# Birth and death process. Total time spent in state $i$.

Question:

Let $X(t)$ be a birth-death process with $\lambda_n = \lambda > 0$ and $\mu_n = \mu > 0,$ where $\lambda > \mu$ and $X(0) = 0$. Show that the total time $T_i$ spent in state $i$ is $\exp(\lambda−\mu)$-distributed.

Solution from the professor:

Writing $q_i$ for the probability of ever visiting $0$ having started at $i$ we have

$q_0 = 1$ and

$$q_i = \frac{\mu}{\lambda+\mu} q_{i-1} + \frac{\lambda}{\lambda+\mu} q_{i+1}$$

for $i ≥ 1$

The zeros of the characteristic polynomial for this difference equation are

$$p(x) = \frac{\lambda}{\lambda+\mu} x^2 - x + \frac{\mu}{\lambda+\mu} = 0$$

$$x = \frac{\mu}{\lambda}$$ or $$x=1$$

so that $q_i = A (\frac{\mu}{\lambda})^i + B1^i$ for some constants $A, B \in \mathbb{R}$. As we must have $q_i → 0$ as $i \rightarrow \infty$ we have $B = 0$ after which $q_0 = 1$ gives $A = 1$, so that $q_i = (\frac{\mu}{\lambda})^i$ for $i ≥ 0$.

To find $T_0$ we note that this time is the sum of the $\exp(\lambda)$-distributed time it takes to leave $0$ plus another independent $\exp(\lambda)$-distributed time added for each revisit of $0$, where the number $N$ of such revisits has PMF

$$P(N = n) = (\frac{\mu}{\lambda})^n(1−\frac{\mu}{\lambda})$$

for $n ≥ 0$

As the CHF of an $\exp(\lambda)$-distributed random variable is

$$E(e^{jω exp(\lambda)}) = \frac{\lambda}{\lambda-j\omega}$$

it follows that (making use of the basic fact that the CHF of a sum of independent random variables is the product of the CHF’s of the individual random variables)

$$E[e^{j\omega T_0}] = \frac{\lambda}{\lambda-j\omega} \sum_{n=0}^\infty (\frac{\lambda}{\lambda - j\omega})^n (\frac{\mu}{\lambda})^n(1−\frac{\mu}{\lambda}) = \frac{\lambda-\mu}{\lambda-\mu-j\omega}$$

To find $T_i$ we note that (by considering what the first state after having left $i$ is $i−1$ or $i+1$) the probability of ever returning to $i$ having started there is

Im having a hard time understanding this part and would appreciate any help

$$\frac{\mu}{\lambda+\mu} + \frac{\lambda}{\lambda+\mu} q_{i} = \frac{\mu}{\lambda+\mu} + \frac{\lambda}{\lambda+\mu}\frac{\mu}{\lambda} = \frac{2\mu}{\lambda+\mu}$$

As the time spent at each visits of $i$ is $\exp(\lambda+\mu)$-distributed it follows as above that

$$E[e^{j\omega T_i}] = \frac{\lambda+\mu}{\lambda+\mu-j\omega} \sum_{n=0}^\infty (\frac{\lambda+\mu}{\lambda +\mu- j\omega})^n (\frac{2\mu}{\lambda+\mu})^n(1−\frac{2\mu}{\lambda + \mu}) = \frac{\lambda-\mu}{\lambda-\mu-j\omega}$$

• I think the new tag is far too localized, and is already covered by similar tags. I strongly suggest posting on meta before adding new tags. – user296602 Feb 14 '16 at 18:40
• There are tags for other types of processes, at least i know there is for poisson processes, levy processes. And birth and death processes are one of the most prominent processes around. But sure i can remove it if its not appropriate. And good you pointed out that thread on meta, didnt know about that. thanks. – JKnecht Feb 14 '16 at 18:49

It uses the transition structure of the chain. Starting from a state $i > 0$, the next state is either $i-1$ or $i+1$.
If it is $i-1$ you will return to state $i$ for sure (since all steps are up or down by $1$); however, if the next state is $i+1$, the probability of eventually returning to $i$ is the same as the probability of ever reaching state $0$, starting from state $1$ (because if we start from state $i+1$ and erase all states < $i$---these have already been accounted for--- the process looks like we want to return to state $0$, starting from state $1$).
So, the probability of returning to state $i$ at some time $t > 0$ is $$P(\text{down step}) \cdot 1 + P(\text{up-step}) \cdot q_1 = \frac{2 \mu}{\lambda + \mu}$$
The professor had a typographical error in the derivation; it should be $q_1$, not $q_i$.