Mathematics of the coalescent: multiple alleles

It's quite easy to extend this approach to multiple alleles.⁵We're interested in seeing how far back in time we have to go before all alleles are descended from a single common ancestor. We'll assume that we have $m$ alleles in our sample. The first thing we have to calculate is the probability that any two of the alleles in our sample are identical by descent from the immediately preceding generation. To make the calculation easier, we assume that the effective size of the population is large enough that the probability of two coalescent events in a single generation is vanishingly small. We already know that the probability of a coalescence in the immediately preceding generation for two randomly chosen alleles is $1/2N_e$. But there are $m(m-1)/2$ different pairs of alleles in our sample. So the probability that one pair of these alleles is involved in a coalescent event in the immediately preceding generation is

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

From this it follows⁶ that the probability that the first coalescent event involving this sample of alleles occurred $t$ generations ago is

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

So the mean time back to the first coalescent event is

$$\frac{2N_e}{m(m-1)/2} = \frac{4N_e}{m(m-1)} \hbox{ generations} \quad .$$

But this is, of course, only the first coalescent event. We were interested in how long we have to wait until all alleles are descended from a single common ancestor. Now is where Kingman's sneaky trick comes in. After the first coalescent event, we now have $m-1$ alleles in our sample, instead of $m$. So the whole process starts over again with $m-1$ alleles instead of $m$. Since the time to the first coalescence depends only on the number of alleles in the sample and not on how long the first coalescence event took, we can calculate the average time until all coalescences have happened as

$$\begin{eqnarray*} \bar t &=& \sum_{k=2}^m \bar t_k \\ &=& \sum_{k=2}^m \frac{4... ...)} \\ &=& 4N_e\left(1 - \frac{1}{m}\right) \\ &\approx& 4N_e \end{eqnarray*}$$