How to calculate variance in a Poisson process with exponential lifetimes of arrivals?

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:

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

Answer 2

I will give a rough sketch. First, let $(X_i)_{i\in\mathbb{N}}$ be a sequence of independent exponentially distributed random variables with parameter $\mu$. Let $(Y_i)_{i\in\mathbb{N}}$ be independent uniform random variables on $[0,s-t]$, and $N$ be Poisson distributed with parameter $\lambda (s-t)$. Everything shall be independent here. I can write $M(s-t) = \sum_{i=1}^N \mathbb{1}_{\{X_i > s-t-Y_i\}}$ in distribution. The argument is then that $M(s)$ given $M(t)=10$ is the sum of those 10 who have been alive at time $t$ and are still at time $s$ plus the sum of those who have been born after $t$ and are still alive at time $s$. Then we can use the independence and memorylessness of the exponential distribution and drop the condition in the variance. In particular, $$ \mathsf{Var}[M(s)\mid M(t) = 10] = \mathsf{Var}\bigg[\sum_{i=1}^N \mathbb{1}_{\{X_i > s-t-Y_i\}}\bigg] + \mathsf{Var}\bigg[\sum_{i=1}^{10} \mathbb{1}_{\{X_i > s-t\}}\bigg] \\ = \sum_{j=0}^\infty \sum_{i=1}^j \mathsf{Var}[\mathbb{1}_{\{N=j\}} \mathbb{1}_{\{X_i > s-t-Y_i\}}] + 10 \mathsf{Var}[ \mathbb{1}_{\{X_1 > s-t\}}] \\ =\sum_{j=0}^\infty j \big(\mathbb{P}(N=j)\mathbb{P}(X_1 > s-t-Y_1) - (\mathbb{P}(N=j)\mathbb{P}(X_1 > s-t-Y_1))^2\big) + 10(\mathbb{P}(X_1 > s-t) - \mathbb{P}(X_1 > s-t)^2). $$ Can you take it from here?