I think I'm missing a fundamental step in regards to how to combine two exponential distributions in the context of this problem.

If we have a birth and death process where birth rate ~ $exp(\lambda_n) $ and the death rate ~ $exp( \mu_n )$, and state can go from n to $n-1$ or to $n+1$, why is it that $v_i=\lambda_i+\mu_i$?

Note that the transition probability(time spent in state i before transition) is $T_i$ ~ $exp(v_i) $ and this is where $v_i$ is used

  • $\begingroup$ Is this from a course or textbook? If so, please add the [self-study] tag & read its wiki. $\endgroup$ – gung Apr 4 '15 at 22:22
  • $\begingroup$ It is but I only use it as an example to understand a particular step of the provided solution. Should I still mark it as such $\endgroup$ – Borat.sagdiyev Apr 4 '15 at 22:29
  • $\begingroup$ Probably, & read the wiki. We will give you hints & help you think through the problem for yourself, but probably not just give you a full answer. $\endgroup$ – gung Apr 4 '15 at 22:31
  • $\begingroup$ it is a very short question. Presumably answerable in 1 sentence. It's a property of exponential variables that I'm just not seeing. Please read what I wrote above about the solution being provided and myself just inquiring about one of the steps $\endgroup$ – Borat.sagdiyev Apr 4 '15 at 22:38
  • $\begingroup$ I've edited the question to make it clear that what I'm asking about is a fundamental property of birth and death processes and not a textbook problem $\endgroup$ – Borat.sagdiyev Apr 4 '15 at 22:48

I found the property I was looking for. IT applies to this problem because in order to change state, either a birth or a death must occur, whichever takes less time.

If $X_1 ... X_n$ are i.i.d exponential R.V's with parameters $\lambda_1...\lambda_n$ then,

$min_i$ $X_i$ is an exponential R.V. with parameter $\lambda_1+...+\lambda_n$


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.