The wright-Fisher model of genetic drift is:

$$p_{ij} = \binom{2N}{j}\left(\frac{i}{2N}\right)^j \left(1- \frac{i}{2N}\right)^{2N-j} $$

,where $\binom{2N}{j}$ is a binomial coefficient.

From this equation one can infer that the expected heterozygostiy should decrease by $1-\frac{1}{2N}$ at each time step because:

$$E[x_{t+1}(1-x_{t+1}) \space|\space x_t] = (1-\frac{1}{2N})x_t(1-x_t)$$

I don't understand this equality. That might be very simple though! Can you help me making sense of it?



2 Answers 2


To derive it, first use that $E[x(1-x)]= E[x-x^2]=E[x]-E[x^2]$ and that $E[x^2]=\text{Var}[x]+E[x]^2$ to rewrite the left-hand side: $$E\left[x_{t+1}(1-x_{t+1})\right] = E\left[x_{t+1}\right](1-E\left[x_{t+1}\right])-\text{Var}\left[x_{t+1}\right].$$ The equation for $p_{ij}$ is just saying that $2Nx_{t+1}$ is binomially distributed with $2N$ trials with success probability $x_t$, so $E[x_{t+1}]=x_t$ and $\text{Var}[x_{t+1}]=x_t(1-x_t)/(2N)$. (Note that $\text{Var}[\alpha x]=\alpha^2\text{Var}[x]$.) Plugging these values into the equation above gives the form you're looking for.

  • $\begingroup$ I feel close to the understanding! Can you help me understanding why $$E\left[x_{t+1}(1-x_{t+1})\right] = E\left[x_{t+1}\right](1-E\left[x_{t+1}\right])-\text{Var}\left[x_{t+1}\right].$$ $\endgroup$
    – Remi.b
    Apr 12, 2014 at 10:45
  • $\begingroup$ @Remi.b: If Var(x)= E(x(x-1)+E(x)-E(x)^2 = E(x(x-1)) + E(x)(1-E(x)) then -E(x(x-1)) = E(x(1-x))= E(x)(1-E(x)) - Var(x). See, e.g., math.ubc.ca/~feldman/m302/binomial.pdf – daniel 11 mins ago $\endgroup$
    – daniel
    Apr 12, 2014 at 12:58
  • $\begingroup$ @Remi.b It should be: http://www.math.ubc.ca/~feldman/m302/binomial.pdf $\endgroup$
    – Chris
    Apr 14, 2014 at 7:06
  • $\begingroup$ You are welcome. Sometimes you don't see a tree in the wood :-) $\endgroup$
    – Chris
    Apr 14, 2014 at 8:02
  • $\begingroup$ My bad. It was just a cite for the variance. Sorry. $\endgroup$
    – daniel
    Apr 15, 2014 at 3:08

The notation at this site resembles that in your question but preserves the $\frac{x_t}{2N}$ notation for probability of selecting an allele.

$$E[\frac{x_{t+1}}{2N})(1 - \frac{x_{t+1}}{2N} )|x_t] = (\frac{x_{t}}{2N})(1 - \frac{x_{t}}{2N}) (1 - \frac{1}{2N}) $$

The expression $(\frac{x_{t}}{2N})(1 - \frac{x_{t}}{2N}) $ is the probability of heterozygosity at step $t.$

Your expression says that the expected heterozygosity at step $(t+1)$ is equal to the heterozygosity at step at $t$ multiplied by $(1-\frac{1}{2N}).$

For a population of size $2N,~(1-\frac{1}{2N})^{2N}\approx 1/e,$ so large populations are insulated from this effect and small ones may lose genetic diversity rapidly.

A typical explanation can be found here. The notation at this site condenses the expression $(\frac{x_{t}}{2N})(1 - \frac{x_{t}}{2N}) $ to $H_n$ and explicitly says it is $1 - F_n, $ in which $F_n$ is the probability of homozygous at step $n.$


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