Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $X(t)$ be a pure death process starting from $X(0)=N$. Assume that the death parameters are $\mu_1, \mu_2,\dots,\mu_N$. Let $T$ be an independent exponentially distributed random variable with parameter $\theta$. Show that $Pr\{X(T)=0\} = \prod_{i=1}^{N} \frac{\mu_i}{\mu_i+\theta}$.

My thoughts are that I should condition on $T=t$ and integrate so we would have

$$Pr\{X(T)=0\} = \int_0^\infty Pr\{X(t) = 0 | T = t\} Pr\{T=t\}dt$$

In the book we are given the formula for $P_n(t) = Pr\{X(t) = n\}$ but it is messy, here it is, $$P_n(t) = \mu_{n+1}\dots\mu_{N} [ A_{n,n} e^{-\mu_n t} + \dots + A_{N,n}e^{-\mu_N t} ]$$ Where, $$A_{k,n} = \frac{1}{(\mu_N - \mu_k) \dots (\mu_{k+1} - \mu_k)(\mu_{k-1} - \mu_k) \dots (\mu_n - \mu_k)}$$ Using my approach leads to some very ugly algebra and some terms that just don't seem to cancel. I tried doing it for the case of $N=2$ but still could not get it in the right form. I feel like I might be approaching this problem wrong but I don't see any other way to go about it.

share|cite|improve this question
Did you complete the indications of my answer? – Did Apr 7 '11 at 7:35
Dear user8043, I am curious: was my answer sufficient to make you solve the question or are you still lost? – Did Apr 25 '11 at 21:08

Going back to a fixed time $t$ only complicates things, on the contrary you should use the randomness of $T$. Here is a hint: $[X(T)=0]=[T\ge S]$ where $S$ is a random time, independent on $T$. It happens that $S=S_1+\cdots+S_N$ where the $S_i$ are independent, not equidistributed, but with explicit simple distributions that you should be able to determine. Now, $P(T\ge t)=\exp(-\theta t)$ for every $t\ge0$, hence $$ P(T\ge S)=E(\exp(-\theta S))=E(\exp(-\theta S_1))E(\exp(-\theta S_2))\ldots E(\exp(-\theta S_N)), $$ and if you know the distribution of each $S_i$, you should be able to compute each $E(\exp(-\theta S_i))$ and to deduce $P(X(T)=0)=P(T\ge S)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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