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I am working though the book of J. Norris, "Markov Chains" as self-study and have difficulty with ex. 2.7.1, part a.

The exercise can be read through Google books. My understanding is that the probability is given by (0,i) matrix element of exp(t*Q). Setting up forward evolution equation leads to differential difference equation which I was hinted admits no closed form solution (see my question on mathoverflow).

I obviously need an nudge in a right direction, and the exercise must admit a simple solution. Any directions are much appreciated.

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Odd, if you are reading it as self-study why do you call it homework? –  Listing May 18 '11 at 16:57
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@user3123 I removed homework tag and added stochastic-processes tag. I initially tagged it as a homework because it would be a homework should I enroll in a course. –  Sasha May 18 '11 at 16:59

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It seems you misread the question. One is not interested in the probability of the event $[X_t=i]$ for a given time $t$ but in the probability of the event that there exists a time $t$ such that $X_t=i$.

Here is a nudge...

One approach is to consider the sequence $(Y_n)_n$ of the successive different states visited by $(X_t)_t$. You could try to show that $(Y_n)_n$ is a discrete Markov chain, to compute its transition probabilities (these are very simple and, hint, they do not depend on the numbers $q_i$) and, finally, to compute the probability that there exists an integer time $n$ such that $Y_n=i$.

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Thanks ! Following the book of Norris, the associate discrete time Markov process is 1D random walk with probabilities`$\lambda$ and $\mu$ of up-moves and down-moves, respectively, and is independent of $q_i$. Let $i=2k$, this state can be reached in $n=2m$ number of steps with $m>abs(k)$. The probability is $\mathbb{P}(Y_n=i) = \binom{n}{m+k} \lambda^{(m+k)}*\mu^{(m-k)}$. So if $i>0$ and $1>\lambda>0$ the probability of reaching $i$ in $n$ steps is non-zero for every $n>=abs(i)$. –  Sasha May 18 '11 at 20:07
    
Sure, but this is not an answer to question (a). –  Did May 18 '11 at 20:20
    
Am I correct in looking at this as a random variate which is the difference of independent geometric distributions ? Then the probability is $\lambda \mu^{i+1}/(1-\lambda \mu)$ for positive $i$, and $\mu \lambda^{1-i}/(1-\lambda \mu)$ for non-positive. –  Sasha May 18 '11 at 21:23
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No this does not. OK, let us go back to my post, shall we? Did you compute the transition probabilities of $(Y_n)$? Next, fix $i$, for example $i\ge1$. Can you compute $P_0(T_i<T_{n})$, for a given $n\le-1$? Next, what happens when $n\to-\infty$? –  Did May 19 '11 at 17:41
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@Didier A-ha. Many thanks for your help. It turns out my previous analysis with first passage times is also correct, as the particular Gauss function simplifies so that $\lambda^i F(i/2,(i+1)/2,i+1,4 \lambda \mu)$ equals $(1/\lambda-1)^i$ for $0<\lambda <1/2$ and equals $1$ for $1>\lambda >1/2$. Hurray! And thanks again! –  Sasha May 19 '11 at 19:04

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