Exercise 2.7.1 of J. Norris, "Markov Chains" 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. 
 A: 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 $Y_n=i$.
A: At first, I misread the question just as Didier said. (Bit of lacking in taste for probability problems. )
To summarize the previous answer and comments, I'll write a brief formularized answer. 
Without loss of generality, we assume $i > 0$,
$$P(\exists t > 0, s.t. X_t = i) = \lim_{n \to \infty} P(T_i < T_{-n})$$
So the problem can be simplified into calculating $P_0(T_i < T_{-n})$. (I add the subscript because it can be helpful later. $P_k$ denotes starting from $X_0 = k$. $T_i$ denotes the hitting time for $i$. )
First of all,
$$P_i(T_i < T_{-n}) = 1, P_{-n}(T_i < T_{-n}) = 0$$
Then, according to the transition matrix, 
$$P_j = \lambda P_{j-1} + \mu P_{j+1}, \quad -n < j < i$$
which can be rewritten into
$$P_{j+1} - P_j = \frac{\lambda}{\mu} (P_j - P_{j-1})$$
Therefore, (and the final form will be slightly different for $\frac{\lambda}{\mu} = 1$)
$$1 = P_{i} - P_{-n} = \sum_{j = -(n-1)}^{i} (P_{j} - P_{j-1})$$
$$ = \sum_{j = -(n-1)}^{i} (\frac{\lambda}{\mu})^{j+(n-1)} (P_{-(n-1)} - P_{-n}) = \frac{1 - (\frac{\lambda}{\mu})^{i+n}}{1 - \frac{\lambda}{\mu}} (P_{-(n-1)} - P_{-n})$$
The rest are simple calculations. 
$$P_0 = P_0 - P_{-n} = \sum_{j=-(n-1)}^0 (P_{j} - P_{j-1}) = \ ... $$
I'll add part b for reference: 
(b) Find for all integers $i$, the expected total time spent in state $i$, starting from $0$. 
1) $\lambda = \mu = \frac{1}{2},$ all expected total time is $\infty$, because the jump chain is recurrent.
2) $\lambda > \mu,$ time spent in each state is positive but not $\infty$, because the jump chain is transient. 
I don't know a closed form yet. For $i \geq 0$,
$$t_0(i) = \sum_{j = i}^{\infty} \binom{2j-i}{i} \lambda^{j} \mu^{(j-i)} $$
For $i \le 0$,
$$t_0(i) = \sum_{j = -i}^{\infty} \binom{2j-(-i)}{-i} \lambda^{(j - (-i))} \mu^{j} $$
is the expected times of visits for the jump chain to i.
And the expected time for each visit is $\frac{1}{q_i}$. 
They are independent.
3) $\lambda < \mu,$ similar to 2). 
