A Leslie matrix approach for viability analysis

Recall our parameter definitions for a Leslie matrix model:⁵

$$\begin{eqnarray*} p_i &=& \hbox{\qquad probability of surviving from age $i$\ to... ...{\qquad number of individuals in age class $i$\ at time $t$} \\ \end{eqnarray*}$$

Given these definitions the probability of surviving from birth to age $x$, $l_x$, is

$$\begin{eqnarray*} l_1 &=& p_0 \\ l_2 &=& p_0p_1 \\ &\vdots& \\ l_x &=& \prod_{k=0}^{x-1}p_k \\ \end{eqnarray*}$$

For a Leslie matrix model, the leading eigenvalue⁶ is given by the unique, positive solution of the Euler-Lotka equation

$$\sum_{x=0}^\infty \lambda^{-x}l_xf_x = 1 \quad .$$

We can use this equation and some observations about avian biology to calculate $\lambda$ from a few simple life-history observations (and assumptions).

1. Constant probability of survival for adults, $s$.

2. Constant fecundity for adults, $f$, where $f$ is the number of offspring successfully fledged.

3. Reproduction begins at age $\alpha$.

4. Probability of survival to adulthood, $l_\alpha$, given that an individual successfully fledges.

Using these assumptions we can rewrite the Euler-Lotka equation and solve for $\lambda$.

$$\sum_{x=\alpha}^\infty \lambda^{-x}(l_\alpha s^{x-\alpha})f = 1 \\$$

because $f = 0$ for $x < \alpha$ and $l_x = l_as^{x-\alpha}$ for $x > \alpha$. Now a little bit of rearranging makes things considerably simpler.

$$\begin{eqnarray*} (l_\alpha f)\lambda^{-\alpha} \sum_{x=\alpha}^\infty \lambda^{... ...mbda^\alpha\left(1 - {s \over \lambda}\right) &=& l_\alpha f \\ \end{eqnarray*}$$

This equation can be solved numerically for $\lambda$ once we have estimates for $s$, $l_\alpha$, and $f$.

Northern spotted owls start breeding at age 3 ($\alpha = 3$); their annual adult survivorship is $s = 0.942$; their annual reproductive rate, i.e., the average number of fledged offspring per individual, is $f = 0.24$; and their probability of survival to age 3 is $l_3 = 0.0722$. Thus,

$$\lambda^3\left(1 - {0.942 \over \lambda}\right) = 0.0722 \times 0.24 \quad .$$

Solving this equation gives us an estimate of $\lambda = 0.961$. After some more manipulations, the standard error of this estimate can be calculated as 0.029. Thus, the approximate 95% confidence interval for $\lambda$ is [0.903,1.019]. It is also possible to do some sophisticated calculations that include the effects of demographic stochasticity and (mild) environmental stochasticity. The long-run population growth rate calculated including these effects is only 2% smaller than the growth rate calculated when ignoring them.

In short, projecting population dynamics based on the data available in 1988 (after almost 15 years of conservation concern) suggests a 4% annual decline in abundance, plus or minus 5.8%. The observed decline based on long-term surveys is only about 1%. Both observations seem to suggest that populations are currently near a demographic equilibrium. Why, then, all the fuss? Why is this species listed as threatened under the Endangered Species Act?

First, Taylor and Gerrodette [7] remind us that what we've just shown is that we cannot reject the null hypothesis of no population decline.⁷ That's not the same thing as saying that the population is ``near demographic equilibrium.'' We could have said just as legitimately that we cannot reject the null hypothesis that the population is declining at a rate of almost 10% per year (3.9% + 5.8%). The calculations Taylor and Gerrodette present suggest that even if the population were actually declining at 4% per year, there might have been as little as a 13% chance of detecting it with the data available in Lande's analysis.

Second, let's look at that leading eigenvalue a bit more closely, especially how it's affected by changes in juvenile survival and individual fecundity.

$l_\alpha$	$f$	$\lambda$
0.0722	0.24	0.961	baseline
0.14	0.24	0.977
0.25	0.24	1.002
0.0722	0.50	0.980
0.0722	0.90	1.006
0.14	0.50	1.011

If fledging success were the only life-history stage amenable to manipulation, it would have to be increased more almost four-fold for the leading eigenvalue to be greater than 1, i.e., for the projection to be for an increasing population. Doubling both the fledging success and survivorship from fledgling to adult, while not easy, may be acheivable. These analyses suggest that we need to pay attention to both life history stages and that the population cannot be overly stable if doubling survivorship and fledging success barely bumps the leading eigenvalue above 1.

The leading eigenvalue is close to 1 (and the population appears to be stable) only because of high adult survivorship - mean adult lifetime is ${1 \over 1-s} \approx 17$ years. Because adults die off so slowly and constitute most of the population, the population size changes relatively little from one year to the next. Even if all reproduction stopped now, the rate of decline would be about $1-s$ per year or 6%.

Third, a leading eigenvalue of 0.961 suggests a slow population decline, 4% per year. In fact, it can't be statistically distinguished from an eigenvalue of 1, which would mean the populatio size is stable. Recall, however, that this is a linear model leading to a geometric rate of population size change.

$\lambda$	Time to 50% population size
0.99	69
0.98	34
0.97	23
0.96	17

Rule of 69: doubling time $\approx {69 \over r}$, where $r = 1-\lambda$.

$$\begin{eqnarray*} N_t &=& N_0 e^{rt} \\ \frac{N_t}{N_0} &=& e^{rt} \\ \ln\left... ...ght) &=& rt \\ t &=& \frac{\ln\left(\frac{N_t}{N_0}\right)}{r} \end{eqnarray*}$$

So the time it takes for population size to double, i.e., for $N_t = 2N_0$ is

$$\begin{eqnarray*} t_d &=& \frac{\ln(2)}{r} \\ &\approx& \frac{0.69}{r} \end{eqnarray*}$$

If we take the linear analysis as a forecast rather than a projection, then we would predict that the population size could be half of what it is now in as little as 7 years (using the lower 95% confidence bound on $\lambda$) or it could be double what it is now in about 35 years (using the upper 95% confidence bound on $\lambda$).