# Probability of exponential growth event

Under the assumption of exponential growth of a population of cells, the population size at time $t$, $N(t)$, is:

$$N(t) = N_0\exp(rt)$$

where $r$ is the rate of division and $t$ is time.

What is the proper way to derive the probability that a cell divides in the time interval $[a, b]$ given a rate $r$?

Intuitively the probability of division in $[a, b]$ should be proportional to the duration of the interval, $\delta t$, and the rate of division $r$, so the probability is:

$$P(\text{division in} [a, b]) = \exp^{-1/r\delta t}$$

Is this right?

• Conditionally on $N_a=n$, the probability that there is no division during $(a,b)$ is $e^{-r(b-a)n}$. Assuming that $N_0=1$, $N_a$ is geometrically distributed with $P(N_a=n)=e^{-ra}(1-e^{-ra})^{n-1}$ for every $n\geqslant1$. Summing on $n$, one gets that the probability of no division during $(a,b)$ is $\sum\limits_{n\geqslant1}e^{-r(b-a)n}e^{-ra}(1-e^{-ra})^{n-1}=\frac1{e^{rb}-e^{ra}+1}$. – Did Apr 7 '16 at 19:45
• Note that the point process of the times when a division occurs is not Poisson since the instantaneous rate at time $t$ is $rN_t$, which depends on the number of events $N_t$ during $(0,t)$. – Did Apr 7 '16 at 19:49
• Surely this is me but I fail to see how the accepted answer addresses the question. – Did Apr 11 '16 at 16:15

If $\lambda$ is the expected number of events in an interval then using a Poisson distribution, the probability of no events in that interval is $$\frac{\lambda^0e^{-\lambda}}{0!}=e^{-\lambda}$$ Thus, the probability that at least one event occurs would be $$1-e^{-\lambda}$$ If you are actually interested in the probability that exactly one event occurs then that is $$\frac{\lambda^1e^{-\lambda}}{1!}=\lambda e^{-\lambda}$$
The expected number of events in that time would be $\lambda=N_0\left(e^{br}-e^{ar}\right)$.
• There is a rate given, but no information about any distribution. It might be that the formula $N=\left\lfloor N_0e^{rt}\right\rfloor$ is followed precisely. In that case, $$t_k=\frac1r\log\left(1+\frac k{N_0}\right)$$ is the time when the $k^{\text{th}}$ event happens. Thus, we just need to look to see if any of the $t_k$'s lie in $[a,b]$. There is no probability involved. However, assuming a Poisson distribution seems reasonable. – robjohn Apr 7 '16 at 16:30