Details For MMT-008 Solved Assignment

IGNOU MMT-008 Assignment Question Paper 2023

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
$N_2(\mu ,\mathrm{\Sigma })$

$N_2(\mu ,\mathrm{\Sigma })$

N_(2)(mu,Sigma)
N_{2}(mu, Sigma)



$N_2(\mu ,\mathrm{\Sigma })$


, 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
$N_2(\mu ,\frac{1}{3}\mathrm{\Sigma })$

$N_2\left(\mu ,\frac{1}{3}\mathrm{\Sigma }\right)$

N_(2)(mu,(1)/(3)Sigma)
N_{2}left(mu, frac{1}{3} Sigmaright)



$N_2(\mu ,\frac{1}{3}\mathrm{\Sigma })$


.
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}



$M_t,\underset{t\to 0}{lim}\frac{M_t}{t}=\frac{1}{\mu }$


.

2. a) Let
   $(X,Y)$
   
   $(X,Y)$
   
   (X,Y)
   (X, Y)
   
   
   
   $(X,Y)$
   
   
   have the joint p.d.f. given by:

$f(x,y)=\{\begin{array}{ll}1,& \text{if}|y|<x;0<x<1\\ 0,& \text{otherwise}\end{array}$

$f(x,y)=\left\{\begin{array}{l}1,\text{ if }|y|<x;0<x<1\\ 0,\text{ otherwise }\end{array}\right.$

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}



$f(x,y)=\{\begin{array}{ll}1,& \text{ if }|y|<x;0<x<1\\ 0,& \text{ otherwise }\end{array}$


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
$\mathrm{E}(\mathrm{X}\mid \mathrm{Y}=\mathrm{y})$

$\mathrm{E}(\mathrm{X}\mid \mathrm{Y}=\mathrm{y})$

E(X∣Y=y)
mathrm{E}(mathrm{X} mid mathrm{Y}=mathrm{y})



$\mathrm{E}(\mathrm{X}\mid \mathrm{Y}=\mathrm{y})$


and
$\mathrm{E}(\mathrm{Y}\mid \mathrm{X}=\mathrm{x})$

$\mathrm{E}(\mathrm{Y}\mid \mathrm{X}=\mathrm{x})$

E(Y∣X=x)
mathrm{E}(mathrm{Y} mid mathrm{X}=mathrm{x})



$\mathrm{E}(\mathrm{Y}\mid \mathrm{X}=\mathrm{x})$


.
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
$V[\frac{Y}{X}=x]$

$V\left[\frac{Y}{X}=x\right]$

V[(Y)/(X)=x]
Vleft[frac{Y}{X}=xright]



$V[\frac{Y}{X}=x]$


.

3. a) Let
   $\mathrm{X}\sim {\mathrm{N}}_3(\mu ,\mathrm{\Sigma })$
   
   $\mathrm{X}\sim {\mathrm{N}}_3(\mu ,\mathrm{\Sigma })$
   
   X∼N_(3)(mu,Sigma)
   mathrm{X} sim mathrm{N}_{3}(mu, Sigma)
   
   
   
   $\mathrm{X}\sim {\mathrm{N}}_3(\mu ,\mathrm{\Sigma })$
   
   
   , where
   $\mu =[5,3,4{]}^{\mathrm{\prime }}$
   
   $\mu =[5,3,4{]}^{\mathrm{\prime }}$
   
   mu=[5,3,4]^(‘)
   mu=[5,3,4]^{prime}
   
   
   
   $\mu =[5,3,4{]}^{\mathrm{\prime }}$
   
   
   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:
$$\left(\begin{array}{c}2X_1+X_2-X_3\\ X_1+X_2+X_3\end{array}\right)$$

$$\left(\begin{array}{c}2X_1+X_2-X_3\\ X_1+X_2+X_3\end{array}\right)$$

([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)



$$\left(\begin{array}{c}2X_1+X_2-X_3\\ X_1+X_2+X_3\end{array}\right)$$


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.

4. 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.

