Stochastic processes: Need step by step solution on problem (c) Problem 1 In P&K

Question

Stochastic processes: Need step by step solution on problem (c)

Problem 1 In P&K; Chapter 6.3 the birth and death process is treated. The forward Kolmogorov differential equations are given as equations (6.24) on page 300. (6.24) The limits limt-too Pj(t) Tj exist and are independent of the initial state i. It may happen that Tj 0 for all states j, f.ex.the Yule-process. When the limits , are strictly positive and satisfy 00Tj-1, they are called the limiting distribution of the process. Since the probabilities converge to constants, the limit of the derivatives must be zero. Passing to the limit in the forward Kolmogorov differential equations (6.24) produces (6.35) (6.35) whose solution is given in equation (6.37). The purpose of this preliminary exercise is to show that if the detailed balance equations Tj ,-B+1+1 or generally qi,-Tj Zn have a solution for all pairs i, j such that /V = 1, then this solution also solves (6.35). The qij are the elements of the infinitesimal generator (6.17), and gj-PHqjk-These are the same as the qi in the next problem

Explanation / Answer

Here we are asked to do part c)

Let the stationary distribution for the given matrix be given as: X, Y and Z for the three states. Then we have here:

The sum of all probabilities should be equal to 1. Therefore

X + Y + Z = 1

From the first column, we get:

X = -3X + Y + 2Z

4X = Y + 2Z

From third column, we get:

Z = X + 2Y - 3Z

4Z = X + 2Y

Also from the second column, we get:

4Y = 2X + Z

Therefore, we get the three equations from the matrix as:

4X = Y + 2Z
4Z = X + 2Y
4Y = 2X + Z

Therefore as all the above 3 equations are symmetric, therefore X = Y = Z satisfies all the equations above.

As X + Y + Z = 1

Therefore,

X = Y = Z = (1/3)

This is the required stationary distribution here:

[ 1/3, 1/3, 1/3 ]

In long term it means that there is an equal probability of being in any of three states.

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