MMT-008 Solved Assignment 2023

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 P P P \mathrm{P} P is a transition matrix of a Markov Chain, then all the rows of lim n → ∞ P n lim n → ∞ P n lim_(nrarr oo)P^(n) \lim _{\mathrm{n} \rightarrow \infty} \mathrm{P}^{\mathrm{n}} lim n → ∞ P 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 ( μ , Σ ) N 2 ( μ , Σ ) N_(2)(mu,Sigma) N_{2}(\mu, \Sigma) N 2 ( μ , Σ ) , then X 1 + X 2 + X 3 3 X 1 + X 2 + X 3 3 (X_(1)+X_(2)+X_(3))/(3) \frac{X_{1}+X_{2}+X_{3}}{3} X 1 + X 2 + X 3 3 follows N 2 ( μ , 1 3 Σ ) N 2 μ , 1 3 Σ N_(2)(mu,(1)/(3)Sigma) N_{2}\left(\mu, \frac{1}{3} \Sigma\right) N 2 ( μ , 1 3 Σ ) .
d) The partial correlation coefficients and multiple correlation coefficients lie between -1 and 1.
e) For a renewal function M t , lim t → 0 M t t = 1 μ M t , lim t → 0 M t t = 1 μ 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 , lim t → 0 M t t = 1 μ .
  1. a) Let ( X , Y ) ( X , Y ) (X,Y) (X, Y) ( X , Y ) have the joint p.d.f. given by:
f ( x , y ) = { 1 , if | y | < x ; 0 < x < 1 0 , otherwise f ( x , y ) = 1 ,  if  | y | < x ; 0 < x < 1 0 ,  otherwise  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 ) = { 1 ,  if  | y | < x ; 0 < x < 1 0 ,  otherwise 
i) Find the marginal p.d.f.’s of X X X \mathrm{X} X and Y Y Y \mathrm{Y} 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 E ( X ∣ Y = y ) E ( X ∣ Y = y ) E(X∣Y=y) \mathrm{E}(\mathrm{X} \mid \mathrm{Y}=\mathrm{y}) E ( X ∣ Y = y ) and E ( Y ∣ X = x ) E ( Y ∣ X = x ) E(Y∣X=x) \mathrm{E}(\mathrm{Y} \mid \mathrm{X}=\mathrm{x}) E ( Y ∣ X = x ) .
b) Let the joint probability density function of two discrete random X X X \mathrm{X} X and Y Y Y \mathrm{Y} Y be given as:
X X X \mathrm{X} 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 X X X \mathrm{X} X and Y Y Y \mathrm{Y} Y . ii) Find the conditional distribution of X X X \mathrm{X} X given Y = 1 Y = 1 Y=1 \mathrm{Y}=1 Y = 1 .
iii) Test the independence of variable s X s X sX \mathrm{s} X s X and Y Y Y \mathrm{Y} Y .
iv) Find V [ Y X = x ] V Y X = x V[(Y)/(X)=x] V\left[\frac{Y}{X}=x\right] V [ Y X = x ] .
  1. a) Let X ∼ N 3 ( μ , Σ ) X ∼ N 3 ( μ , Σ ) X∼N_(3)(mu,Sigma) \mathrm{X} \sim \mathrm{N}_{3}(\mu, \Sigma) X ∼ N 3 ( μ , Σ ) , where μ = [ 5 , 3 , 4 ] ′ μ = [ 5 , 3 , 4 ] ′ mu=[5,3,4]^(‘) \mu=[5,3,4]^{\prime} μ = [ 5 , 3 , 4 ] ′ and
∑ = ( 2 1 1 1 1 0.5 1 0.5 1 ) ∑ = 2 1 1 1 1 0.5 1 0.5 1 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) ∑ = ( 2 1 1 1 1 0.5 1 0.5 1 )
Find the distribution of:
( 2 X 1 + X 2 − X 3 X 1 + X 2 + X 3 ) 2 X 1 + X 2 − X 3 X 1 + X 2 + X 3 ([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) ( 2 X 1 + X 2 − X 3 X 1 + X 2 + X 3 )
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:
∑ = ( 1 − 2 0 − 2 5 0 0 0 1 ) ∑ = 1 − 2 0 − 2 5 0 0 0 1 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) ∑ = ( 1 − 2 0 − 2 5 0 0 0 1 )
Also calculate the proportion of total population variance for the first principal component.
