Probability – How to Calculate Variance in a Poisson Process with Exponential Lifetimes

Question

I am trying to understand how to combine the concepts of Poisson process and the birth and death process.

I have a Poisson process where people arrive with rate $\lambda$ — so when an event occurs, a new person arrives. Suppose the lifetime of each person is independent Exponential $($$\mu$$)$.

Further, let $M$$($$t$$)$ be the number of people at time $t$. I take that $M(t)$ $=$ $10$. So, at time $t$, there are $10$ people.

Now, I would like to calculate $Var[$$M$$(t + s)$ $|$ $M(t)$ $=$ $10$$]$, where $0$ $<$ $t$ $<$ $s$.

I have this question because I have been studying about independent increments in Poisson processes. Here, we have $2$ disjoint intervals: $[$$0$, $t$$)$ and $[$$t$, $s$$)$. If this was just a standard Poisson process, we can say that the number of events until time $t$ is independent of the number of events in the interval $[$$t$, $t+s$$)$.

But the case at hand is a bit different, because the lifetimes of the events/people are $iid$, and distributed as $L$ ~ Exponential $($$\mu$$)$. This means each event/person lives for an exponential $($$\mu$$)$ amount of time.

Given this, how can I proceed here? I tried working out a few things but I am stuck while trying to calculate conditional variance, and any advice on the direction to follow would be very helpful. Thank you so much.

Answer 1

There are two parts to the calculation:

• The ten alive at time $t$: each has an independent probability of being alive at time $t+s$ of $e^{-\mu s}$ and so their binomial contribution to the expectation is $10 e^{-\mu s}$ and to the variance is $10 e^{-\mu s}(1-e^{-\mu s})$
• Any individuals who might arrive at time $a$ with $t < a \le t+s$ (at a rate $\lambda$) and still be alive at time $t+s$ with probability $e^{-\mu(t+s-a)}$. In effect we have a non-homogeneous Poisson process with parameter varying over time $\lambda e^{-\mu(t+s-a)}$ amounting to a combined Poisson distribution with parameter $\int_t^{t+s} \lambda e^{-\mu(t+s-a)}\, da = \frac{\lambda}{\mu}\left(1-e^{-\mu s}\right)$ and this is their contribution to both the mean and the variance.

Putting these together:

• $\mathbb E[T(t+s) \mid T(t)=10] = 10 e^{-\mu s} + \frac{\lambda}{\mu}\left(1-e^{-\mu s}\right)$
• $\mathrm{Var}(T(t+s) \mid T(t)=10) = 10 e^{-\mu s}\left(1-e^{-\mu s}\right) + \frac{\lambda}{\mu}\left(1-e^{-\mu s}\right) = \left(10 e^{-\mu s}+\frac{\lambda}{\mu} \right) \left(1-e^{-\mu s}\right)$