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?
[Math] Exponential growth and exponential distribution
probability
Related Solutions
if you are using $e^{rt}$, then since doubling time is (in your chosen unit of time) equal to $1$, we get $e^{(r)(1)}=2$. Taking logarithms to the base $e$, we get $r=\ln 2$.
It simplifies things to use the doubling time as the unit of time. However, in a more complicated situation, where we have say two kinds of cell, what unit shall we use? In the long run, the conventional units of time are more useful.
So the population at time $x$ is $e^{(\ln 2)(x)}$. Since $e^{\ln 2}=2$, this is just a fancy way of writing $2^x$.
Both models can be considered as continuous models, since $x$ is not restricted to be an integer. They are in fact the same model. And yes, we do get results that are not integers, but that is not important. No mathematical model will represent a complex biological phenomenon exactly. If the numbers involved are large, an "error" of half a cell has no practical significance. Indeed, the error is likely to be far larger than that.
The function $e^t$ has nice technical properties. For example, its derivative is $e^t$. the derivative of $2^t$ is the more messy $(\ln 2)2^t$. But anything one can do by using base $e$ can also be done using base $2$, or $10$. There will be small differences of detail, that's all.
I did a monte carlo simulation to model the problem because I didn't know how to do it analytically. Graphs at the bottom.
Python code: https://github.com/jschaf/cellarch
The simulation works as follows:
Partition NUM_MACHINES into NUM_PARTITIONS separate groups. This simulation uses partition to refer to separate groups instead of
cellused by the math StackExchange question.Uniformly distribute NUM_DATA pieces of data to all subsets of machines such that:
- Every subset of machines reside in the same partition.
- Every subset has exactly a size of NUM_REPLICAS.
Generate machine failure start times pulling samples from an exponential distribution.
Get the cumulative sum of the failure times to generate subsequent failure times for a machine. Meaning, turn [1, 3, 2, 7] into [1, 4, 6, 13].
Create an outage for a machine by adding the time to repair to the failure start time. The time to repair is drawn from a normal distribution. Meaning, assuming we draw time
Rfrom the normal distribution turn[1, 4, 6, 13]into[(1, 1 + R1), (4, 4 + R2), (6, 6 + R3), (13, 13 + R4)]Find all outages where N machines are down at the same time. This is an outage clique. When N == NUM_REPLICAS, this means we might have an outage for some subset of data.
Find all outage cliques where each machine in the clique hosts the same piece of data. The found cliques mean some data is completely unavailable.
Run the above steps many times to get the time to first data unavailability.
Best Answer
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 $1-\mathrm e^{-sr}$.