# How to derive the Equilibrium value $\hat F$ equation

In population genetics, there is a relationship between $$\hat F$$ and $$Nm$$.

$$\hat F=\frac{1}{1+4Nm}$$

In population genetics textbook, (i.e. Hartl, D. L., and A. G. Clark. 2007. Principles of population genetics. Fourth edition. Sinauer Associates Incorporated, Sunderland, Massachusetts.) it reads that using the equation below:

$$F_t= \bigg(\frac{1}{2N}\bigg)(1-m)^2+\bigg(1-\frac{1}{2N}\bigg)(1-m)^2F_{t-1}$$

Set $$\hat F = F_t = F_{t-1}$$. After expanding the squared terms on the right-hand side, and assuming that $$m$$ is small enough and $$N$$ large enough, and that terms in $$m^2$$ and $$m/N$$ can be ignored, some rearrangement leads to:

$$\hat F=\frac{1}{1+4Nm}$$

I tried redoing it algebraically, without success. So do you know how to get to "$$\hat F=\frac{1}{1+4Nm}$$"

$$(1-m)^2 = 1-2m+m^2$$

$$F_t= \bigg(\frac{1}{2N}\bigg)(1-2m+m^2)+\bigg(1-\frac{1}{2N}\bigg)(1-2m+m^2)F_{t}$$

$$F_t-\bigg((1-2m+m^2)-\frac{(1-2m+m^2)}{2N}\bigg)F_{t}= \frac{(1-2m+m^2)}{2N}$$

$$F_t(1-(1-2m+m^2)-\frac{(1-2m+m^2)}{2N})= \frac{(1-2m+m^2)}{2N}$$

$$F_t(-2m+m^2-\frac{(1-2m+m^2)}{2N})= \frac{(1-2m+m^2)}{2N}$$

$$F_t(\frac{2N(-2m+m^2)}{2N}-\frac{(1-2m+m^2)}{2N})= \frac{(1-2m+m^2)}{2N}$$

$$F_t(\frac{2N(-2m+m^2)-(1-2m+m^2)}{2N})= \frac{(1-2m+m^2)}{2N}$$

$$F_t= \frac{(1-2m+m^2)}{2N} * (\frac{2N}{2N(-2m+m^2)-(1-2m+m^2)})$$

$$F_t= \frac{1-2m+m^2}{(-4Nm+2Nm^2)-1+2m-m^2}$$

Then $$m^2= 0$$

$$F_t= \frac{1-2m}{(-4Nm)-1+2m}$$

$$F_t= \frac{1}{(-4Nm)-1+2m} -\frac{2m}{(-4Nm)-1+2m}$$

??

$$F_t= \frac{1}{(-4Nm)-1+2m}$$

Magic

$$\hat F=\frac{1}{1+4Nm}$$

Sign problem

When you go from

$$F_t(1-(1-2m+m^2)-\frac{(1-2m+m^2)}{2N})$$

to

$$F_t(-2m+m^2-\frac{(1-2m+m^2)}{2N})$$

You make a mistake in the sign. It should be

$$F_t(2m-m^2-\frac{(1-2m+m^2)}{2N})$$

Approximations

I will assume this error of sign propagate to inverse your end result. You hence should get

$$F_t= \frac{1}{4Nm+1-2m} -\frac{2m}{(-4Nm)-1+2m}$$

$$\frac{2m}{4Nm+1+2m}$$ is negligible compared to $$\frac{1}{4Nm+1+2m}$$, so you can indeed remove it. You get

$$F_t= \frac{1}{4Nm+1-2m}$$

Also, $$2m$$ is negligible compared to $$4Nm$$, so again you can remove it. You end up with

$$F_t= \frac{1}{4Nm+1}$$