5. 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:
$$\begin{array}{c}\mu =\left[\begin{array}{l}9\\ 7\end{array}\right]\\ \text{and}\sum =\left[\begin{array}{ll}13& 9\\ 9& 7\end{array}\right]\text{.}\end{array}$$

$$\begin{array}{c}\mu =\left[\begin{array}{l}9\\ 7\end{array}\right]\\ \text{ and }\sum =\left[\begin{array}{ll}13& 9\\ 9& 7\end{array}\right]\text{. }\end{array}$$

{:[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}



$$\begin{array}{c}\mu =\left[\begin{array}{l}9\\ 7\end{array}\right]\\ \text{ and }\sum =\left[\begin{array}{ll}13& 9\\ 9& 7\end{array}\right]\text{. }\end{array}$$


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:
${\chi }_{2,0.05}^2=10.60,{\chi }_{3,0.05}^2=12.83,{\chi }_{4,0.05}^2=14.89,{\mathrm{F}}_{2,10,0.05}=4.10$

${\chi }_{2,0.05}^2=10.60,{\chi }_{3,0.05}^2=12.83,{\chi }_{4,0.05}^2=14.89,{\mathrm{F}}_{2,10,0.05}=4.10$

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



${\chi }_{2,0.05}^2=10.60,{\chi }_{3,0.05}^2=12.83,{\chi }_{4,0.05}^2=14.89,{\mathrm{F}}_{2,10,0.05}=4.10$


]

6. a) Let the random vector
   $X^{\mathrm{\prime }}=(X_1,X_2,X_3)$
   
   $X^{\mathrm{\prime }}=\left(X_1,X_2,X_3\right)$
   
   X^(‘)=(X_(1),X_(2),X_(3))
   X^{prime}=left(X_{1}, X_{2}, X_{3}right)
   
   
   
   $X^{\mathrm{\prime }}=(X_1,X_2,X_3)$
   
   
   has mean vector
   $[-2,3,4]$
   
   $[-2,3,4]$
   
   [-2,3,4]
   [-2,3,4]
   
   
   
   $[-2,3,4]$
   
   
   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 & 9end{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
   $[{\mathrm{X}}_1,{\mathrm{X}}_2]$
   
   $\left[{\mathrm{X}}_1,{\mathrm{X}}_2\right]$
   
   [X_(1),X_(2)]
   left[mathrm{X}_{1}, mathrm{X}_{2}right]
   
   
   
   $[{\mathrm{X}}_1,{\mathrm{X}}_2]$
   
   
   .

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



$\mathrm{E}\left({\mathrm{X}}_1\right)<1$


.

7. a) If the random vector
   $\mathrm{Z}$
   
   $\mathrm{Z}$
   
   Z
   mathrm{Z}
   
   
   
   $\mathrm{Z}$
   
   
   be
   ${\mathrm{N}}_4(\mu ,\mathrm{\Sigma })$
   
   ${\mathrm{N}}_4(\mu ,\mathrm{\Sigma })$
   
   N_(4)(mu,Sigma)
   mathrm{N}_{4}(mu, Sigma)
   
   
   
   ${\mathrm{N}}_4(\mu ,\mathrm{\Sigma })$
   
   
   , 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 & 7end{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
$(0,3)$

$(0,3)$

(0,3)
(0,3)



$(0,3)$


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
$\mathrm{C}(\mathrm{T})$

$\mathrm{C}(\mathrm{T})$

C(T)
mathrm{C}(mathrm{T})



$\mathrm{C}(\mathrm{T})$


and which is the better policy in the long-run in terms of cost.

8. a) Consider the Markov chain with three states,
   $S=\{1,2,3\}$
   
   $S=\{1,2,3\}$
   
   S={1,2,3}
   S={1,2,3}
   
   
   
   $S=\{1,2,3\}$
   
   
   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}=2left[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
$\mathrm{P}({\mathrm{X}}_1=1)=\mathrm{P}({\mathrm{X}}_1=2)=\frac{1}{4}$

$\mathrm{P}\left({\mathrm{X}}_1=1\right)=\mathrm{P}\left({\mathrm{X}}_1=2\right)=\frac{1}{4}$

P(X_(1)=1)=P(X_(1)=2)=(1)/(4)
mathrm{P}left(mathrm{X}_{1}=1right)=mathrm{P}left(mathrm{X}_{1}=2right)=frac{1}{4}



$\mathrm{P}({\mathrm{X}}_1=1)=\mathrm{P}({\mathrm{X}}_1=2)=\frac{1}{4}$


, then find
$\mathrm{P}({\mathrm{X}}_1=3,{\mathrm{X}}_2=2,{\mathrm{X}}_3=1)$

$\mathrm{P}\left({\mathrm{X}}_1=3,{\mathrm{X}}_2=2,{\mathrm{X}}_3=1\right)$

P(X_(1)=3,X_(2)=2,X_(3)=1)
mathrm{P}left(mathrm{X}_{1}=3, mathrm{X}_{2}=2, mathrm{X}_{3}=1right)



