Jeremy Fisher

Let us model the likes of J. Fisher, a species of cricket frog celebrited in literature. Imagine we know survival probability of eggs, larvae, and overwintering metamorphs and juveniles. These are the estimates of survival through one stage to another, e.g., from just-hatched tadpole to metamorph.

Survivals, \(\sigma\)

Fecundity of the average adult: \(f=100\)

Matrix misrepresentation

Let us first model this population using a matrix representation, where we combine egg and larvae into one stage and use metamorphs/juveniles as the other. The matrix is incorrect because there is no point in time when these two stages stably co-occur, but what the heck.

s_e = 0.3
s_l = 0.3
s_m = 0.1
f = 100

## three different ways to do it wrong.
A4 <- matrix(c(0,   0,  0, f,
              s_e, 0,  0, 0,
              0,  s_l, 0, 0,
              0,  0,  s_m, 0), nr=4, byrow=TRUE)
A3 <- matrix(c(0,       0, f,
              s_e*s_l, 0, 0,
              0,       s_m, 0), nr=3, byrow=TRUE)
A2 <- matrix(c(0,        f,
              s_e*s_l*s_m, 0), nr=2, byrow=TRUE)
A3
##      [,1] [,2] [,3]
## [1,] 0.00  0.0  100
## [2,] 0.09  0.0    0
## [3,] 0.00  0.1    0
Re(eigen(A3)$values[1])
## [1] 0.9654894

A simpler, and correct approach.

In which we find the correct way is simpler.

Population growth rate, \(\lambda\), is actually just the product of the relevant survival probabilities and fecundity,

\[\lambda = \sigma_e \sigma_l \sigma_mf\].

(lambda = f*s_e*s_l*s_m)
## [1] 0.9

We see that the estimate we get from the matrix differs from what we get when we do it correctly.

Estimating population growth rate this way makes it very easy to investigate effects of evironmental stressors on particular life history stages.

Think of it as a matrix model with one stage.

Elasticity, e, is proportional change in \(\lambda\) resulting from a proportional change in a matrix element. It can be defined (cf. Grant and Benton 2000, Vonesh and de la Cruz 2002), as \[ \mathrm{e}_i = \frac{\delta \log \lambda}{\delta \log p_i}=1\] where \(p_i\) is survival or fecundity. If we walk through the algebra, we would see that the elasticity of each parameter (survival or fecundity) equals 1, that is, the elasticities of each is the same. This might seem crazy, but it is not. Recall that elasticities of all matrix elements must sum to one, and we have only one ‘matrix’ element. Consider the proportional change in \(\lambda\) in response to a 1% drop in either egg vs. metamorph survival – they are identical.

\[\lambda_e = (0.99) \sigma_e \sigma_l \sigma_mf\] \[\lambda_m = \sigma_e \sigma_l (0.99)\sigma_mf\]

The upsides

The above simplicity may give you greater confidence in discussing the biology of each life history stage and comparing their relative importances.

The complicated stochastic modeling that you have already done can still be used to investigate stochastic population dynamics - you can even add density-dependence if you like.

Appendix

Proof by induction

Let \(\sigma\) be the product of all survivals, and \(f\) be fecundity. Show that elasticities of each parameter equal 1. Let \(p_i = \sigma\).

\[\begin{align*} \frac{\delta \log \lambda}{\delta \log p_i}&=\frac{\log(\sigma f) - \log( (\sigma+h) f)}{log(\sigma) - \log (\sigma + h)}\\ &=\frac{\log(\sigma) + \log(f) - \log(\sigma+h) - \log(f)}{\log(\sigma) - \log(\sigma+h)}\\ &=\frac{\log(\sigma) - \log(\sigma+h)}{\log(\sigma) - \log(\sigma+h)}=1 \end{align*}\]