# IGNOU MMT-008 Solved Assignment 2023 | M.Sc. MACS

Solved By – Narendra Kr. Sharma – M.Sc (Mathematics Honors) – Delhi University

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## IGNOU MMT-008 Assignment Question Paper 2023

mmt-008-qp-ea0e2efe-a177-495c-a706-872c870c0b28
1. State whether the following statements are True or False. Justify your answer with a short proof or a counter example:
a) If $\mathrm{P}$ $\mathrm{P}$ P \mathrm{P} $\mathrm{P}$ is a transition matrix of a Markov Chain, then all the rows of $\underset{\mathrm{n}\to \mathrm{\infty }}{lim}{\mathrm{P}}^{\mathrm{n}}$ $\underset{\mathrm{n}\to \mathrm{\infty }}{lim}{\mathrm{P}}^{\mathrm{n}}$ lim_(nrarr oo)P^(n) \lim _{\mathrm{n} \rightarrow \infty} \mathrm{P}^{\mathrm{n}} $\underset{\mathrm{n}\to \mathrm{\infty }}{lim}{\mathrm{P}}^{\mathrm{n}}$ are identical.
b) In a variance-covariance matrix all elements are always positive.
c) If ${X}_{1},{X}_{2},{X}_{3}$ ${X}_{1},{X}_{2},{X}_{3}$ X_(1),X_(2),X_(3) X_{1}, X_{2}, X_{3} ${X}_{1},{X}_{2},{X}_{3}$ are iid from ${N}_{2}\left(\mu ,\mathrm{\Sigma }\right)$ ${N}_{2}\left(\mu ,\mathrm{\Sigma }\right)$ N_(2)(mu,Sigma) N_{2}(\mu, \Sigma) ${N}_{2}\left(\mu ,\mathrm{\Sigma }\right)$ , then $\frac{{X}_{1}+{X}_{2}+{X}_{3}}{3}$ $\frac{{X}_{1}+{X}_{2}+{X}_{3}}{3}$ (X_(1)+X_(2)+X_(3))/(3) \frac{X_{1}+X_{2}+X_{3}}{3} $\frac{{X}_{1}+{X}_{2}+{X}_{3}}{3}$ follows ${N}_{2}\left(\mu ,\frac{1}{3}\mathrm{\Sigma }\right)$ ${N}_{2}\left(\mu ,\frac{1}{3}\mathrm{\Sigma }\right)$ N_(2)(mu,(1)/(3)Sigma) N_{2}\left(\mu, \frac{1}{3} \Sigma\right) ${N}_{2}\left(\mu ,\frac{1}{3}\mathrm{\Sigma }\right)$ .
d) The partial correlation coefficients and multiple correlation coefficients lie between -1 and 1.
e) For a renewal function ${M}_{t},\underset{t\to 0}{lim}\frac{{M}_{t}}{t}=\frac{1}{\mu }$ ${M}_{t},\underset{t\to 0}{lim}\frac{{M}_{t}}{t}=\frac{1}{\mu }$ M_(t),lim_(t rarr0)(M_(t))/(t)=(1)/(mu) M_{t}, \lim _{t \rightarrow 0} \frac{M_{t}}{t}=\frac{1}{\mu} ${M}_{t},\underset{t\to 0}{lim}\frac{{M}_{t}}{t}=\frac{1}{\mu }$ .
1. a) Let $\left(X,Y\right)$ $\left(X,Y\right)$ (X,Y) (X, Y) $\left(X,Y\right)$ have the joint p.d.f. given by:
$f\left(x,y\right)=\left\{\begin{array}{ll}1,& \text{if}|y| f(x,y)={[1″,”,” if “|y| < x;0 < x < 1],[0″,”,” otherwise “]:} f(x, y)= \begin{cases}1, & \text { if }|y|<x ; 0<x<1 \\ 0, & \text { otherwise }\end{cases}
i) Find the marginal p.d.f.’s of $\mathrm{X}$ $\mathrm{X}$ X \mathrm{X} $\mathrm{X}$ and $\mathrm{Y}$ $\mathrm{Y}$ Y \mathrm{Y} $\mathrm{Y}$ .
ii) Test the independence of $X$ $X$ X X $X$ and $Y$ $Y$ Y Y $Y$ .
