[markov chain] prove that probability equals 6/pi²

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Homework Statement


attachment.php?attachmentid=42880&stc=1&d=1327088638.gif


Homework Equations


N/A

The Attempt at a Solution


I'll shortly explain what my reasoning is so far, but please ignore it if it comes across too jumbled:
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Let P denote the markov matrix associated with this problem, then I think I was able to argue that the probability that is asked for is equal to 1- \sum_{n=1}^{+\infty} P^n(0,0) where P^n(0,0) denotes the element in the first row of the first column of the n-th power of the Markov matrix.

And I then wanted to calculate P^n(0,0) for every n by trying to find a pattern in P^1(0,0), P^2(0,0), P^3(0,0), etc. I think I found one: define t_n = \left(t_{n-1} + \left( \frac{n+1}{n} \right)^2 \prod_{k=1}^{n+1} x_k \right) x_n (with t_0 = x_1) with x_n = p_{n,n-1}, then I think \sum_{n=1}^{+\infty} P^n(0,0) = \sum_{n=0}^{+\infty} t_n. But it seems near impossible to prove that 1 minus this expression equals \frac{6}{\pi^2} so I'm probably way off track...
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Any better suggestions? How would you approach this problem instead?
 

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Just an idea. In some circumstances (if \|P\|\leq 1) we have

\sum_{n=0}^{+\infty}{P^n}=(I-P)^{-1}

So all you need to do know is to invert I-P...
 
*tries to invert infinite-dimensional matrix*
 
mr. vodka said:

Homework Statement


attachment.php?attachmentid=42880&stc=1&d=1327088638.gif


Homework Equations


N/A

The Attempt at a Solution


I'll shortly explain what my reasoning is so far, but please ignore it if it comes across too jumbled:
---
Let P denote the markov matrix associated with this problem, then I think I was able to argue that the probability that is asked for is equal to 1- \sum_{n=1}^{+\infty} P^n(0,0) where P^n(0,0) denotes the element in the first row of the first column of the n-th power of the Markov matrix.

And I then wanted to calculate P^n(0,0) for every n by trying to find a pattern in P^1(0,0), P^2(0,0), P^3(0,0), etc. I think I found one: define t_n = \left(t_{n-1} + \left( \frac{n+1}{n} \right)^2 \prod_{k=1}^{n+1} x_k \right) x_n (with t_0 = x_1) with x_n = p_{n,n-1}, then I think \sum_{n=1}^{+\infty} P^n(0,0) = \sum_{n=0}^{+\infty} t_n. But it seems near impossible to prove that 1 minus this expression equals \frac{6}{\pi^2} so I'm probably way off track...
---

Any better suggestions? How would you approach this problem instead?

I found the notation confusing at first, but now I see that what you want to prove is that the chain is transient, with Pr{return to 0} = 1-6/pi^2.

The first order of business is to establish that the chain is transient, with P{return to zero} < 1. Since you have a discrete-time birth-death process, quite a lot is known or knowable about your system. For example, look at the 1995 paper by Van Doorn, freely downloadable from
http://journals.cambridge.org/downl...21a.pdf&code=f73435367827c0b479829c68c457666d
(Google search on "birth-death process+discrete time", and look at the 4th entry "Geometric Ergodicity ... "). In particular, Theorem 2.1 may be a starting point. As for the problem of actually computing the return probability, all I can suggest is that you look at the first-passage-probability equations (for end-state 0) and try to solve them, perhaps using z-transform techniques, or something similar.

Good luck.

RGV
 
Last edited:
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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