by mikejf, Sept. 13, 2017, 9:26 p.m., 0 comments

This is another post using my notes from a Caltech "bioscience bootcamp". One of the cooler things I did yesterday was building a total-internal-reflection-fluorescence microscope and an optical tweezer, which happen to differ only by one lens! I won't go over the principles here because I unfortunately don't have the time, but I might plan a post on at least one of them tomorrow.
For this post I'm focusing on interesting ways of deriving the 1D diffusion equation and Poisson distribution from discrete models of random processes. I remember learning some of these in statistical mechanics, but this was a quick and neat refresher.
The diffusion equation comes from the "master equation" stating that the total probability mass at a certain displacement after one time stamp (\\(p(x, t + \Delta t) \\)) is equal to the intial probability mass (\\(p(x,t)\\)), plus contributions from mass at adjacent displacements and "jumped" over with rate \\(k\\) per \\(\Delta t\\) (\\(k \Delta t p(x+a,t) \\) and \\(k \Delta t p(x-a,t)\\)) minus the mass that jumps out to either direction (the 2 \\(k\Delta t p(x,t) \\) terms):
$$ p(x, t + \Delta t) = p(x,t) + k \Delta t p(x+a,t) + k \Delta t p(x-a,t) - k \Delta t p(x,t) - k \Delta t p(x,t). $$
We can subtract by \\(p(x,t)\\) and divide by \\(\Delta t\\) on both sides and multiply the right by \\(a^2 /a^2\\) (or 1) to put the equation in the following suggestive form:
$$ \frac{p(x, t + \Delta t) - p(x,t)}{\Delta t} = k a^2 \left ( \frac{ \frac{p(x+a) - p(x)}{a} - \frac{p(x) - p(x-a)}{a} }{a} \right ) .$$
These look like derivatives. The left side is fairly simple, the time derivative of \\(p\\) as \\(\Delta t \\) goes to 0. The right hand side is the difference between two similarly derivative-looking things, one at \\(x\\), one at \\(x-a\\). That is a second derivative. I'm going to put \\(ka^2 = D\\) and
$$\frac{\partial p(x,t)}{\partial t} = D \frac{\partial^2 p(x,t)}{\partial x^2}.$$
This is the 1D diffusion equation.
### The Poisson distribution from mRNA production
Here we're going to model the production and degradation of mRNA. We assume the probability that an mRNA molecule is transcribed in time \\(\Delta t\\) is \\(r\\) and the rate at which mRNA degrades in time \\(\Delta t\\) per molecule is \\(\gamma\\). The master equation describes the evolution of the probability distribution for the total number of mRNA molecules. The probability that the number is \\(x\\) at time \\( t + \Delta t\\), \\(p(x, t + \Delta t) \\), is equal to the number at the previous time step \\(p(x,t)\\), plus the case where there were \\(x-1\\) at the previous time step and a molecule was produced (\\(r p(x-1,t) \\)) and the case where there were \\(x+1\\) at the previous time step and a molecule degraded (\\(\gamma (x+1) \Delta t p(x+1,t) \\)), minus both cases where the number was \\(x\\) at the previous time step and a molecule was produced (\\(r \Delta t p(x,t)\\)) or degraded (\\(\gamma x \Delta t p(x,t) \\)):
$$p(x, t + \Delta t) = p(x,t) + r \Delta t p(x-1,t) + \gamma (x+1) \Delta t p(x+1, t) - r \Delta t p(x,t) - \gamma x \Delta t p(x,t)$$
We can subtract by \\(p(x,t) \\) and divide by \\(\Delta t\\) to get the time derivative of \\(p\\):
$$ \frac{p(x, t + \Delta t) - p(x,t)}{ \Delta t} = r p(x-1,t) + \gamma (x+1) p(x+1, t) - rp(x,t) - \gamma x p(x,t).$$
We don't want to solve the dynamics of this equation just yet, just look at where it will go in equilibrium. At equilibrium \\(\frac{\partial p}{\partial t}=0\\), and at \\(x=0\\) there is no \\(rp(x-1,t)\\). Therefore for \\(x=0\\):
$$0 = \gamma p(1) - r p(0),$$
or
$$ p(1) = \frac{r}{\gamma} p(0).$$
In general if it's true that
$$ 0 = \gamma N p(N) - r p(N-1),$$
then
$$0 = r p(N-1) + \gamma (N+1) p(N+1) - rp(N) - \gamma (N) p(N)$$
$$ = \gamma (N+1) p(N+1) - r p(N), $$
or
$$ P(N+1) = \frac{r}{\gamma (N+1)} p(N). $$
Since it is true for \\(P(0)\\) then it is true, inductively, for all \\(N\\).
This equation is satisfied if:
$$ P(N) = \left( \frac{r}{\gamma} \right )^N \frac{1}{N!} P(0).$$
To solve for \\(P(0)\\), we use the fact that the sum of the probabilities must be 1, or
$$ \sum_{N=0}^{\infty} \frac{\left ( \frac{r}{\gamma}\right )^N}{N!}P(0) = 0.$$
Using the Taylor series definition of \\(e^x\\), we know that
$$ e^{r/\gamma}P(0) = 1 \ \rightarrow \ P(0) = e^{-r/\gamma}.$$
Therefore we get that the equilibrium number of RNA is somewhere on the distribution
$$P(N) = \frac{\left( \frac{r}{\gamma} \right )^N e^{-r/\gamma}}{N!},$$
which is a Poisson distribution with \\(\lambda = r/\gamma\\), a very common and useful distribution.