$\mathrm{P}({\mathrm{X}}_1=3,{\mathrm{X}}_2=2,{\mathrm{X}}_3=1)$


.
iii) Check whether the chain is irreducible and a periodic.
iv) Find the stationary distribution for the chain.
b) If
${\mathrm{N}}_1(\mathrm{t}),{\mathrm{N}}_2(\mathrm{t})$

${\mathrm{N}}_1(\mathrm{t}),{\mathrm{N}}_2(\mathrm{t})$

N_(1)(t),N_(2)(t)
mathrm{N}_{1}(mathrm{t}), mathrm{N}_{2}(mathrm{t})



${\mathrm{N}}_1(\mathrm{t}),{\mathrm{N}}_2(\mathrm{t})$


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
$$\mathrm{P}({\mathrm{N}}_1(\mathrm{t})=\mathrm{k}[{\mathrm{N}}_1(\mathrm{t})+{\mathrm{N}}_2(\mathrm{t})=\mathrm{n}]={}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{k}}{\mathrm{p}}^{\mathrm{k}}{\mathrm{q}}^{\mathrm{n}-\mathrm{k}}\text{, where}\mathrm{p}=\frac{{\lambda }_1}{{\lambda }_1+{\lambda }_2},\mathrm{q}=\frac{{\lambda }_2}{{\lambda }_1+{\lambda }_2}.$$

$$\mathrm{P}\left({\mathrm{N}}_1(\mathrm{t})=\mathrm{k}\left[{\mathrm{N}}_1(\mathrm{t})+{\mathrm{N}}_2(\mathrm{t})=\mathrm{n}\right]={}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{k}}{\mathrm{p}}^{\mathrm{k}}{\mathrm{q}}^{\mathrm{n}-\mathrm{k}}\text{, where }\mathrm{p}=\frac{{\lambda }_1}{{\lambda }_1+{\lambda }_2},\mathrm{q}=\frac{{\lambda }_2}{{\lambda }_1+{\lambda }_2}.\right.$$

P(N_(1)(t)=k[N_(1)(t)+N_(2)(t)=n]=^(n)C_(k)p^(k)q^(n-k)”, where “p=(lambda_(1))/(lambda_(1)+lambda_(2)),q=(lambda_(2))/(lambda_(1)+lambda_(2)).:}
mathrm{P}left(mathrm{N}_{1}(mathrm{t})=mathrm{k}left[mathrm{N}_{1}(mathrm{t})+mathrm{N}_{2}(mathrm{t})=mathrm{n}right]={ }^{mathrm{n}} mathrm{C}_{mathrm{k}} mathrm{p}^{mathrm{k}} mathrm{q}^{mathrm{n}-mathrm{k}} text {, where } mathrm{p}=frac{lambda_{1}}{lambda_{1}+lambda_{2}}, mathrm{q}=frac{lambda_{2}}{lambda_{1}+lambda_{2}} .right.



$$\mathrm{P}({\mathrm{N}}_1(\mathrm{t})=\mathrm{k}[{\mathrm{N}}_1(\mathrm{t})+{\mathrm{N}}_2(\mathrm{t})=\mathrm{n}]={}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{k}}{\mathrm{p}}^{\mathrm{k}}{\mathrm{q}}^{\mathrm{n}-\mathrm{k}}\text{, where }\mathrm{p}=\frac{{\lambda }_1}{{\lambda }_1+{\lambda }_2},\mathrm{q}=\frac{{\lambda }_2}{{\lambda }_1+{\lambda }_2}.$$

9. a) Let
   $X=\left[\begin{array}{l}{X}_1\\ X_2\end{array}\right]$
   
   $X=\left[\begin{array}{l}{X}_1\\ X_2\end{array}\right]$
   
   X=[[X_(1)],[X_(2)]]
   X=left[begin{array}{l}X_{1} \ X_{2}end{array}right]
   
   
   
   $X=\left[\begin{array}{l}{X}_1\\ X_2\end{array}\right]$
   
   
   be a normal random vector with the mean vector
   $\mu =\left[\begin{array}{l}0\\ 1\end{array}\right]$
   
   $\mu =\left[\begin{array}{l}0\\ 1\end{array}\right]$
   
   mu=[[0],[1]]
   mu=left[begin{array}{l}0 \ 1end{array}right]
   
   
   
   $\mu =\left[\begin{array}{l}0\\ 1\end{array}\right]$
   
   
   and covariance matrix
   $\left[\begin{array}{cc}1& -1\\ -1& 2\end{array}\right]$
   