iii) Find the conditional distribution of $X$ $X$ X X $X$ given $Y=y$ $Y=y$ Y=y Y=y $Y=y$ .
iv) Compute $\mathrm{E}\left(\mathrm{X}\mid \mathrm{Y}=\mathrm{y}\right)$ $\mathrm{E}\left(\mathrm{X}\mid \mathrm{Y}=\mathrm{y}\right)$ E(X∣Y=y) \mathrm{E}(\mathrm{X} \mid \mathrm{Y}=\mathrm{y}) $\mathrm{E}\left(\mathrm{X}\mid \mathrm{Y}=\mathrm{y}\right)$ and $\mathrm{E}\left(\mathrm{Y}\mid \mathrm{X}=\mathrm{x}\right)$ $\mathrm{E}\left(\mathrm{Y}\mid \mathrm{X}=\mathrm{x}\right)$ E(Y∣X=x) \mathrm{E}(\mathrm{Y} \mid \mathrm{X}=\mathrm{x}) $\mathrm{E}\left(\mathrm{Y}\mid \mathrm{X}=\mathrm{x}\right)$ .
b) Let the joint probability density function of two discrete random $\mathrm{X}$ $\mathrm{X}$ X \mathrm{X} $\mathrm{X}$ and $\mathrm{Y}$ $\mathrm{Y}$ Y \mathrm{Y} $\mathrm{Y}$ be given as:
 $\mathrm{X}$ $\mathrm{X}$ X \mathrm{X} $\mathrm{X}$ 2 3 4 5 $Y$ $Y$ Y Y $Y$ 0 0 0.03 0 0 1 0.34 0.30 0.16 0 2 0 0 0.03 0.14
X 2 3 4 5 Y 0 0 0.03 0 0 1 0.34 0.30 0.16 0 2 0 0 0.03 0.14 | | | $\mathrm{X}$ | | | | | :—: | :—: | :—: | :—: | :—: | :—: | | | | 2 | 3 | 4 | 5 | | $Y$ | 0 | 0 | 0.03 | 0 | 0 | | | 1 | 0.34 | 0.30 | 0.16 | 0 | | | 2 | 0 | 0 | 0.03 | 0.14 |
i) Find the marginal distribution of $\mathrm{X}$ $\mathrm{X}$ X \mathrm{X} $\mathrm{X}$ and $\mathrm{Y}$ $\mathrm{Y}$ Y \mathrm{Y} $\mathrm{Y}$ . ii) Find the conditional distribution of $\mathrm{X}$ $\mathrm{X}$ X \mathrm{X} $\mathrm{X}$ given $\mathrm{Y}=1$ $\mathrm{Y}=1$ Y=1 \mathrm{Y}=1 $\mathrm{Y}=1$ .
iii) Test the independence of variable $\mathrm{s}X$ $\mathrm{s}X$ sX \mathrm{s} X $\mathrm{s}X$ and $\mathrm{Y}$ $\mathrm{Y}$ Y \mathrm{Y} $\mathrm{Y}$ .
iv) Find $V\left[\frac{Y}{X}=x\right]$ $V\left[\frac{Y}{X}=x\right]$ V[(Y)/(X)=x] V\left[\frac{Y}{X}=x\right] $V\left[\frac{Y}{X}=x\right]$ .
