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The following is an exercise in Karlin and Pinsky's An Introduction to Stochastic Modeling:

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Here is my question:

This problem seems to assume that the Markov Chain is time-homogeneous. But I cannot make sense of it. Let $(X_n)$ be the Markov chain in the model. How would one expect that the following is true? $$ P(X_4=0\mid X_3=1)=P(X_2=0\mid X_1=1) $$


[Added:] To make my question clearer, all I concern now are

  • (1)how to calculate this particular transition probability

$$P(X_4=0|X_3=1)$$

  • (2) why is $$ P(X_4=0\mid X_3=1)=P(X_2=0\mid X_1=1) $$ true.

For (1), denote $A$ as the event that "no customer arrive in a period" and $B$ as the event that "a taxi comes at the 4th period" (namely "service for the people at the 3rd period is done at the 4th period"). By independence of $A$ and $B$, one has $$ P(X_4=0|X_3=1)=P(AB|X_3=1)=P(A|X_3=1)P(B|X_3=1)=P(A)P(B|X_3=1) \tag{*} $$ where the last equality is given by the independence of the events $A$ and $X_3=1$. Now $P(A)=1-\beta$ is given. One needs $P(B|X_3=1)$ in order to calculate ($*$). What I was really worrying about is the question — "when did this people get on the bus?" — which should determine the "service time" $Z$. More generally, how should one expect that $P(B|X_n=1)$ is independent of $n$, in the case of which one might have $$ P(B|X_n=1)=P(Z=1)? $$

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I'm not sure I understand your problem. In the Markov process the queue length increases by $1$ with probability $\beta$ at each step. If the queue length isn't $0$ then it decreases as specified with $\alpha$. That should be sufficient information to write the matrix.

For example, $$ P(X_4 = 0 | P(X_3 = 1) = \alpha (1-\beta) $$ since this state transition can happen just one way: no customer arrives and the one there departs.

That happens to be the same computation as $P(X_2 = 0 | P(X_1 = 1)$, not because of the previous history (which can't matter in a Markov process) but because it's the only way to empty the queue when there's just one customer.

In general, the queue increases by $1$ with probability $\beta$ independent of its current value, but the decrease is proportional to the number present, since each customer gets independent service.

Time isn't continuous; it's measured in discrete time periods. (You know that.)

I haven't addressed your expression using explicit conditional probabilities because that's not how I think. I'm sure it's possible to frame the argument that way too.

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    Thank you for your answer. Would you elaborate why the queue length decreases with probability $P(Z=1)$? This is exactly what confuses me and I have edited the post accordingly. For a simple example, how would one calculate $P(X_4=0|X_3=1)$?2017-02-02
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    Thanks for your editing. How do you get $\beta-\alpha$? Do you actually mean $\alpha(1-\beta)$?2017-02-02
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    @Jack Yes thanks fixed.2017-02-02