How to calculate probabilities in a Poisson process with exponential lifetime of arrivals?

Question

I have a Poisson process where people arrive at the rate of $λ$ -- so when an event occurs, a new person arrives. This means that the times between successive arrivals are $T_i$ ~ Exponential $(\lambda)$.

And the lifetime of each person is independent Exponential $(μ)$. That is, $L$ ~ Exponential $(\mu)$.

Further, let $M(t)$ be the number of people at time $t$.

Suppose at time $t$, we just have $1$ person.

I am trying to calculate the probability that there will be no people remaining when this $1$ person dies.

My attempt until now is as follows:

$P($there will be no people remaining when this $1$ person dies$)$

$=$ $P($no new person arrives during the lifetime of this $1$ person$)$ $+$ $P($someone or more than one person arrives after time $t$ and dies within the lifetime of this $1$ person$)$

$=$ $P(L < T)$ $+$ $P($one or more than one person arrives after time $t$ and dies before this $1$ person dies$)$

I calculated that $P($$L < T$$)$ $=$ $\frac{\mu}{\mu + \lambda}$,

because $L$ ~ Exponential $(\mu)$, and $T$ ~ Exponential $(\lambda)$.

Now, I am trying to calculate:

$P($one or more than one person arrives after time $t$ and dies before this $1$ person dies$)$

Could someone please point me to how I can find this probability? Thank you very much!

Answer 1

The following is for the combined $P($there will be no people remaining when this $1$ person dies$)$:

Let's use $\ell$ for the lifetime of the $1$ person.

If there is another arrival at $t$ then the probability it survives to $\ell$ is $ e^{-\mu(\ell-t)}$.

Then, given $\ell$, the rate at which others arrive in $[0,\ell]$ and survive beyond $\ell$ is $\lambda e^{-\mu(\ell-t)}$

If you integrate this, you get $\int_{t=0}^\ell \lambda e^{-\mu(\ell-t)}\, dt =\frac{\lambda}{\mu}(1-e^{-\mu \ell})$

and this is the parameter of the Poisson distribution for the number of others alive at time $\ell$

so the probability there are no others alive at $\ell$ is $e^{-\frac{\lambda}{\mu}(1-e^{-\mu \ell})}$.

But $\ell$ is itself a random variable with an exponential distribution rate $\mu$

so we want $$\int_{\ell=0}^\infty e^{-\frac{\lambda}{\mu}(1-e^{-\mu \ell})} \, \mu e^{-\mu \ell} \, d\ell \\ = \frac{\mu}{\lambda}\left( 1- e^{-\lambda/\mu} \right). $$

• For $\lambda = \mu$ this gives $1-e^{-1} \approx 0.6321$, the answer to several other probability questions
• For $\lambda \gg \mu$ this gives almost $\frac{\mu}{\lambda}$ but still more than $\frac{\mu}{\lambda+\mu}$
• For $\lambda \ll \mu$ this gives a little more than $1-\frac{\lambda}{2\mu}$, also more than $\frac{\mu}{\lambda+\mu}$