  1. a) Consider a Markov chain with transition probability matrix:
P = ( 0 0 1 0 0 0 0 1 0 1 0 0 1 4 1 8 1 8 1 2 ) P = 0 0 1 0 0 0 0 1 0 1 0 0 1 4 1 8 1 8 1 2 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 = ( 0 0 1 0 0 0 0 1 0 1 0 0 1 4 1 8 1 8 1 2 )
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 ( x 1 ) x 1 (x_(1)) \left(\mathrm{x}_{1}\right) ( x 1 ) and demands ( x 2 ) x 2 (x_(2)) \left(\mathrm{x}_{2}\right) ( x 2 ) and the following information was obtained:
∑ X = 96 , ∑ Y = 72 , ∑ X 2 = 780 , ∑ Y 2 = 480 , ∑ X Y = 588 ∑ X = 96 , ∑ Y = 72 , ∑ X 2 = 780 , ∑ Y 2 = 480 , ∑ X Y = 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 ∑ X = 96 , ∑ Y = 72 , ∑ X 2 = 780 , ∑ Y 2 = 480 , ∑ X Y = 588
The expected mean vector and variance covariance matrix for the factories in the population are:
μ = [ 9 7 ] and ∑ = [ 13 9 9 7 ] . μ = 9 7  and  ∑ = 13 9 9 7 .  {:[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} μ = [ 9 7 ]  and  ∑ = [ 13 9 9 7 ] . 
Test whether the sample confirms its truthness of mean vector at 5 % 5 % 5% 5 \% 5 % level of significance, if:
i) Σ Σ Sigma \Sigma Σ is known,
ii) Σ Σ Sigma \Sigma Σ is unknown.
[You may use: χ 2 , 0.05 2 = 10.60 , χ 3 , 0.05 2 = 12.83 , χ 4 , 0.05 2 = 14.89 , F 2 , 10 , 0.05 = 4.10 χ 2 , 0.05 2 = 10.60 , χ 3 , 0.05 2 = 12.83 , χ 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,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 χ 2 , 0.05 2 = 10.60 , χ 3 , 0.05 2 = 12.83 , χ 4 , 0.05 2 = 14.89 ,   F 2 , 10 , 0.05 = 4.10 ]
  1. a) Let the random vector X ′ = ( X 1 , X 2 , X 3 ) X ′ = X 1 , X 2 , X 3 X^(‘)=(X_(1),X_(2),X_(3)) X^{\prime}=\left(X_{1}, X_{2}, X_{3}\right) X ′ = ( 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 = ( 1 1 1 1 2 3 1 3 9 ) = 1 1 1 1 2 3 1 3 9 =([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) = ( 1 1 1 1 2 3 1 3 9 ) . 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 X 3 X 3 X_(3) \mathrm{X}_{3} X 3 and [ X 1 , X 2 ] X 1 , X 2 [X_(1),X_(2)] \left[\mathrm{X}_{1}, \mathrm{X}_{2}\right] [ X 1 , X 2 ] .
