A (slightly) mathematical digression

You've probably got the idea, but let's be a little more formal about this.³ Let $N_t$ be the size of a population at time $t$. Then no matter how complicated the population dynamics actually are, it is always possible to write the population size at some later time, $t+1$, as

$$N_{t+1} = (1+R_t)N_t \quad .$$

You should recognize that $R_t$ could vary from one time period to the next for many different reasons, e.g., ``fluctuations in the seasons or in the number of $\dots$ enemies''. The weather might be different in different years. A disease might sweep through the population. The number of individuals added to or subtracted from the population might depend on the number already there. Chance events may cause the population to increase or decline.

Suppose, however, that we're dealing with an annual plant, one that doesn't have a seed bank, and that we've counted the number of individuals present for many years. As a result, we have a whole list of $R_t$'s. Suppose we take the average of these $R_t$'s, call it $\bar R$. If $\bar R > 0$, then the population is growing on average. Another way of saying that is to say that $N_{t+1} > N_t$, on average. If $\bar R < 0$, then the population is declining on average (and $N_{t+1} < N_t$, on average). Deterministic threats are those, like the ones we've talked about so far in this course, that make $\bar R < 0$. Deterministic threats cause population sizes to decline, on average, year after year. So what are stochastic threats?

Well, let's think about that annnual plant again. We have a list of $R_t$'s from when we started to count up to the present. In every generation

$$N_{t+1} = (1+R_t)N_t \quad .$$

But if you think about it a little longer, you realize that this isn't the whole story. Since that equation applies in every generation, we can do the following:

$$\begin{eqnarray*} N_{t+1} &=& (1+R_t)N_t \\ &=& (1+R_t)(1+R_{t-1})N_{t-1} \\ ... ...})N_{t-2} \\ &=& (1+R_t)(1+R_{t-1})(1+R_{t-2})\cdots(1+R_0)N_0 \end{eqnarray*}$$

So what's the big deal? Well, the population will have grown over this period only if the product $(1+R_t)(1+R_{t-1})(1+R_{t-2})\cdots(1+R_0)$ is greater than one. It turns out that this product can be less than one even if $\bar R > 0$.In other words, the population can decline over the long-run even if it increases, on average, every generation.

That undoubtedly sounds paradoxical, but there's a mathematical theorem showing that it's true.⁴ We'll go into the details next time, but for now just consider this: Suppose there's one generation in which $1+R_t = 0$. Then the population in the next year will have zero individuals in it, i.e., it will be extinct. No matter what the average of $1+R_t$ is, that one zero is enough to guarantee population extinction.

As we'll see next time, this theorem also tells us that it's more likely that a long-term decline will happen when $\bar R > 0$ if there's lots of variability among the $R_t$. In fact, if the variation in $R_t$ is big enough, a population is guaranteed to decline even if $\bar R > 0$. Stochastic threats are those that arise from the variability of $R_t$.

• Deterministic threats - Those that cause $\bar R < 0$.

• Stochastic threats - Those that arise because of the variability of $R_t$.