Haploid selection recurrence ($p^{\prime }=\frac{wp}{wp+(1-p)}$)

Consider an explicitly idealized haploid population with two inherited types $A,B$, one allele per individual at the tracked genetic locus, and discrete, nonoverlapping generations. Assume faithful type transmission, no mutation or gene flow, a constant environment, and reproductive contributions independent of type frequency. Let the expected values of offspring surviving to the next census be $m_A,m_B>0$, with relative genotype fitness $w=m_A/m_B$.

For $A$ frequency $p$, expected offspring contributions are proportional to $p{m}_A$ and $(1-p)m_B$. Normalizing gives the deterministic large-population mathematical modelThe denominator is mean relative genotype fitness. For $w=1.2$ and $p_0=0.2$, $p_1=0.24/1.04=3/13\approx 0.23077$. Equivalently, 200 $A$ and 800 $B$ parents with $m_A=1.2,m_B=1$ contribute expected counts 240 and 800. Fractional per-parent expected values are averages, not literal fractional offspring. The ratio of these expectations is a deterministic approximation, not generally the expectation of a random ratio.

The odds multiply particularly simply:Thus $p_{10}\approx 0.60753$: repeated modest advantage can substantially change composition. For $w>1$ and $0<p<1$, $p^{\prime }-p=(w-1)p(1-p)/(1+(w-1)p)>0$. The deterministic boundary $p=0$ remains unchanged: natural selection needs existing variation.

A distinct stochastic extension fixes $N$ offspring and samples their types independently using a binomial distribution: $K\sim \mathrm{Binomial}(N,q)$, where $q=wp/(wp+1-p)$, then $p^{\prime }=K/N$. Its conditional expected value is $q$ and conditional variance is $q(1-q)/N$. With $w=1$, the conditional expected value remains $p$, yet realized frequencies wander through genetic drift. With $w>1$, genetic drift can still lower $p$ or lose $A$ entirely. Over several generations, the deterministic curve need not equal the mean stochastic trajectory, because the recurrence is nonlinear.