b) Define ultimate extinction in a branching process. Let p k = b c k − 1 , k = 1 , 2 , … p k = b c k − 1 , k = 1 , 2 , … p_(k)=bc^(k-1),k=1,2,dots \mathrm{p}_{\mathrm{k}}=\mathrm{bc}^{\mathrm{k}-1}, \mathrm{k}=1,2, \ldots p k = b c k − 1 , k = 1 , 2 , … ;
0 < b < c < b + c < 1 0 < b < c < b + c < 1 0 < b < c < b+c < 1 0<\mathrm{b}<\mathrm{c}<\mathrm{b}+\mathrm{c}<1 0 < b < c < b + c < 1 and p 0 = 1 − ∑ k = 1 ∞ p k p 0 = 1 − ∑ k = 1 ∞ p k p_(0)=1-sum_(k=1)^(oo)p_(k) \mathrm{p}_{0}=1-\sum_{\mathrm{k}=1}^{\infty} \mathrm{p}_{\mathrm{k}} p 0 = 1 − ∑ k = 1 ∞ p k . Then discuss the probability of extinction in different cases for E ( X 1 ) ≥ 1 E X 1 ≥ 1 E(X_(1)) >= 1 \mathrm{E}\left(\mathrm{X}_{1}\right) \geq 1 E ( X 1 ) ≥ 1 or E ( X 1 ) < 1 E X 1 < 1 E(X_(1)) < 1 \mathrm{E}\left(\mathrm{X}_{1}\right)<1 E ( X 1 ) < 1 .
  1. a) If the random vector Z Z Z \mathrm{Z} Z be N 4 ( μ , Σ ) N 4 ( μ , Σ ) N_(4)(mu,Sigma) \mathrm{N}_{4}(\mu, \Sigma) N 4 ( μ , Σ ) , where:
μ = [ 1 2 5 − 2 ] μ = 1 2 5 − 2 mu=[[1],[2],[5],[-2]] \mu=\left[\begin{array}{c} 1 \\ 2 \\ 5 \\ -2 \end{array}\right] μ = [ 1 2 5 − 2 ]
and ∑ = [ 3 3 0 9 3 2 − 1 1 0 − 1 6 − 3 9 1 − 3 7 ] ∑ = 3 3 0 9 3 2 − 1 1 0 − 1 6 − 3 9 1 − 3 7 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] ∑ = [ 3 3 0 9 3 2 − 1 1 0 − 1 6 − 3 9 1 − 3 7 ] .
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 , … . X 1 , X 2 , … . X_(1),X_(2),dots. X_{1}, X_{2}, \ldots . X 1 , X 2 , … . . 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 C 2 = 8 C 2 = 8 C_(2)=8 \mathrm{C}_{2}=8 C 2 = 8 . Find:
i) μ T μ T mu^(T) \mu^{\mathrm{T}} μ T
ii) T T T \mathrm{T} T which minimizes C ( T ) C ( T ) C(T) \mathrm{C}(\mathrm{T}) C ( T ) and which is the better policy in the long-run in terms of cost.
  1. 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
p = 2 [ 1 2 3 1 2 1 4 1 4 1 3 0 2 3 1 2 1 2 0 ] p = 2 1 2 3 1 2 1 4 1 4 1 3 0 2 3 1 2 1 2 0 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] p = 2 [ 1 2 3 1 2 1 4 1 4 1 3 0 2 3 1 2 1 2 0 ]
i) Draw the state transition diagram for this chain.
ii) If P ( X 1 = 1 ) = P ( X 1 = 2 ) = 1 4 P X 1 = 1 = P X 1 = 2 = 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} P ( X 1 = 1 ) = P ( X 1 = 2 ) = 1 4 , then find P ( X 1 = 3 , X 2 = 2 , X 3 = 1 ) P X 1 = 3 , X 2 = 2 , X 3 = 1 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) P ( X 1 = 3 , X 2 = 2 , X 3 = 1 ) .
iii) Check whether the chain is irreducible and a periodic.
iv) Find the stationary distribution for the chain.
b) If N 1 ( t ) , N 2 ( t ) N 1 ( t ) , N 2 ( t ) N_(1)(t),N_(2)(t) \mathrm{N}_{1}(\mathrm{t}), \mathrm{N}_{2}(\mathrm{t}) N 1 ( t ) , N 2 ( t ) are two independent Poisson process with parameters λ 1 λ 1 lambda_(1) \lambda_{1} λ 1 and λ 2 λ 2 lambda_(2) \lambda_{2} λ 2 respectively, then show that