# Biology: Wright-Fisher model of genetic drift

In evolutionary biology (in population genetics to be more accurate) exists the concept of genetic drift. It describes how an allele (gene variant) (that has no advantage or disadvantage in terms of reproductive succes) vary through time. Below is a classical model to describe this process of genetic drift. This model is called the Wright-Fisher model of genetic drift:

$$\frac{(2N)!}{k!(2N-k)!}p^kq^{2N-k} \Leftrightarrow \binom{2N}{k}p^kq^{2N-k}$$

where $\binom{2N}{k}$ is the binomial coefficient.

This formula gives the probability of obtaining $k$ copies of an allele at generation $t+1$ given that there is a frequency of $p$ of this allele in the population at generation $t$. $N$ is the population size and $2N$ is the number of copies of each gene (this model applies to diploid population only).

My questions are:

• 1) From this formula, how can we calculate the probability of extinction of an allele in say 120 generations starting at a given frequency, let's say 0.2?

and

• 2) How can we calculate the probability of extinction rather than fixation of an allele starting at frequency $p$ if we wait an infinite amount of time?

In 1969, Kimura and Ohta showed that assuming an initial frequency of $p$, the mean time to fixation $\bar t_1(p)$ is:

$$\bar t_1(p)=-4N\left(\frac{1-p}{p}\right)\ln(1-p)$$

similarly they showed that the mean time to loss $\bar t_0(p)$ is

$$\bar t_0(p)=-4N\left(\frac{p}{1-p}\right)\ln(p)$$

Combining the two, they found that the mean persistence time of an allele $\bar t(p)$ is given by $\bar t(p) = (1-p)\bar t_0(p) + p\bar t_1(p)$ which equals

$$\bar t(p)=-4N\cdot \left((1-p)\cdot \ln(1-p)+p\cdot \ln(p)\right)$$

It does not answer my questions though!

I asked this same question on Biology.SE (here) a month ago but did not get an answer yet. I hope mathematician will be able to help me with that. On this biology.SE post, @GriffinEvo found via simulations that the probability of extinction of an allele starting at frequency $p$ if we wait an infinite amount of time is $1-p$ (which answer to question 2)). How can we mathematically demonstrate that result? And don't forget question 1) :D

(Question 1) Let $X_n$ denote the population at time $n$, hence the step $n\to n+1$ uses the frequency $p_n=X_n/(2N)$. The transition probabilities yield, for every parameter $|s|\leqslant1$, $$E[s^{X_{n+1}}]=\sum_k{2N\choose k}p_n^kq_n^{2N-k}s^k=(p_ns+q_n)^{2N}=E\left[\left(1-(1-s)p_n\right)^{2N}\right],$$ thus, $$P[X_{n}=0]=\lim_{s\to0}E[s^{X_n}]=E\left[\left(1-p_{n-1}\right)^{2N}\right].$$ It is notoriously impossible to iterate exactly these recursions. However, $1-u\leqslant\mathrm e^{-u}$ for every $u$, hence $$\left(1-(1-s)p_{n-1}\right)^{2N}\leqslant\mathrm e^{-(1-s)X_{n-1}},$$ which yields $$E[s^{X_n}]\leqslant E\left[\mathrm e^{-(1-s)X_{n-1}}\right].$$ This can be rewritten as the fact that, for every $t\geqslant0$, $$E[\mathrm e^{-tX_n}]\leqslant E[\mathrm e^{-a(t)X_{n-1}}],\qquad a(t)=1-\mathrm e^{-t}.$$ In particular, $$P[X_n=0]\leqslant E[\mathrm e^{-X_{n-1}}]\leqslant \mathrm e^{-a^{n-1}(1)X_0}.$$ To iterate approximately $a$ near $0$, note that $\frac1{a(t)}=\frac1t+\frac12+o(1)$ when $t\to0$ hence $\frac1{a^n(1)}\sim\frac{n}2$ when $n\to\infty$. Finally, $X_0=2Np$ hence $$P[X_n=0]\leqslant\mathrm e^{-4Np/n+o(1/n)}.$$
Question 2 is pretty straightforward. The Markov chain $X_n$ that tracks the frequency of the allele is a martingale, so that $\mathbb{E}(X_T)=\mathbb{E}(X_0)$ where $T$ is the fixation time. But $\mathbb{E}(X_T)=\mathbb{P}(X_T=1)$ which is the probability that this allele fixes. And $\mathbb{E}(X_0)$ is just the initial frequency $p$. Therefore, the chance of ultimate fixation is just the initial frequency.