   $\left[\begin{array}{cc}1& -1\\ -1& 2\end{array}\right]$
   
   [[1,-1],[-1,2]]
   left[begin{array}{cc}1 & -1 \ -1 & 2end{array}right]
   
   
   
   $\left[\begin{array}{cc}1& -1\\ -1& 2\end{array}\right]$
   
   
   . Suppose
   $\mathrm{Y}=\mathrm{A}\mathrm{X}+\mathrm{b}$
   
   $\mathrm{Y}=\mathrm{A}\mathrm{X}+\mathrm{b}$
   
   Y=AX+b
   mathrm{Y}=mathrm{AX}+mathrm{b}
   
   
   
   $\mathrm{Y}=\mathrm{A}\mathrm{X}+\mathrm{b}$
   
   
   , where

$$A=\left[\begin{array}{ll}1& 2\\ 2& 1\\ 1& 1\end{array}\right],\mathrm{b}=\left[\begin{array}{l}0\\ 1\\ 2\end{array}\right]\text{and}\mathrm{Y}\sim {\mathrm{N}}_3$$

$$A=\left[\begin{array}{ll}1& 2\\ 2& 1\\ 1& 1\end{array}\right],\mathrm{b}=\left[\begin{array}{l}0\\ 1\\ 2\end{array}\right]\text{ and }\mathrm{Y}\sim {\mathrm{N}}_3$$

A=[[1,2],[2,1],[1,1]],b=[[0],[1],[2]]” and “Y∼N_(3)
A=left[begin{array}{ll}
1 & 2 \
2 & 1 \
1 & 1
end{array}right], mathrm{b}=left[begin{array}{l}
0 \
1 \
2
end{array}right] text { and } mathrm{Y} sim mathrm{N}_{3}













































































$$A=\left[\begin{array}{ll}1& 2\\ 2& 1\\ 1& 1\end{array}\right],\mathrm{b}=\left[\begin{array}{l}0\\ 1\\ 2\end{array}\right]\text{ and }\mathrm{Y}\sim {\mathrm{N}}_3$$


i) Find
$\mathrm{P}(0\le {\mathrm{X}}_2\le 1)$

$\mathrm{P}\left(0\le {\mathrm{X}}_2\le 1\right)$

P(0 <= X_(2) <= 1)
mathrm{P}left(0 leq mathrm{X}_{2} leq 1right)



$\mathrm{P}(0\le {\mathrm{X}}_2\le 1)$


.

ii) Compute
$\mathrm{E}(\mathrm{Y})$

$\mathrm{E}(\mathrm{Y})$

E(Y)
mathrm{E}(mathrm{Y})



$\mathrm{E}(\mathrm{Y})$


.

iii) Find the covariance matrix of
$\mathrm{Y}$

$\mathrm{Y}$

Y
mathrm{Y}



$\mathrm{Y}$


.

iv) Find
$\mathrm{P}({\mathrm{Y}}_3\le 4)$

$\mathrm{P}\left({\mathrm{Y}}_3\le 4\right)$

P(Y_(3) <= 4)
mathrm{P}left(mathrm{Y}_{3} leq 4right)



$\mathrm{P}({\mathrm{Y}}_3\le 4)$


.
b) A box contains two coins: a regular coin and one fake two-headed coin. One coin is chosen at random and tossed twice. The following events are defined:

A: first coin toss results in a head.

B: second coin toss results in a head.

C: coin 1 (regular) has been selected.
Find
$\mathrm{P}(\mathrm{A}\mid \mathrm{C}),\mathrm{P}(\mathrm{B}\mid \mathrm{C}),\mathrm{P}(\mathrm{A}\cap \mathrm{B})\mid \mathrm{C}),\mathrm{P}(\mathrm{A}),\mathrm{P}(\mathrm{B})$

$\mathrm{P}(\mathrm{A}\mid \mathrm{C}),\mathrm{P}(\mathrm{B}\mid \mathrm{C}),\mathrm{P}(\mathrm{A}\cap \mathrm{B})\mid \mathrm{C}),\mathrm{P}(\mathrm{A}),\mathrm{P}(\mathrm{B})$

P(A∣C),P(B∣C),P(AnnB)∣C),P(A),P(B)
mathrm{P}(mathrm{A} mid mathrm{C}), mathrm{P}(mathrm{B} mid mathrm{C}), mathrm{P}(mathrm{A} cap mathrm{B}) mid mathrm{C}), mathrm{P}(mathrm{A}), mathrm{P}(mathrm{B})



