ and $Var[X_i]=\sigma^2$ for $i=1,\cdots,n$
\begin{align*}
E[\bar{X}]&= E\left[\frac{\sum_{i=1}^m X_i}{n}\right]= \frac{\sum_{i=1}^n E[X_i]}{n}= \frac{n\mu}{n}=\mu
\end{align*}
\begin{align*}
Var\left[\bar{X}\right]&=E\left[\left(\bar{X}-\mu\right)^2\right]= Var\left[\frac{\sum_{i=1}^nX_i}{n}\right]\\&= \frac{\sum_{i=1}^n Var[X_i]}{n^2} = \frac{n\sigma^2}{n^2}=\frac{\sigma^2}{n}
\end{align*}
\begin{align*}
E\left[V\right]& = E\left[\sum_{i=1}^n\left(X_i-\bar{X}\right)^2\right]=E\left[\sum_{i=1}^n\left((X_i-\mu)-(\bar{X}-\mu)\right)^2\right]\\
& = E\left[ \sum_{i=1}^n \left(X_i-\mu\right)^2 \right]+ n E\left[\left(\bar{X}-\mu\right)^2\right]-2E\left[ \sum_{i=1}^n (X_i-\mu)(\bar{X}-\mu)\right]\\& = E\left[ \sum_{i=1}^n \left(X_i-\mu\right)^2 \right]- n E\left[\left(\bar{X}-\mu\right)^2\right]\\
&= n Var[X_i]- nVar[\bar{X}]\\
& = n\sigma^2 - n\cdot\frac{\sigma^2}{n}=(n-1)\sigma^2
\end{align*}
Or since
\begin{align*}
E\left[\bar{X}^2\right]& =Var[\bar{X}]+E[\bar{X}^2]\\
& = \frac{\sigma^2}{n}+\mu^2
\end{align*}
\begin{align*}
E\left[V\right]& = E\left[\sum_{i=1}^n\left(X_i-\bar{X}\right)^2\right]=E\left[ \sum_{i=1}^n X_i^2-2\sum_{i=1}X_i\bar{X}^2+ n\bar{X}^2\right]\\
& = E\left[\sum_{i=1}^nX_i^2-n\bar{X}^2\right]\\
& =n\cdot\left[\sigma^2+\mu^2\right]-n\cdot\left[ \frac{\sigma^2}{n}+\mu^2\right]\\
& = (n-1)\sigma^2
\end{align*}
\item
\begin{enumerate}[label=(\roman*)]
\item Due to the orthogonality of $A$, $A'A=AA'=I_{n\times n}$ ($I_{n\times n}$ is diagonal matrix of 1's. Let $A=(a_{1\cdot},\cdots,a_{n\cdot })'$ where $a_{j\cdot}$ is the $j^{th}$ row vector. Then we have for $i,j=1,\cdots,n$
$$a_{i\cdot }a_{i\cdot}'=1 \ and \ a_{i\cdot }a_{j\cdot }'=0$$
\begin{align*}
a_{1\cdot}a_{1\cdot}'&=\sum_{k=1}^n \frac{1}{\sqrt{n}}\cdot \frac{1}{\sqrt{n}}=\frac{\sum_{j=1}^na_{1j}}{\sqrt{n}}=1
\\
a_{i\cdot}a_{1\cdot}'&=\sum_{j=1}^na_{ij}\cdot\frac{1}{\sqrt{n}}=\frac{\sum_{j=1}^na_{ij}}{\sqrt{n}}=0
\end{align*}
Hence $\sum_{j=1}^n a_{ij}=\sqrt{n}$ for $j=1$ and $\sum_{j=1}^n a_{ij}=0$ for $j=2,\cdots,n$.
