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When recording change in allele frequency in diploid, bi-allelic, infinite and panmixic population we usually use this kind of equation:

$\delta_p = \frac{p * q *( p (w11 - w12) + q * (w12 - w22))}{\bar{w}}$

$\bar{w} = p^2 * w11 + 2*p*q*w12+q^2*w22$

$\delta_p$ = change of $p$ (frequency of one of the allele) from one time step to another

$w11$ is the mean fitness of individuals of genotype 11. $p$ and $q$ are the allele frequencies.

The only indicators for the fitness distribution is the arithmetic mean. Why don't we include other indicator of the probability distribution of fitness? The skew, the sd, the median for examples. Could you argue why we don't need to care about the probability distribution of fitness of individuals with genotypes 11 (for example)? In other words, why is the mean fitness (=w11) a sufficient statistics?

I wouldn't be able to answer if one asks me:

  1. Why don't you take the median instead of the arithmetic mean?"

  2. Why don't you care about the variance, the skew (or any other moment) of your distribution?

  3. What if the traits were not countinuous but discrete (sex is a discrete trait for example)?

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Could you point us to where you have seen the statement "we don't care about the prob... the mean fitness is a sufficient statistic" –  GriffinEvo Jul 31 '13 at 14:57
    
@GriffinEvo I am stating that we don't care about the probability distribution, distribution skew, etc... as only the mean fitness per genotype and overall mean fitness (pondered by their frequency) appear in the equation. –  Remi.b Jul 31 '13 at 15:13
    
Thanks kmm. I didn't Mathjax. Can we make the overline over w for mean.w ? –  Remi.b Jul 31 '13 at 15:16
    
@GriffinEvo To add to my last comment. Does it make sense? Do you know any mathematical formulation that take into account other indications (sd, normal/poisson, skew, etc...) of the probability distribution of fitness when describing a change in allele frequency over time due to natural selection? –  Remi.b Jul 31 '13 at 15:19
    
I doubt it is going to be the case where evaluating the mean is going to be adequate for describing the population fitness. Its just a common compromise in experimental and mathematical fitness... this would presuppose that the distribution of fitness is gaussian, extrapolating to a large population. even then you'd need to have an estimate of standard deviation. –  shigeta Jul 31 '13 at 17:13

1 Answer 1

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It's simplest to think about a haploid population with size $N$. Say there are two alleles, $A$ and $a$, with $A$ having frequency $p$ in this generation. Each $A$ individual will have some realized fitness (i.e., number of offspring) in this generation; let's call this number $w_A^{(i)}$ for the $i^\text{th}$ individual. The total number of $A$ individuals in the next generation is then $\sum_{i=1}^{Np}w_A^{(i)}=Npw_A$, where $w_A$ is the arithmetic mean of the fitness distribution, $w_A\equiv\sum_{i=1}^{Np}w_A^{(i)}/(Np)$. So regardless of how weird the distribution of the $w_A^{(i)}$ was, all that matters (in the large-$N$ limit) is the arithmetic mean.

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