$\mathrm{P}(\mathrm{A}\mid \mathrm{C}),\mathrm{P}(\mathrm{B}\mid \mathrm{C}),\mathrm{P}(\mathrm{A}\cap \mathrm{B})\mid \mathrm{C}),\mathrm{P}(\mathrm{A}),\mathrm{P}(\mathrm{B})$


and
$\mathrm{P}(\mathrm{A}\cap \mathrm{B})$

$\mathrm{P}(\mathrm{A}\cap \mathrm{B})$

P(AnnB)
mathrm{P}(mathrm{A} cap mathrm{B})



$\mathrm{P}(\mathrm{A}\cap \mathrm{B})$


.

10. a) Consider three random variables
    $X_1,X_2,X_3$
    
    $X_1,X_2,X_3$
    
    X_(1),X_(2),X_(3)
    X_{1}, X_{2}, X_{3}
    
    
    
    $X_1,X_2,X_3$
    
    
    having the covariance matrix

$$\left[\begin{array}{ccc}1& 0.12& 0.08\\ 0.12& 1& 0.06\\ 0.08& 0.06& 1\end{array}\right].$$

$$\left[\begin{array}{ccc}1& 0.12& 0.08\\ 0.12& 1& 0.06\\ 0.08& 0.06& 1\end{array}\right].$$

[[1,0.12,0.08],[0.12,1,0.06],[0.08,0.06,1]].
left[begin{array}{ccc}
1 & 0.12 & 0.08 \
0.12 & 1 & 0.06 \
0.08 & 0.06 & 1
end{array}right] .














































$$\left[\begin{array}{ccc}1& 0.12& 0.08\\ 0.12& 1& 0.06\\ 0.08& 0.06& 1\end{array}\right].$$


Write the factor model, if number of variables and number of factors are 3 and 1 respectively.
b) A particular component in a machine is replaced instantaneously on failure. The successive component lifetimes are uniformly distributed over the interval [2,5] years. Further, planned replacements take place every 3 years.
Compute
i) long-terms rate of replacements.
ii) long-terms rate of failures.


[stray-random categories=trigonometry]

MMT-008 Sample Solution 2023

Question:-01

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.

Answer:
The statement “If
$P$

$P$

P
P



$P$


is a transition matrix of a Markov Chain, then all the rows of
$\underset{n\to \mathrm{\infty }}{lim}{P}^n$

$\underset{n\to \mathrm{\infty }}{lim}{P}^n$

lim_(n rarr oo)P^(n)
lim_{{n to infty}} P^n



$\underset{n\to \mathrm{\infty }}{lim}{P}^n$


are identical” is generally
not true for all Markov Chains
. It is true under certain conditions, such as if the Markov Chain is irreducible, aperiodic, and positive recurrent (i.e., it is ergodic).

Counterexample:

Consider a simple Markov Chain with two states
$A$

$A$

A
A



$A$


and
$B$

$B$

B
B



$B$


and the following transition matrix:
$$P=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$$

$$P=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$$

P=([1,0],[0,1])
P = begin{pmatrix}
1 & 0 \
0 & 1
end{pmatrix}



$$P=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$$


In this case, the Markov Chain is not irreducible (it consists of two disconnected states). The limit
$\underset{n\to \mathrm{\infty }}{lim}{P}^n$

$\underset{n\to \mathrm{\infty }}{lim}{P}^n$

lim_(n rarr oo)P^(n)
lim_{{n to infty}} P^n



$\underset{n\to \mathrm{\infty }}{lim}{P}^n$


exists and is:
$$\underset{n\to \mathrm{\infty }}{lim}{P}^n=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$$

$$\underset{n\to \mathrm{\infty }}{lim}{P}^n=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$$

lim_(n rarr oo)P^(n)=([1,0],[0,1])
lim_{{n to infty}} P^n = begin{pmatrix}
1 & 0 \
0 & 1
end{pmatrix}



$$\underset{n\to \mathrm{\infty }}{lim}{P}^n=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$$


Here, the rows are not identical, contradicting the statement.