1. a) Let $\mathrm{X}\sim {\mathrm{N}}_{3}\left(\mu ,\mathrm{\Sigma }\right)$ $\mathrm{X}\sim {\mathrm{N}}_{3}\left(\mu ,\mathrm{\Sigma }\right)$ X∼N_(3)(mu,Sigma) \mathrm{X} \sim \mathrm{N}_{3}(\mu, \Sigma) $\mathrm{X}\sim {\mathrm{N}}_{3}\left(\mu ,\mathrm{\Sigma }\right)$ , where $\mu =\left[5,3,4{\right]}^{\mathrm{\prime }}$ $\mu =\left[5,3,4{\right]}^{\mathrm{\prime }}$ mu=[5,3,4]^(‘) \mu=[5,3,4]^{\prime} $\mu =\left[5,3,4{\right]}^{\mathrm{\prime }}$ and
$\sum =\left(\begin{array}{ccc}2& 1& 1\\ 1& 1& 0.5\\ 1& 0.5& 1\end{array}\right)$ $\sum =\left(\begin{array}{ccc}2& 1& 1\\ 1& 1& 0.5\\ 1& 0.5& 1\end{array}\right)$ sum=([2,1,1],[1,1,0.5],[1,0.5,1]) \sum=\left(\begin{array}{ccc} 2 & 1 & 1 \\ 1 & 1 & 0.5 \\ 1 & 0.5 & 1 \end{array}\right) $\sum =\left(\begin{array}{ccc}2& 1& 1\\ 1& 1& 0.5\\ 1& 0.5& 1\end{array}\right)$
Find the distribution of:
$\left(\begin{array}{c}2{X}_{1}+{X}_{2}-{X}_{3}\\ {X}_{1}+{X}_{2}+{X}_{3}\end{array}\right)$ $\left(\begin{array}{c}2{X}_{1}+{X}_{2}-{X}_{3}\\ {X}_{1}+{X}_{2}+{X}_{3}\end{array}\right)$ ([2X_(1)+X_(2)-X_(3)],[X_(1)+X_(2)+X_(3)]) \left(\begin{array}{c} 2 X_{1}+X_{2}-X_{3} \\ X_{1}+X_{2}+X_{3} \end{array}\right) $\left(\begin{array}{c}2{X}_{1}+{X}_{2}-{X}_{3}\\ {X}_{1}+{X}_{2}+{X}_{3}\end{array}\right)$
b) Determine the principal components ${Y}_{1},{Y}_{2}$ ${Y}_{1},{Y}_{2}$ Y_(1),Y_(2) Y_{1}, Y_{2} ${Y}_{1},{Y}_{2}$ and ${Y}_{3}$ ${Y}_{3}$ Y_(3) Y_{3} ${Y}_{3}$ for the covariance matrix:
$\sum =\left(\begin{array}{ccc}1& -2& 0\\ -2& 5& 0\\ 0& 0& 1\end{array}\right)$ $\sum =\left(\begin{array}{ccc}1& -2& 0\\ -2& 5& 0\\ 0& 0& 1\end{array}\right)$ sum=([1,-2,0],[-2,5,0],[0,0,1]) \sum=\left(\begin{array}{ccc} 1 & -2 & 0 \\ -2 & 5 & 0 \\ 0 & 0 & 1 \end{array}\right) $\sum =\left(\begin{array}{ccc}1& -2& 0\\ -2& 5& 0\\ 0& 0& 1\end{array}\right)$
Also calculate the proportion of total population variance for the first principal component.
1. a) Consider a Markov chain with transition probability matrix:
$P=\left(\begin{array}{cccc}0& 0& 1& 0\\ 0& 0& 0& 1\\ 0& 1& 0& 0\\ \frac{1}{4}& \frac{1}{8}& \frac{1}{8}& \frac{1}{2}\end{array}\right)$ $P=\left(\begin{array}{cccc}0& 0& 1& 0\\ 0& 0& 0& 1\\ 0& 1& 0& 0\\ \frac{1}{4}& \frac{1}{8}& \frac{1}{8}& \frac{1}{2}\end{array}\right)$ P=([0,0,1,0],[0,0,0,1],[0,1,0,0],[(1)/(4),(1)/(8),(1)/(8),(1)/(2)]) P=\left(\begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ \frac{1}{4} & \frac{1}{8} & \frac{1}{8} & \frac{1}{2} \end{array}\right) $P=\left(\begin{array}{cccc}0& 0& 1& 0\\ 0& 0& 0& 1\\ 0& 1& 0& 0\\ \frac{1}{4}& \frac{1}{8}& \frac{1}{8}& \frac{1}{2}\end{array}\right)$
i) Whether the chain is irreducible? If irreducible classify the states of a Markov chain i.e., recurrent, transient, periodic and mean recurrence time.
ii) Find the limiting probability vector.
b) At a certain filling station, customers arrive in a Poisson process with an average time of 12 per hour. The time interval between service follows exponential distribution and as such the mean time taken to service to a unit is 2 minutes. Evaluate:
i) Probability that there is no customer at the counter.
ii) Probability that there are more than two customers at the counter.
iii) Average number of customers in a queue waiting for service.
iv) Expected waiting time of a customer in the system.
v) Probability that a customer wait for 0.11 minutes in a queue.
