Mathematics of the coalescent: two alleles

J. F. C. Kingman developed a convenient and powerful way to describe how the time to common ancestry is related to effective population size [2,3]. The process he describes is referred to as the coalescent, because it is based on describing the probability of coalescent events, i.e., those points in the genealogy of a sample of alleles where two alleles are descended from the same allele in the immediately preceding generation.¹ Let's consider a simple case, one that we've already seen, first, i.e., two alleles drawn at random from a single populations.

The probability that two alleles drawn at random from a population are copies of the same allele in the preceding generation is also the probability that two alleles drawn at random from that population are identical by descent with respect to the immediately preceding generation. We know what that probability is,² namely

$$\frac{1}{2N_e^{(f)}} \quad .$$

I'll just use $N_e$ from here on out, but keep in mind that the appropriate population size for use with the coalescent is the inbreeding effective size. Of course, this means that the probability that two alleles drawn at random from a population are not copies of the same allele in the preceding generation is

$$1 - \frac{1}{2N_e} \quad .$$

We'd like to calculate the probability that a coalescent event happened at a particular time $t$, in order to figure out how far back in the ancestry of these two alleles we have to go before they have a common ancestor. How do we do that?

Well, in order for a coalescent event to occur at time $t$, the two alleles must have not have coalesced in the generations preceding that.³ The probability that they did not coalesce in the first $t-1$ generations is simply

$$\left(1 - \frac{1}{2N_e}\right)^{t-1} \quad .$$

Then after having remained distinct for $t-1$ generations, they have to coalesce in generation $t$, which they do with probability $1/2N_e$. So the probability that two alleles chosen at random coalesced $t$ generations ago is

$$P(T=t) = \left(1 - \frac{1}{2N_e}\right)^{t-1}\left(\frac{1}{2N_e}\right) \quad .$$ (1)

It's not too hard to show, once we know the probability distribution in equation (1), that the average time to coalescence for two randomly chosen alleles is $2N_e$.⁴