Revision as of 04:33, 5 February 2016

Coalescence of Two Lineages

The PDF of an exponential distribution of the coalescence of two lineages in a diploid population of size N.

[math]P(\text{coalescence at time }g)=\frac{1}{2N}e^{-g/2N}[/math]

For example, the probability of two lineages coalescing in a small population of 20 individuals in exactly the ninth generations is 2%.

Integrate to get the CDF.

[math]F(\text{coalescence at time }g)=\int_0^g\frac{1}{2N}e^{-g/2N}[/math]

[math]F(\text{coalescence at time }g)=\frac{1}{2N}\int_0^g e^{-g\frac{1}{2N}}[/math]

[math]F(\text{coalescence at time }g)=\frac{1}{2N} \frac{-e^{-g\frac{1}{2N}}}{\frac{1}{2N}} + C[/math]

[math]F(\text{coalescence at time }g)=-e^{-g/2N} + C[/math]

Because the CDF must [math]\lim_{g \to \infty}\left( -e^{-g/2N} + C \right)= 1[/math] and [math]\lim_{g \to \infty} -e^{-g/2N}= 0[/math] then [math]C = 1[/math].

[math]F(\text{coalescence at time }g)=-e^{-g/2N} + 1[/math]

[math]F(\text{coalescence at time }g)=1-e^{-g/2N}[/math]

For example, there is a 95% probability that two lineages will coalesce within 6N generations.

[math]F(\text{coalescence at time }g)=0.95=1-e^{-g/2N}[/math]

[math]1-0.95=e^{-g/2N}[/math]

[math]\log_e 0.05=-\frac{g}{2N}[/math]

[math]-2N\log_e 0.05=g[/math]

[math]-2N\times-3=g[/math]

[math]6N=g[/math]