1. a) A service station has 5 mechanics each of whom can service a scooter in 2 hours on the average. The scooters are registered at a single counter and then sent for servicing to different mechanics. Scooters arrive at a service station at an average rate of 2 scooters per hour. Assuming that the scooter arrivals are Poisson and service times are exponentially distributed, determine:
i) Identify the model.
ii) The probability that the system shall be idle.
iii) The probability that there shall be 3 scooters in the service centre.
iv) The expected number of scooters waiting in a queue.
v) The expected number of scooters in the service centre.
vi) The average waiting time in a queue.
b) A random sample of 12 factories was conducted for the pairs of observations on sales $\left({\mathrm{x}}_{1}\right)$ $\left({\mathrm{x}}_{1}\right)$ (x_(1)) \left(\mathrm{x}_{1}\right) $\left({\mathrm{x}}_{1}\right)$ and demands $\left({\mathrm{x}}_{2}\right)$ $\left({\mathrm{x}}_{2}\right)$ (x_(2)) \left(\mathrm{x}_{2}\right) $\left({\mathrm{x}}_{2}\right)$ and the following information was obtained:
$\sum X=96,\sum Y=72,\sum {X}^{2}=780,\sum {Y}^{2}=480,\sum XY=588$ $\sum X=96,\sum Y=72,\sum {X}^{2}=780,\sum {Y}^{2}=480,\sum XY=588$ sum X=96,sum Y=72,sumX^(2)=780,sumY^(2)=480,sum XY=588 \sum X=96, \sum Y=72, \sum X^{2}=780, \sum Y^{2}=480, \sum X Y=588 $\sum X=96,\sum Y=72,\sum {X}^{2}=780,\sum {Y}^{2}=480,\sum XY=588$
The expected mean vector and variance covariance matrix for the factories in the population are:
$\begin{array}{c}\mu =\left[\begin{array}{l}9\\ 7\end{array}\right]\\ \text{and}\sum =\left[\begin{array}{ll}13& 9\\ 9& 7\end{array}\right]\text{.}\end{array}$ {:[mu=[[9],[7]]],[” and “sum=[[13,9],[9,7]]”. “]:} \begin{gathered} \mu=\left[\begin{array}{l} 9 \\ 7 \end{array}\right] \\ \text { and } \sum=\left[\begin{array}{ll} 13 & 9 \\ 9 & 7 \end{array}\right] \text {. } \end{gathered}
Test whether the sample confirms its truthness of mean vector at $5\mathrm{%}$ $5\mathrm{%}$ 5% 5 \% $5\mathrm{%}$ level of significance, if:
i) $\mathrm{\Sigma }$ $\mathrm{\Sigma }$ Sigma \Sigma $\mathrm{\Sigma }$ is known,
ii) $\mathrm{\Sigma }$ $\mathrm{\Sigma }$ Sigma \Sigma $\mathrm{\Sigma }$ is unknown.
