Survivorship measures both relative and absolute fitness

I recently reviewed a very good paper that (in the original version) claimed there was no way to turn surivorship (i.e. the proportion of a genotype or whatever that survives in a selection experiment) into relative fitness. This struck me as odd, but the authors are very knowledgeable in this area, so I did some digging. I came to realize that the probability of survival is both a measure of absolute and relative fitness. I’m sure this has been proven before, but I wasn’t aware of it, so I wrote up the proof here with a mini-example, just in case it’s useful to others. If you know of someone who has shown this previously, I’d appreciate the reference. Thanks!

The probability of survival is a measure of relative fitness (and absolute fitness!).

In their original submission (since revised), the authors stated that they are only measuring absolute fitness for viability because “[i]t was not possible to evaluate relative fitness in a logistic regression framework.” Here I want to show that the probability of survivorship, which is the parameter fit with logistic regression in this case, is a measure of both absolute and relative fitness.

The key property of absolute fitness ($W_i$) is that it is proportional to the change in abundance from $t$ to $t+1$:

$$N_i(t+1)=W_i{N}_i(t)$$

The key property of relative fitness ($w_i$) is that it is proportional to the change in genotype frequency from $t$ to $t+1$:

$$p_i(t+1)=w_i{p}_i(t)$$

The probability of survivorship fulfills both of these relationships, as I’ll show with a toy example (haploid, two-allele, one-locus system), but it should generalize to any number of genotypes or change in probability density if you’re talking about a quantitative trait.

Let’s imagine two alleles $a$ and $A$ with proportions at time $t$, $p_a(t)$ and $p_A(t)=1-p_a(t)$. Assume that the proportion surviving from $t$ to $t+1$ is $W_a$ and $W_A$. This is absolute fitness because if $W_i=0.5$ and $N_i(t)=100$, then $N_i(t+1)=W_i{N}_i(t)=50$.

The average fitness is $\overline{W}=p_a{W}_a+p_A{W}_A$ and the relative fitnesses are $w_a=W_a/\overline{W}$ and $w_A=W_A/\overline{W}$.

If there are $N=100$ individuals and 40 of them have $a$ and 60 have $A$, then $p_a=0.4$, $p_A=0.6$. Assume that 4/5 of $a$ survive and 3/5 of $A$ survive ($W_a=0.8,W_A=0.6$), then the abundances in the next generation are $N{p}_a{W}_a=100\times 0.4\times 0.8=32$ and $N{p}_A{W}_A=100\times 0.6\times 0.6=36$. Hence, the allele frequencies in the next generation are $p_a(t+1)=32/(32+36)=0.47$ and $p_A(t+1)=36/(32+36)=0.53$. This fulfills the definition of relative fitness: $w_a=W_a/\overline{W}=0.8/(0.4\times 0.8+0.6\times 0.6)=1.18$ and $w_A=W_A/\overline{W}=0.6/(0.4\times 0.8+0.6\times 0.6)=0.88$, therefore it’s true that $p_a(t+1)=w_{a}p(t)=1.18\times 0.4=0.47$ and $p_A(t+1)=w_{A}p(t)=0.88\times 0.6=0.53$

Below is some R code demonstrating these calculations.

# Libraries
library(magrittr)
library(tibble)

# Parameters
N <- 100
N_a <- 40
N_A <- N - N_a
p_a <- N_a / N
p_A <- 1 - p_a

W_a <- 0.8
W_A <- 0.6

W_bar <- p_a * W_a + p_A * W_A
w_a <- W_a / W_bar
w_A <- W_A / W_bar

# Calculations
tibble::tibble(
  Genotype = c("$a$", "$A$", "Total"),
  `$N(t)$` = c(N_a, N_A, N),
  `$p(t)$` = c(p_a, p_A, 1),
  `$W$` = c(W_a, W_A, p_a * W_a + p_A * W_A),
  `$\\bar{W}$` = p_a * W_a + p_A * W_A,
  `$w$` = `$W$` / `$\\bar{W}$`,
  `$N(t + 1)$` = `$N(t)$` * `$W$`,
  `$p(t + 1)$` = `$N(t + 1)$` / `$N(t + 1)$`[3]
) %>%
  
  # Print table
  knitr::kable(digits = 2)