From this textbook

Under the wright-Fisher model of genetic drift and under the assumption that all alleles are neutral, the probability that $k$ alleles had $k$ distinct parent alleles the previous generation is

$$Pr(k) = \prod_{i=1}^{k-1}1-\frac{i}{2N} ≈ 1-\frac{{k \choose 2}}{2N} $$

The chance of two alleles not coalescing for $t$ generations is $\left(1-\frac{1}{2N}\right)^t$ , and the chance that they coalesce in the next generation is $\frac{1}{2N}$. Therefore, the probability that 2 alleles had a common ancestor t+1 generations ago is

$$\frac{1}{2N}\left( 1-\frac{1}{2N} \right)^t ≈ \frac{1}{2N}e^{\frac{-t}{2N}}$$

I understand up to this point!

The probability that the $k$ alleles do not coalesce for $t$ generations, and then one pair coalesce to give $k-1$ alleles at $t+1$ generations ago is as follows:

$$Pr(k)^t \left[ 1-Pr(k) \right] ≈ \frac{{k \choose 2}}{2N}exp\left[ -\frac{{k \choose 2}}{2N}t \right] $$

Can you help me to understand this last part? (both the left and right part of the equation)


1 Answer 1


Is this the exact text from the book? The left side seems to represent the probability for

"No coalescence in $k$ lines in $t$ generations (i.e. the $Pr(k)^t$ term), and at least one coalescence among those lines in generation $t+1$ (the $1-Pr(k)$ term)"

which is the same event as

"First coalescence event in $k$ lines is exactly in generation $t+1$".

The right hand side is derived analogous to the second equation, with $1-\frac{1}{2N}$ ("no coalescence in two lines") replaced by $1-\frac{k\choose2}{2N}$ (approximation for "no coalescence in $k$ lines):

$$ \begin{align} Pr(k)^t \left[ 1-Pr(k) \right] &≈ \left( 1-\frac{k\choose2}{2N} \right)^t \left [1 - 1 + \frac{k\choose2}{2N} \right] \\ &\approx \exp\left( - \frac{k\choose2}{2N} t \right) \frac{k\choose2}{2N} \end{align} $$

were is first approximation is due to your first calculation. The second one seems to use a first order Tailor approximation of $\exp(x)$:

$$ \exp(-xt) = exp(-x)^t \approx (1-x)^t $$

  • $\begingroup$ +1 Ok, I got the idea. No, that is not exactly the same text than in the book. I summarized it. Thanks a lot for you answer. The answer would be perfect if you could just show that $$\frac{1}{2N}\left( \prod_{i=1}^{k-1}1-\frac{i}{2N} \right)^t ≈ \frac{{k \choose 2}}{2N}exp\left[ -\frac{{k \choose 2}}{2N}t \right]$$ $\endgroup$
    – Remi.b
    Commented Apr 15, 2014 at 14:37
  • $\begingroup$ I extended my answer to show the equation. I have never seen this approximation of $\exp$ before... $\endgroup$
    – Paul Staab
    Commented Apr 15, 2014 at 15:13

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