# Time Until Extinction for a Pure Death Process Where the Time is Exponentially Distributed

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.

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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 $P(X(T)=0)=P(T\ge S)$.