Question

Consider a machine whose condition at any time can be classified as being in one of the following three states: State 1: Good operating order State 2: Deteriorated operating order State 3: In repair The condition of the machine is observed at the end of each period in a sequence of periods. Let Xn denote the condition of the machine at the end of period n for n = 1, 2, . . . Let X0 be the condition of the machine at the start. The sequence of machine conditions Xn is a Markov chain with transition probabilities p11 = 0.9 p12 = 0.1 p13 = 0 p21 = 0 p22 = 0.9 p23 = 0.1 p31 = 1 p32 = 0 p33 = 0 and the process starts in state X0 = 1. a) Find P(X4 = 1). b) Calculate the limiting distribution. c) What is the long run rate of repairs per unit time.

          Consider a machine whose condition at any time can be classified as being in one of the following three states:
State 1: Good operating order
State 2: Deteriorated operating order
State 3: In repair
The condition of the machine is observed at the end of each period in a sequence of periods. Let Xn denote the condition of the machine at the end of period n for n = 1, 2, . . . Let X0 be the condition of the machine at the start. The sequence of machine conditions Xn is a Markov chain with transition probabilities
p11 = 0.9 p12 = 0.1 p13 = 0
p21 = 0 p22 = 0.9 p23 = 0.1
p31 = 1 p32 = 0 p33 = 0
and the process starts in state X0 = 1.
a) Find P(X4 = 1).
b) Calculate the limiting distribution.
c) What is the long run rate of repairs per unit time.

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Consider a machine whose condition at any time can be classified as being in one of the following three states:
State 1: Good operating order
State 2: Deteriorated operating order
State 3: In repair
The condition of the machine is observed at the end of each period in a sequence of periods. Let Xn denote the condition of the machine at the end of period n for n = 1, 2, . . . Let X0 be the condition of the machine at the start. The sequence of machine conditions Xn is a Markov chain with transition probabilities
p11 = 0.9 p12 = 0.1 p13 = 0
p21 = 0 p22 = 0.9 p23 = 0.1
p31 = 1 p32 = 0 p33 = 0
and the process starts in state X0 = 1.
a) Find P(X4 = 1).
b) Calculate the limiting distribution.
c) What is the long run rate of repairs per unit time.

Added by Kelsey B.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

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Solved by Expert Shaiju T

06/16/2022

Step 1

9 & 0.1 & 0 \\ 0 & 0.9 & 0.1 \\ 1 & 0 & 0 \end{bmatrix} \] Show more…

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Consider a machine whose condition at any time can be classified as being in one of the following three states: State 1: Good operating order State 2: Deteriorated operating order State 3: In repair The condition of the machine is observed at the end of each period in a sequence of periods. Let Xn denote the condition of the machine at the end of period n for n = 1, 2, ... Let X0 be the condition of the machine at the start. The sequence of machine conditions Xn is a Markov chain with transition probabilities p11 = 0.9 p12 = 0.1 p13 = 0 p21 = 0 p22 = 0.9 p23 = 0.1 p31 = 1 p32 = 0 p33 = 0 and the process starts in state X0 = 1. a) Find P(X4 = 1). b) Calculate the limiting distribution. c) What is the long run rate of repairs per unit time.

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Transcript

00:01 Hey everyone in this video.

00:02 Firstly, the computed the transition probability matrix is shown and now in the a part we have the probability of state x4 is equal to 1 is the probability x4 is equal to 0 .9 raise to power 4 plus 0 .1 multiply 0 .9 multiply 0 .1 multiply 1 plus 0 .9 multiply 0 .1 multiply 0 .1 multiply 0 .1 multiply 0 .1 multiply 1.

00:28 So, this is equal to 0 .6561 plus 0 .009 plus 0 .009.

00:37 So, after simplifying, this is equal to 0 .6741.

00:43 Thus, the probability of x4 is equal to 1 is equal to 0 .6741.

00:48 So now, in the b part we have the equations, the equations for calculating the limiting distribution are.

00:56 We have first equation, second, third and 4.

00:58 Equations as you can see on the screen.

01:01 Now to satisfy the limiting constraint that is submission k p i k is equal to 1.

01:07 Now simplify the any two equations as shown below.

01:11 So from equation 1, 0 .9 pi 1 plus pi 3 is equal to 0 .0 .9 pi 1 plus pi 3 minus pi 1 is equal to 0 .9 minus 1, multiply pi 1 plus pi 3 is equal to 0.

01:29 So minus 0 .1 pi 1 plus pi 3 is equal to 0 .1.

01:33 So this is our 5th equation.

01:35 Minus 0 .1 pi 1 plus pi 3 is equal to 0 .1.

01:39 So now the value of pi 3 is equal to 0 .1 pi 1 .5 .5 .5...

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