\begin{align*}
\sum_{i=1}^n Y_i ^2& =Y'Y = X'A'AX=X'(A'A)X\\
& = X'X=\sum_{i=1}^n X_i^2
\end{align*}
\item
Note that $Y_1= \sum_{i=1}^n\frac{1}{\sqrt{n}}\cdot X_i= \frac{\sum_{i=1}^nX_i}{\sqrt{n}}=\sqrt{n}\cdot\bar{X}$
\begin{align*}
\sum_{i=2}^nY_i^2&= Y'Y-Y_1^2=\sum_{i=1}^n X_i^2-\left(\sqrt{n}\bar{X}\right)^2\\
& = \sum_{i=1}^nX_i-n\bar{X}^2 \\
& =\sum_{i=1}^n \left(X_i-\bar{X}\right)^2
\end{align*}
Therefore $\bar{X}=\frac{Y_1}{\sqrt{n}}$ and $\sum_{i=1}^n \left(X_i-\bar{X}\right)^2 = \sum_{i=2}^n Y_i^2$
\item since $X_1,\cdots,X_n\stackrel{i.i.d.}{\sim}N\left(\mu,\sigma^2\right)$
\begin{align*}
f_{X_1,\cdots,X_n}\left(x_1,\cdots,x_n\right)& = \prod_{i=1}^n \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x_i-\mu)^2}{2\sigma^2}}\\
& = \left(2\pi\right)^{-\frac{n}{2}}\sigma^{-n}e^{-\frac{\sum_{i=1}^n\left(x_i-\mu\right)^2}{2\sigma^2}}\\
& = \left(2\pi\right)^{-\frac{n}{2}}\sigma^{-n}e^{-\frac{(X-\mathbf{1}\mu)'(X-\mathbf{1}\mu)}{2\sigma^2}}
\end{align*}
where $\mathbf{1}=(1,\cdots,1)'$. Let $A=(a_{\cdot1},\cdots,a_{\cdot n})$, $A'=(a_{\cdot 1},\cdots,a_{\cdot n})'$ where $a_{\cdot j}$ is the $j^{th}$ column vector,
since $Y=AX$, $X=A'AX=A'Y$, $\frac{d}{dY}X= A$, $\left|\frac{d}{dY}X\right|=\det(A)=\det(A'A)=1$, then we have
\begin{align*}
f_{Y_1,\cdots,Y_n}\left(y_1,\cdots,y_n\right)& =\left. f_{X_1,\cdots,X_n}(x_1,\cdots,x_n)\left|\frac{d}{dY}X\right|\right|_{X=A'Y}\\
& =\left. \left(2\pi\right)^{-\frac{n}{2}}\sigma^{-n}e^{-\frac{(X-\mu)'(X-\mu)}{2\sigma^2}}\right|_{X=A'Y}\\
& = \left(2\pi\right)^{-\frac{n}{2}}\sigma^{-n}e^{-\frac{(A'Y-\mathbf{1}\mu)'(A'Y-\mathbf{1}\mu)}{2\sigma^2}}\\
& = \left(2\pi\right)^{-\frac{n}{2}}\sigma^{-n}e^{-\frac{(Y-A\mathbf{1}\mu)'AA'(Y-A\mathbf{1}\mu)}{2\sigma^2}}\\
& = \left(2\pi\right)^{-\frac{n}{2}}\sigma^{-n}e^{-\frac{(Y-A\mathbf{1}\mu)'(Y-A\mathbf{1}\mu)}{2\sigma^2}}\\
& = \left(2\pi\right)^{-\frac{n}{2}}\sigma^{-n}e^{-\frac{\sum_{i=1}^n(y_i-a_{i\cdot }\mathbf{1}\mu)^2}{2\sigma^2}}\\
& = \left(2\pi\right)^{-\frac{n}{2}}\sigma^{-n}e^{-\frac{\sum_{i=1}^n\left(y_i-\sum_{j=1}^na_{ij}\mu\right)^2}{2\sigma^2}}\\
& = \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{(y_1-\sqrt{n}\mu)^2}{2\sigma^2}}\cdot\prod_{i=2}^n \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{y_i^2}{2\sigma^2}}
\end{align*}
The last equation is due to (b) (i). Note that $E[Y]= AE[X]=A\mathbf{1}\mu$,$Var[Y]=A'Var[X]A=A'A\sigma^2=I\cdot\sigma^2$, therefore $Y_1\sim N(\sqrt{n}\mu,\sigma^2 )\perp Y_2,\cdots,Y_n\stackrel{i.i.d.}{\sim}N(0,\sigma^2)$
Or
\begin{align*}
\sum_{i=1}^n \left(X_i-\mu\right)^2 & = \sum_{i=1}^nX_i^2-2\mu\sum_{i=1}^nX_i+n\mu^2\\
& = \sum_{i=1}^n Y_i^2-2\sqrt{n} Y_i+n\mu^2\\
& = \sum_{i=2}^nY_i^2+(Y_1-\sqrt{n}\mu)^2
\end{align*}
The second equation is due to (b) (i). Hence
\begin{align*}