[You may use: ${\chi }_{2,0.05}^{2}=10.60,{\chi }_{3,0.05}^{2}=12.83,{\chi }_{4,0.05}^{2}=14.89,{\text{}\mathrm{F}}_{2,10,0.05}=4.10$ ${\chi }_{2,0.05}^{2}=10.60,{\chi }_{3,0.05}^{2}=12.83,{\chi }_{4,0.05}^{2}=14.89,{\text{}\mathrm{F}}_{2,10,0.05}=4.10$ chi_(2,0.05)^(2)=10.60,chi_(3,0.05)^(2)=12.83,chi_(4,0.05)^(2)=14.89,F_(2,10,0.05)=4.10 \chi_{2,0.05}^{2}=10.60, \chi_{3,0.05}^{2}=12.83, \chi_{4,0.05}^{2}=14.89, \mathrm{~F}_{2,10,0.05}=4.10 ]
1. a) Let the random vector ${X}^{\mathrm{\prime }}=\left({X}_{1},{X}_{2},{X}_{3}\right)$ ${X}^{\mathrm{\prime }}=\left({X}_{1},{X}_{2},{X}_{3}\right)$ X^(‘)=(X_(1),X_(2),X_(3)) X^{\prime}=\left(X_{1}, X_{2}, X_{3}\right) ${X}^{\mathrm{\prime }}=\left({X}_{1},{X}_{2},{X}_{3}\right)$ has mean vector $\left[-2,3,4\right]$ $\left[-2,3,4\right]$ [-2,3,4] [-2,3,4] $\left[-2,3,4\right]$ and variance covariance matrix $=\left(\begin{array}{lll}1& 1& 1\\ 1& 2& 3\\ 1& 3& 9\end{array}\right)$ $=\left(\begin{array}{lll}1& 1& 1\\ 1& 2& 3\\ 1& 3& 9\end{array}\right)$ =([1,1,1],[1,2,3],[1,3,9]) =\left(\begin{array}{lll}1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 3 & 9\end{array}\right) $=\left(\begin{array}{lll}1& 1& 1\\ 1& 2& 3\\ 1& 3& 9\end{array}\right)$ . Fit the equation $Y={b}_{0}+{b}_{1}X+{b}_{2}{X}_{2}$ $Y={b}_{0}+{b}_{1}X+{b}_{2}{X}_{2}$ Y=b_(0)+b_(1)X+b_(2)X_(2) Y=b_{0}+b_{1} X+b_{2} X_{2} $Y={b}_{0}+{b}_{1}X+{b}_{2}{X}_{2}$ . Also obtain the multiple correlation coefficient between ${\mathrm{X}}_{3}$ ${\mathrm{X}}_{3}$ X_(3) \mathrm{X}_{3} ${\mathrm{X}}_{3}$ and $\left[{\mathrm{X}}_{1},{\mathrm{X}}_{2}\right]$ $\left[{\mathrm{X}}_{1},{\mathrm{X}}_{2}\right]$ [X_(1),X_(2)] \left[\mathrm{X}_{1}, \mathrm{X}_{2}\right] $\left[{\mathrm{X}}_{1},{\mathrm{X}}_{2}\right]$ .
b) Define ultimate extinction in a branching process. Let ${\mathrm{p}}_{\mathrm{k}}={\mathrm{b}\mathrm{c}}^{\mathrm{k}-1},\mathrm{k}=1,2,\dots$ ${\mathrm{p}}_{\mathrm{k}}={\mathrm{b}\mathrm{c}}^{\mathrm{k}-1},\mathrm{k}=1,2,\dots$ p_(k)=bc^(k-1),k=1,2,dots \mathrm{p}_{\mathrm{k}}=\mathrm{bc}^{\mathrm{k}-1}, \mathrm{k}=1,2, \ldots ${\mathrm{p}}_{\mathrm{k}}={\mathrm{b}\mathrm{c}}^{\mathrm{k}-1},\mathrm{k}=1,2,\dots$ ;
$0<\mathrm{b}<\mathrm{c}<\mathrm{b}+\mathrm{c}<1$ $0<\mathrm{b}<\mathrm{c}<\mathrm{b}+\mathrm{c}<1$ 0 < b < c < b+c < 1 0<\mathrm{b}<\mathrm{c}<\mathrm{b}+\mathrm{c}<1 $0<\mathrm{b}<\mathrm{c}<\mathrm{b}+\mathrm{c}<1$ and ${\mathrm{p}}_{0}=1-\sum _{\mathrm{k}=1}^{\mathrm{\infty }}{\mathrm{p}}_{\mathrm{k}}$ ${\mathrm{p}}_{0}=1-\sum _{\mathrm{k}=1}^{\mathrm{\infty }}{\mathrm{p}}_{\mathrm{k}}$ p_(0)=1-sum_(k=1)^(oo)p_(k) \mathrm{p}_{0}=1-\sum_{\mathrm{k}=1}^{\infty} \mathrm{p}_{\mathrm{k}} ${\mathrm{p}}_{0}=1-\sum _{\mathrm{k}=1}^{\mathrm{\infty }}{\mathrm{p}}_{\mathrm{k}}$ . Then discuss the probability of extinction in different cases for $\mathrm{E}\left({\mathrm{X}}_{1}\right)\ge 1$ $\mathrm{E}\left({\mathrm{X}}_{1}\right)\ge 1$ E(X_(1)) >= 1 \mathrm{E}\left(\mathrm{X}_{1}\right) \geq 1 $\mathrm{E}\left({\mathrm{X}}_{1}\right)\ge 1$ or $\mathrm{E}\left({\mathrm{X}}_{1}\right)<1$ $\mathrm{E}\left({\mathrm{X}}_{1}\right)<1$ E(X_(1)) < 1 \mathrm{E}\left(\mathrm{X}_{1}\right)<1 $\mathrm{E}\left({\mathrm{X}}_{1}\right)<1$ .
