Let's consider that some particles get created at rate $\beta$ following a Poisson distribution over a period $\tau$ i.e. if X is the number of created particles, then $X \sim Poisson(\beta)$.
A particle gets transformed from state A to state B at rate $\alpha$. i.e. transforming $\sim Exp(\alpha)$ and that the number of transformed particles follows a poisson process.
We are only looking at these particles during this lifetime $\tau$.
I want to determine the number of particles that are still in a state A at time $\tau$.
What I only got to determine is that $\beta \, \tau$ particles are created and the probability of a particle reaching age $\tau$ without transforming is $1 - e^{- \alpha \tau}$ but the probability of reaching age $\tau$ is only valid if the particle got created at time 0, right ?
I know that if $X \sim Poisson(Y)$ and $Y \sim Exp(\alpha)$, then, $X \sim Geom(\frac{1}{1+\alpha})$. How can I use this in this situation ?
How can I interpret the Geometric distribution ? What are the failures/successes ?
Also I imagine it's not possible to have more transformed particles than created ones, how do we account for that ?
Thank you very much for your help !
Let $\Big\{[t_{i-1},t_i),s_i\Big\}_{i=1}^n$ be a uniform tagged partition of $[0,\tau)$.
If $X_i$ is the number of individuals born in $[t_{i-1},t_i)$ that remain in state $A$ at time $\tau$, then $X_i$ is approximately $\text{Poisson}\left(\frac{\beta(t_i-t_{i-1})}{\tau}e^{-\alpha(\tau-s_i)}\right)$ when $n$ is large.
If we put $X=\sum_{i=1}^{n}X_i$ then $X$ is the total number of individuals born in $[0,\tau)$ that remain in state $A$ at time $\tau$ which, from independence, is approximately $\text{Poisson}\left(\sum_{i=1}^{n}\frac{\beta(t_i-t_{i-1})}{\tau}e^{-\alpha(\tau-s_i)}\right)$.
Taking $n$ to infinity tells us the distribution you seek is $\text{Poisson}\left(\int_0^\tau \frac{\beta}{\tau}e^{-\alpha(\tau-x)}\mathrm{d}x\right)$