Conditions for the Statement to be True:

For an ergodic Markov Chain, the statement is true. In such a case, the Markov Chain has a unique stationary distribution
$\pi$

$\pi$

pi
pi



$\pi$


, and:
$$\underset{n\to \mathrm{\infty }}{lim}{P}^n=\left(\begin{array}{c}\pi \\ \pi \\ ⋮\\ \pi \end{array}\right)$$

$$\underset{n\to \mathrm{\infty }}{lim}{P}^n=\left(\begin{array}{c}\pi \\ \pi \\ ⋮\\ \pi \end{array}\right)$$

lim_(n rarr oo)P^(n)=([pi],[pi],[vdots],[pi])
lim_{{n to infty}} P^n = begin{pmatrix}
pi \
pi \
vdots \
pi
end{pmatrix}


































$$\underset{n\to \mathrm{\infty }}{lim}{P}^n=\left(\begin{array}{c}\pi \\ \pi \\ ⋮\\ \pi \end{array}\right)$$


Here, all rows are identical and equal to the stationary distribution
$\pi$

$\pi$

pi
pi



$\pi$


.
So, the statement is not universally true for all Markov Chains but holds under specific conditions.

---

Page Break

---

b) In a variance-covariance matrix all elements are always positive.
Answer:
The statement “In a variance-covariance matrix all elements are always positive” is
false
.

Counterexample:

Consider a simple dataset with two variables
$X$

$X$

X
X



$X$


and
$Y$

$Y$

Y
Y



$Y$


, where
$X=[1,2,3]$

$X=[1,2,3]$

X=[1,2,3]
X = [1, 2, 3]



$X=[1,2,3]$


and
$Y=[3,2,1]$

$Y=[3,2,1]$

Y=[3,2,1]
Y = [3, 2, 1]



$Y=[3,2,1]$


.
The variance-covariance matrix for this dataset would be:
$$\left(\begin{array}{cc}\text{Var}(X)& \text{Cov}(X,Y)\\ \text{Cov}(Y,X)& \text{Var}(Y)\end{array}\right)=\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)$$

$$\left(\begin{array}{cc}\text{Var}(X)& \text{Cov}(X,Y)\\ \text{Cov}(Y,X)& \text{Var}(Y)\end{array}\right)=\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)$$

([“Var”(X),”Cov”(X”,”Y)],[“Cov”(Y”,”X),”Var”(Y)])=([1,-1],[-1,1])
begin{pmatrix}
text{Var}(X) & text{Cov}(X, Y) \
text{Cov}(Y, X) & text{Var}(Y)
end{pmatrix}
= begin{pmatrix}
1 & -1 \
-1 & 1
end{pmatrix}



$$\left(\begin{array}{cc}\text{Var}(X)& \text{Cov}(X,Y)\\ \text{Cov}(Y,X)& \text{Var}(Y)\end{array}\right)=\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)$$


Here, the covariance between
$X$

$X$

X
X



$X$


and
$Y$

$Y$

Y
Y



$Y$


is
$-1$

$-1$

-1
-1



$-1$


, which is not positive. Therefore, the statement is false.

Additional Notes:

1. The diagonal elements of a variance-covariance matrix, which represent variances, are always non-negative because variance cannot be negative.
2. Off-diagonal elements, which represent covariances, can be negative, zero, or positive, depending on the relationship between the variables involved.

---

Page Break

---

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
$N_2(\mu ,\mathrm{\Sigma })$

$N_2(\mu ,\mathrm{\Sigma })$

N_(2)(mu,Sigma)
N_2(mu, Sigma)



$N_2(\mu ,\mathrm{\Sigma })$


, 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
$N_2(\mu ,\frac{1}{3}\mathrm{\Sigma })$

$N_2\left(\mu ,\frac{1}{3}\mathrm{\Sigma }\right)$

N_(2)(mu,(1)/(3)Sigma)
N_2left(mu, frac{1}{3} Sigmaright)



$N_2(\mu ,\frac{1}{3}\mathrm{\Sigma })$


.
Answer:
The statement “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
$N_2(\mu ,\mathrm{\Sigma })$

$N_2(\mu ,\mathrm{\Sigma })$

N_(2)(mu,Sigma)
N_2(mu, Sigma)



$N_2(\mu ,\mathrm{\Sigma })$


, 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
$N_2(\mu ,\frac{1}{3}\mathrm{\Sigma })$

$N_2\left(\mu ,\frac{1}{3}\mathrm{\Sigma }\right)$

N_(2)(mu,(1)/(3)Sigma)
N_2left(mu, frac{1}{3} Sigmaright)



$N_2(\mu ,\frac{1}{3}\mathrm{\Sigma })$


” is
true
.