1. a) If the random vector $\mathrm{Z}$ $\mathrm{Z}$ Z \mathrm{Z} $\mathrm{Z}$ be ${\mathrm{N}}_{4}\left(\mu ,\mathrm{\Sigma }\right)$ ${\mathrm{N}}_{4}\left(\mu ,\mathrm{\Sigma }\right)$ N_(4)(mu,Sigma) \mathrm{N}_{4}(\mu, \Sigma) ${\mathrm{N}}_{4}\left(\mu ,\mathrm{\Sigma }\right)$ , where:
$\mu =\left[\begin{array}{c}1\\ 2\\ 5\\ -2\end{array}\right]$ $\mu =\left[\begin{array}{c}1\\ 2\\ 5\\ -2\end{array}\right]$ mu=[[1],[2],[5],[-2]] \mu=\left[\begin{array}{c} 1 \\ 2 \\ 5 \\ -2 \end{array}\right] $\mu =\left[\begin{array}{c}1\\ 2\\ 5\\ -2\end{array}\right]$
and $\sum =\left[\begin{array}{cccc}3& 3& 0& 9\\ 3& 2& -1& 1\\ 0& -1& 6& -3\\ 9& 1& -3& 7\end{array}\right]$ $\sum =\left[\begin{array}{cccc}3& 3& 0& 9\\ 3& 2& -1& 1\\ 0& -1& 6& -3\\ 9& 1& -3& 7\end{array}\right]$ sum=[[3,3,0,9],[3,2,-1,1],[0,-1,6,-3],[9,1,-3,7]] \sum=\left[\begin{array}{cccc}3 & 3 & 0 & 9 \\ 3 & 2 & -1 & 1 \\ 0 & -1 & 6 & -3 \\ 9 & 1 & -3 & 7\end{array}\right] $\sum =\left[\begin{array}{cccc}3& 3& 0& 9\\ 3& 2& -1& 1\\ 0& -1& 6& -3\\ 9& 1& -3& 7\end{array}\right]$ .
Find ${r}_{34},{r}_{34.21}$ ${r}_{34},{r}_{34.21}$ r_(34),r_(34.21) r_{34}, r_{34.21} ${r}_{34},{r}_{34.21}$ .
b) Suppose life times ${X}_{1},{X}_{2},\dots .$ ${X}_{1},{X}_{2},\dots .$ X_(1),X_(2),dots. X_{1}, X_{2}, \ldots . ${X}_{1},{X}_{2},\dots .$ . are i.i.d. uniformly distributed on $\left(0,3\right)$ $\left(0,3\right)$ (0,3) (0,3) $\left(0,3\right)$ and ${C}_{1}=2$ ${C}_{1}=2$ C_(1)=2 C_{1}=2 ${C}_{1}=2$ and ${\mathrm{C}}_{2}=8$ ${\mathrm{C}}_{2}=8$ C_(2)=8 \mathrm{C}_{2}=8 ${\mathrm{C}}_{2}=8$ . Find:
i) ${\mu }^{\mathrm{T}}$ ${\mu }^{\mathrm{T}}$ mu^(T) \mu^{\mathrm{T}} ${\mu }^{\mathrm{T}}$
ii) $\mathrm{T}$ $\mathrm{T}$ T \mathrm{T} $\mathrm{T}$ which minimizes $\mathrm{C}\left(\mathrm{T}\right)$ $\mathrm{C}\left(\mathrm{T}\right)$ C(T) \mathrm{C}(\mathrm{T}) $\mathrm{C}\left(\mathrm{T}\right)$ and which is the better policy in the long-run in terms of cost.
