Under scenarios of stabilizing or disruptive selection, we can add a quadratic component to our model of phenotype and fitness like so.

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Specifically, I am not clear on where the 1/2 comes from nor do I understand the derivation of gamma, which is equal to as follows:

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x is the phenotype of interest, w(x) is the fitness of said phenotypic value. uBS is the mean phenotype before selection. Any help would be greatly appreciated.

  • $\begingroup$ Hi, do you have a link to where those equations are from? Thanks. $\endgroup$
    – user438383
    May 17, 2021 at 8:57
  • $\begingroup$ Those two equations come from Lande and Arnold 1983 per the comment below this one :) Sorry! $\endgroup$ May 17, 2021 at 16:51

1 Answer 1


Coefficient $1/2$ is a matter of definition/convenience. One could have written $\beta x_i + \gamma x_i^2+\bar{w}$, but then a factor of $2$ would surface in the other forumals.

The other two equations are adapted to one-dimensional case (one trait) from the Lande and Arnold's paper The measurement fo selection on correlated characters. In particular, the equation with covariance is their equation (13a), which follows from their more general equation (4) for quadratic selection. The last equation is the restatement of their (14b).

Here I retake the notation of the above-mentioned article by Lande, but adopt it to the case of a single trait. Then the variance matrix $P$ is just a number (variance). Equations (14a) and (13b) define $$\gamma=\frac{1}{P^2}C, (A)\\ C=Cov\left[w, (z-\bar{z})^2\right], $$ which is the definition of $\gamma$ in the OP. It is essential here to use the correct expression for the covariance: $$ Cov\left[f(z), g(z)\right] = \int \left[f(z)-\bar{f}\right]\left[g(z)-\bar{g}\right]p(z)dz= \int f(z)g(z)p(z)dz -\bar{f}\bar{g} $$ In our case we thus have: $$ C = \int w(z)(z-\bar{z})^2p(z)dz - \bar{w}\overline{(z-\bar{z})^2}, $$ where $\bar{w}=1$ by definition, whereas $\overline{(z-\bar{z})^2}=P$ is the variance of the trait. We thus have $$ C = \int w(z)(z-\bar{z})^2p(z)dz - P, (B) $$ where $$ p(z)=\frac{1}{\sqrt{2\pi P}}e^{-\frac{(z-\bar{z})^2}{2P}}. $$

Let us now consider the integral $$ \int \frac{\partial^2w(z)}{\partial z^2}p(z)dz= \frac{\partial w(z)}{\partial z}p(z)|_{-\infty}^{+\infty} - \int \frac{\partial w(z)}{\partial z }\frac{\partial p(z)}{\partial z }dz=\\ \frac{\partial w(z)}{\partial z}p(z)|_{-\infty}^{+\infty} - w(z)\frac{\partial p(z)}{\partial z }|_{-\infty}^{+\infty}+ \int w(z)\frac{\partial^2 p(z)}{\partial z^2 }dz $$ The first two terms vanish, since $w(z)$ is bounded, whereas $p(z)\rightarrow 0$ as $z\rightarrow \pm \infty$. We thus have $$ \int \frac{\partial^2w(z)}{\partial z^2}p(z)dz= \int w(z)\frac{\partial^2 p(z)}{\partial z^2 }dz= \int w(z)\frac{\partial^2 }{\partial z^2 }\frac{1}{\sqrt{2\pi P}}e^{-\frac{(z-\bar{z})^2}{2P}}dz=\\ \int w(z)\left[\frac{(z-\bar{z})^2}{P^2}-\frac{1}{P}\right]\frac{1}{\sqrt{2\pi P}}e^{-\frac{(z-\bar{z})^2}{2P}}dz= \int w(z)\left[\frac{(z-\bar{z})^2}{P^2}-\frac{1}{P}\right]p(z)dz=\\ \frac{1}{P^2}\int w(z)(z-\bar{z})^2p(z)dz - \frac{\bar{w}}{P}=\\ \frac{1}{P^2}\left[\int w(z)(z-\bar{z})^2p(z)dz - P\right]=\frac{C}{P^2}=\gamma, $$ where we used $\bar{w}=1$, and equations (A) and (B) above.

  • $\begingroup$ The notation in the Lande and Arnold paper is incredibly difficult for me. If you look at 14b, could you help me understand the math that describes why the first part of 14b is equal-to the second part of 14b. I run a population genetics club and this will be incredibly helpful for us. $\endgroup$ May 17, 2021 at 17:30
  • $\begingroup$ Essentially the same question as above in the sense that am confused about making the connection between the two equations for gamma above in this thread. $\endgroup$ May 17, 2021 at 18:46
  • 1
    $\begingroup$ @user6817734 I have added the details of calculations. I hope it helps. $\endgroup$
    – Roger V.
    May 18, 2021 at 8:06
  • $\begingroup$ This is absolutely incredible. Thank you so very much. $\endgroup$ May 18, 2021 at 19:19

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