Justification:

1. Mean
   : The mean of
   $\frac{X_1+X_2+X_3}{3}$
   
   $\frac{X_1+X_2+X_3}{3}$
   
   (X_(1)+X_(2)+X_(3))/(3)
   frac{X_1+X_2+X_3}{3}
   
   
   
   $\frac{X_1+X_2+X_3}{3}$
   
   
   is
   $\frac{\mu +\mu +\mu }{3}=\mu$
   
   $\frac{\mu +\mu +\mu }{3}=\mu$
   
   (mu+mu+mu)/(3)=mu
   frac{mu + mu + mu}{3} = mu
   
   
   
   $\frac{\mu +\mu +\mu }{3}=\mu$
   
   
   .
2. Covariance Matrix
   : The covariance matrix of
   $X_1+X_2+X_3$
   
   $X_1+X_2+X_3$
   
   X_(1)+X_(2)+X_(3)
   X_1+X_2+X_3
   
   
   
   $X_1+X_2+X_3$
   
   
   is
   $\mathrm{\Sigma }+\mathrm{\Sigma }+\mathrm{\Sigma }=3\mathrm{\Sigma }$
   
   $\mathrm{\Sigma }+\mathrm{\Sigma }+\mathrm{\Sigma }=3\mathrm{\Sigma }$
   
   Sigma+Sigma+Sigma=3Sigma
   Sigma + Sigma + Sigma = 3Sigma
   
   
   
   $\mathrm{\Sigma }+\mathrm{\Sigma }+\mathrm{\Sigma }=3\mathrm{\Sigma }$
   
   
   because
   $X_1,X_2,X_3$
   
   $X_1,X_2,X_3$
   
   X_(1),X_(2),X_(3)
   X_1, X_2, X_3
   
   
   
   $X_1,X_2,X_3$
   
   
   are independent. Therefore, the covariance matrix of
   $\frac{X_1+X_2+X_3}{3}$
   
   $\frac{X_1+X_2+X_3}{3}$
   
   (X_(1)+X_(2)+X_(3))/(3)
   frac{X_1+X_2+X_3}{3}
   
   
   
   $\frac{X_1+X_2+X_3}{3}$
   
   
   is
   $\frac{1}{3^2}(3\mathrm{\Sigma })=\frac{1}{3}\mathrm{\Sigma }$
   
   $\frac{1}{3^2}(3\mathrm{\Sigma })=\frac{1}{3}\mathrm{\Sigma }$
   
   (1)/(3^(2))(3Sigma)=(1)/(3)Sigma
   frac{1}{3^2}(3Sigma) = frac{1}{3}Sigma
   
   
   
   $\frac{1}{3^2}(3\mathrm{\Sigma })=\frac{1}{3}\mathrm{\Sigma }$
   
   
   .

Since both the mean and the covariance matrix match the given distribution
$N_2(\mu ,\frac{1}{3}\mathrm{\Sigma })$

$N_2\left(\mu ,\frac{1}{3}\mathrm{\Sigma }\right)$

N_(2)(mu,(1)/(3)Sigma)
N_2left(mu, frac{1}{3} Sigmaright)



$N_2(\mu ,\frac{1}{3}\mathrm{\Sigma })$


, the statement is true.

---

Page Break

---

d) The partial correlation coefficients and multiple correlation coefficients lie between -1 and 1.
Answer:
The statement “The partial correlation coefficients and multiple correlation coefficients lie between -1 and 1” is
true
.

Justification:

1. Partial Correlation Coefficients
   : The partial correlation coefficient measures the strength and direction of the relationship between two variables while controlling for the effect of one or more other variables. Mathematically, it is defined in a way that ensures its value lies between -1 and 1, inclusive. Specifically, it is computed as the correlation between the residuals resulting from the linear regression of each variable against the control variables. Since residuals are uncorrelated with the predicted values, the partial correlation coefficient is constrained to be between -1 and 1.
2. Multiple Correlation Coefficients
   : The multiple correlation coefficient
   $R$
   
   $R$
   
   R
   R
   
   
   
   $R$
   
   
   is defined as the square root of the coefficient of determination
   $R^2$
   
   $R^2$
   
   R^(2)
   R^2
   
   
   
   $R^2$
   
   
   , which is the proportion of the variance in the dependent variable that is predictable from the independent variables in a multiple regression model. Since
   $R^2$
   
   $R^2$
   
   R^(2)
   R^2
   
   
   
   $R^2$
   
   
   is between 0 and 1,
   $R$
   
   $R$
   
   R
   R
   
   
   
   $R$
   
   
   must be between 0 and 1. However,
   $R$
   
   $R$
   
   R
   R
   
   
   
   $R$
   
   
   can be negative if the model includes a constant term and the slope is negative, but its absolute value will still be between 0 and 1.