1. a) Consider the Markov chain with three states, $S=\left\{1,2,3\right\}$ $S=\left\{1,2,3\right\}$ S={1,2,3} S=\{1,2,3\} $S=\left\{1,2,3\right\}$ following the transition matrix
$\mathrm{p}=2\left[\begin{array}{ccc}1& 2& 3\\ \frac{1}{2}& \frac{1}{4}& \frac{1}{4}\\ \frac{1}{3}& 0& \frac{2}{3}\\ \frac{1}{2}& \frac{1}{2}& 0\end{array}\right]$ $\mathrm{p}=2\left[\begin{array}{ccc}1& 2& 3\\ \frac{1}{2}& \frac{1}{4}& \frac{1}{4}\\ \frac{1}{3}& 0& \frac{2}{3}\\ \frac{1}{2}& \frac{1}{2}& 0\end{array}\right]$ p=2[[1,2,3],[(1)/(2),(1)/(4),(1)/(4)],[(1)/(3),0,(2)/(3)],[(1)/(2),(1)/(2),0]] \mathrm{p}=2\left[\begin{array}{ccc} 1 & 2 & 3 \\ \frac{1}{2} & \frac{1}{4} & \frac{1}{4} \\ \frac{1}{3} & 0 & \frac{2}{3} \\ \frac{1}{2} & \frac{1}{2} & 0 \end{array}\right] $\mathrm{p}=2\left[\begin{array}{ccc}1& 2& 3\\ \frac{1}{2}& \frac{1}{4}& \frac{1}{4}\\ \frac{1}{3}& 0& \frac{2}{3}\\ \frac{1}{2}& \frac{1}{2}& 0\end{array}\right]$
i) Draw the state transition diagram for this chain.
ii) If $\mathrm{P}\left({\mathrm{X}}_{1}=1\right)=\mathrm{P}\left({\mathrm{X}}_{1}=2\right)=\frac{1}{4}$ $\mathrm{P}\left({\mathrm{X}}_{1}=1\right)=\mathrm{P}\left({\mathrm{X}}_{1}=2\right)=\frac{1}{4}$ P(X_(1)=1)=P(X_(1)=2)=(1)/(4) \mathrm{P}\left(\mathrm{X}_{1}=1\right)=\mathrm{P}\left(\mathrm{X}_{1}=2\right)=\frac{1}{4} $\mathrm{P}\left({\mathrm{X}}_{1}=1\right)=\mathrm{P}\left({\mathrm{X}}_{1}=2\right)=\frac{1}{4}$ , then find $\mathrm{P}\left({\mathrm{X}}_{1}=3,{\mathrm{X}}_{2}=2,{\mathrm{X}}_{3}=1\right)$ $\mathrm{P}\left({\mathrm{X}}_{1}=3,{\mathrm{X}}_{2}=2,{\mathrm{X}}_{3}=1\right)$ P(X_(1)=3,X_(2)=2,X_(3)=1) \mathrm{P}\left(\mathrm{X}_{1}=3, \mathrm{X}_{2}=2, \mathrm{X}_{3}=1\right) $\mathrm{P}\left({\mathrm{X}}_{1}=3,{\mathrm{X}}_{2}=2,{\mathrm{X}}_{3}=1\right)$ .
iii) Check whether the chain is irreducible and a periodic.
iv) Find the stationary distribution for the chain.
b) If ${\mathrm{N}}_{1}\left(\mathrm{t}\right),{\mathrm{N}}_{2}\left(\mathrm{t}\right)$ ${\mathrm{N}}_{1}\left(\mathrm{t}\right),{\mathrm{N}}_{2}\left(\mathrm{t}\right)$ N_(1)(t),N_(2)(t) \mathrm{N}_{1}(\mathrm{t}), \mathrm{N}_{2}(\mathrm{t}) ${\mathrm{N}}_{1}\left(\mathrm{t}\right),{\mathrm{N}}_{2}\left(\mathrm{t}\right)$ are two independent Poisson process with parameters ${\lambda }_{1}$ ${\lambda }_{1}$ lambda_(1) \lambda_{1} ${\lambda }_{1}$ and ${\lambda }_{2}$ ${\lambda }_{2}$ lambda_(2) \lambda_{2} ${\lambda }_{2}$ respectively, then show that