Fluctuating Population Size and Genetic Drift

In general the variance from a binomial process is [math]\sigma^2 = n p (1-p)[/math], where n is the number of trials and p is the probability of one of the two outcomes.

The variance for the sampling of two alleles each generation is [math]2N p (1-p)[/math], where N is the diploid population size and p is the allele frequency.

We like to work with allele frequencies rather than allele counts in a population. The variation is scaled by 2N so the underlying standard deviation is a fraction between zero and one.

[math]\frac{\sigma}{2N} = \frac{\sqrt{2N p (1-p)}}{2N}[/math]

[math]\sigma^2 = \left(\frac{\sqrt{2N p (1-p)}}{2N}\right)^2 = \frac{2N p (1-p)}{4N^2} = \frac{p (1-p)}{2N}[/math]

Imagine the population fluctuating between two sizes, N₁ and N₂, so that it spends half of its time at N₁ and half at N₂. There is an effective population size, N_e, of constant size that has the same average variance in allele frequency change each generation as the fluctuating population.

[math]\frac{p(1-p)}{2N_e}=\frac{1}{2}\times\frac{p(1-p)}{2N_1}+\frac{1}{2}\times\frac{p(1-p)}{2N_2}[/math]

We can cancel out a lot of components to get

[math]\frac{1}{N_e}=\frac{1}{2}\times\frac{1}{N_1}+\frac{1}{2}\times\frac{1}{N_2}[/math]

and solve for N_e.

[math]N_e=\frac{1}{\frac{1}{2}\times\frac{1}{N_1}+\frac{1}{2}\times\frac{1}{N_2}}[/math]

This is a harmonic mean, the inverse of the average of the inverses of the individual population sizes.

This simple example illustrated two population sizes with half of the time spent at each. More generally

[math]N_e=\frac{1}{\sum\limits_{i}{\frac{t_i}{N_i}}}[/math]

to be continued ...