1
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – user296602 Feb 14 '16 at 18:40
  • $\begingroup$ 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. $\endgroup$ – JKnecht Feb 14 '16 at 18:49
0
$\begingroup$

I have got a solution now for the part that was hardest to understand.

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 $q_i$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.