Therefore, both the partial correlation coefficients and multiple correlation coefficients are bounded between -1 and 1, making the statement true.

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---

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}



$M_t,\underset{t\to 0}{lim}\frac{M_t}{t}=\frac{1}{\mu }$


.
Answer:
The statement “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}



$M_t,\underset{t\to 0}{lim}\frac{M_t}{t}=\frac{1}{\mu }$


” is
generally true under certain conditions
.

Justification:

A renewal function
$M_t$

$M_t$

M_(t)
M_t



$M_t$


is defined as the expected number of renewals (or arrivals, or events) that have occurred by time
$t$

$t$

t
t



$t$


. Mathematically, it is often defined as:
$$M_t=\mathbb{E}[N(t)]$$

$$M_t=\mathbb{E}[N(t)]$$

M_(t)=E[N(t)]
M_t = mathbb{E}[N(t)]



$$M_t=\mathbb{E}[N(t)]$$


where
$N(t)$

$N(t)$

N(t)
N(t)



$N(t)$


is the number of renewals by time
$t$

$t$

t
t



$t$


.
The mean inter-arrival time (or mean time between renewals) is denoted by
$\mu$

$\mu$

mu
mu



$\mu$


and is defined as:
$$\mu =\mathbb{E}[X]$$

$$\mu =\mathbb{E}[X]$$

mu=E[X]
mu = mathbb{E}[X]



$$\mu =\mathbb{E}[X]$$


where
$X$

$X$

X
X



$X$


is the random variable representing the time between renewals.
Under the assumption that
$\mu <\mathrm{\infty }$

$\mu <\mathrm{\infty }$

mu < oo
mu < infty



$\mu <\mathrm{\infty }$


and the distribution of inter-arrival times has a finite variance, it is generally true in renewal theory that:
$$\underset{t\to 0}{lim}\frac{M_t}{t}=\frac{1}{\mu }$$

$$\underset{t\to 0}{lim}\frac{M_t}{t}=\frac{1}{\mu }$$

lim_(t rarr0)(M_(t))/(t)=(1)/(mu)
lim_{{t rightarrow 0}} frac{M_t}{t} = frac{1}{mu}



$$\underset{t\to 0}{lim}\frac{M_t}{t}=\frac{1}{\mu }$$


This result is often derived using advanced methods in renewal theory and stochastic processes, and it provides a way to relate the renewal function to the mean inter-arrival time
$\mu$

$\mu$

mu
mu



$\mu$


.
So, the statement is true under the conditions that
$\mu <\mathrm{\infty }$

$\mu <\mathrm{\infty }$

mu < oo
mu < infty



$\mu <\mathrm{\infty }$


and the distribution of inter-arrival times has a finite variance.


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Is this Complete Solution for IGNOU Assignments?

Yes, It is Complete Solution, a comprehensive solution to the assignments for IGNOU. Valid from January 1, 2023 to December 31, 2023.

Is the Complete Solution aligned with the IGNOU requirements?

Yes, the Complete Solution is aligned with the IGNOU requirements and has been solved accordingly.

Is the Complete Solution guaranteed error-free?

Yes, the Complete Solution is guaranteed to be error-free.The solutions are thoroughly researched and verified by subject matter experts to ensure their accuracy.

Can I access the Complete Solution anytime?

As of now, you have access to the Complete Solution for a period of 6 months after the date of purchase, which is sufficient to complete the assignment. However, we can extend the access period upon request. You can access the solution anytime through our app.

What if I need help with a specific assignment question?

The app provides complete solutions for all assignment questions. If you still need help, you can contact the support team for assistance at Whatsapp +91-9958288900

Can I access the educational materials on multiple devices?

No, access to the educational materials is limited to one device only, where you have first logged in. Logging in on multiple devices is not allowed and may result in the revocation of access to the educational materials.

How do I make a payment for accessing the solutions?

Payments can be made through various secure online payment methods available in the app.Your payment information is protected with industry-standard security measures to ensure its confidentiality and safety. You will receive a receipt for your payment through email or within the app, depending on your preference.

What are the instructions for formatting my assignments?

The instructions for formatting your assignments are detailed in the Assignment Booklet, which includes details on paper size, margins, precision, and submission requirements. It is important to strictly follow these instructions to facilitate evaluation and avoid delays.

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