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Given a continuous exponential growth model, say $P\left(t\right) =P_{0}e^{rt}$, is it immediate that we have an exponential distribution? An example I have in mind is the division of cells: Assuming continuous division with exponential growth, do we automatically know that the probability of division between $t=a$ and $t=b$ is $e^{-ra}-e^{-rb}$ or is the exponential distribution of cell division an additional assumption one must impose?

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The division rate at time $t$ being $\lambda(t)$ means that $P(t+\mathrm dt)=P(t)+(\lambda(t)\mathrm dt)P(t)$. Hence, for every $t\geqslant0$, $P'(t)=\lambda(t)P(t)$ and $P(t)=\mathrm e^{\Lambda(t)}P_0$ with $\Lambda(t)=\int\limits_0^t\lambda(s)\mathrm ds$.

In your case, $\Lambda(t)=rt$ hence $\lambda(t)=r$ for every $t$.

In the general case, the probability that an individual cell present at time $t$ stays undivided until time $t+s$ is $\mathrm e^{-\Lambda(t+s)+\Lambda(t)}$. Thus, assuming that $\lambda(t)=r$ for every $t$, the probability that an individual cell present at time $t$ divides before time $t+s$ is $1-\mathrm e^{-sr}$.

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This assumption of exponentially distributed division times, is a consequence of ignoring the age-structure of the population. See publications of Glenn Webb for age-structured models of cell division. This models go from an ODE to a PDE, fortunately there is a characteristic line that still describes it which simplifies the solution. However the mathematics are more complicated, and do not fit in a simple e^x type model.

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