f_{Y_1,\cdots,Y_n}\left(y_1,\cdots,y_n\right)& =\left. f_{X_1,\cdots,X_n}(x_1,\cdots,x_n)\left|\frac{d}{dY}X\right|\right|_{X=A'Y}\\
& = \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{(y_1-\sqrt{n}\mu)^2}{2\sigma^2}}\cdot\prod_{i=2}^n \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{y_i^2}{2\sigma^2}}
\end{align*}
\item Since $Y_1\sim N(\sqrt{n}\mu,\sigma^2 )\perp Y_2,\cdots,Y_n\stackrel{i.i.d.}{\sim}N(0,\sigma^2)$, $\bar{X}=\frac{Y_1}{\sqrt{n}}\sim N\left(\mu,\frac{\sigma^2}{\sqrt{n}}\right)\perp Y_2,\cdots,Y_n$ and $\frac{Y_2^2}{\sigma^2},\cdots,\frac{Y_n^2}{\sigma^2}\stackrel{i.i.d.}{\sim}\chi^2_1$, then $\sum_{i=2}^n \frac{Y_i^2}{\sigma^2}\sim\chi^2_{n-1}$. Therefore $\bar{X}\sim N\left(\mu,\frac{\sigma^2}{\sqrt{n}}\right)\perp V=\sum_{i=2}^n Y_i^2\sim \sigma^2\cdot\chi^2_{n-1}$
\end{enumerate}
\end{enumerate}
\end{proof}
\section*{4.} Consider a large population of individuals and let $\theta$ denote the (unknown) proportion of the population belonging to a sensitive group A (e.g. drug users).
Suppose, we randomly select $n$ individuals from the population and ask each person to select a card from a deck and answer the question written on the card. Each card in the deck has one of the two questions: $Q_1$: Do you belong to A? and $Q_2$: Do you not belong to A? Also, 85\% percent of the cards ask $Q_1$ and the remaining 15\% ask $Q_2$.
Assume that each person answers Yes or No truthfully to the selected question. For
$i = 1,\cdots,n$, let $X_i = 1$ if the $i^{th}$ person answers 'Yes' otherwise $X_i = 0$. So, the data are
the observed values of $X_1, \cdots,X_n$.
Give the joint distribution of $X_1,\cdots,X_n$ and the distribution of the total number of
Yes responses.
\begin{proof}
We first consider to calculate the probability for the $i^{th}$ person to answer 'Yes'
\begin{align*}
P(X_i=1)&= \ \ \ P(answer \ Q_1)\cdot P\left(\left.'Yes' \ as\ response\right| answer \ Q_1 \right)\\ & \ \ \ \ + P(answer \ Q_2)\cdot P\left(\left.'Yes' \ as \ response\right| answer \ Q_2 \right)\\
& = 0.85\times\theta + 0.15\times(1-\theta)\\
& = 0.15+ 0.7\theta
\end{align*}
Then we have $X_1,\cdots,X_n\stackrel{i.i.d.}{\sim}Bernoulli \left( 0.15+ 0.7\theta\right)$, therefore
\begin{align*}
P\left(X_1=x_1,\cdots,X_n=x_n\right)& = \prod_{i=1}^n \left(0.15+0.7\theta\right)^{x_i}\left(1-\left(0.15+0.7\theta\right)\right)^{1-x_i}\\& = \left(0.15+0.7\theta\right)^{\sum_{i=1}^nx_i}\left(0.85-0.7\theta\right)^{n-\sum_{i=1}^nx_i}
\end{align*}
Let $Y_n =\sum_{i=1}^nX_i$ be the total number of 'Yes' response, then $Y_n\sim Binomial\left(n, 0.15+0.7\theta\right)$
\begin{align*}
P\left(Y_n=y\right)= \left(\begin{array}{c}
n\\
y
\end{array}\right)(0.15+0.7\theta)^y(0.85-0.7\theta)^{n-y}\ \ for \ y=0,\cdots,n
\end{align*}
\end{proof}
\end{document}