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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

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

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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

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

Tadaa!

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  • $\begingroup$ Yup! Good old sign problems. Thanks for the help. $\endgroup$ – M. Beausoleil Mar 29 at 21:35

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