Introduction

Stochastic threats arise from the simple fact that it is always possible for a population to decline over a series of generations even if it tends to increase from generation to generation on average. Consider an annual plant. For such an organism, the following very simple model provides an excellent model of the dynamics:

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

Let's put some numbers into this equation:

$$N_{t+1} = (1+0.02)(1-0.02)(1+0.01)(1-0.01)(10)$$

$$= 9.995 \quad .$$ (1)

Now if I were to ask you to calculate the average growth rate over these four generations, you'd probably do something like

$$\frac{(1+0.02) + (1-0.02) + (1+0.01) + (1-0.01)}{4} = 1 \quad .$$

That suggests that, on average, the population is neither growing nor declining. But the calculation in equation (1) clearly shows that the population has declined - not by much, admittedly, but it's still declined. What gives?

Well, instead of calculating the average growth rate as the arithmetic mean,¹ which is what we just did, suppose we calculated the geometric mean, i.e.,

$$\left((1+0.02)(1-0.02)(1+0.01)(1-0.01)\right)^{(1/4)} = 0.999875 \quad .$$

Now $(0.999875)^4 = 9.995$, so the geometric mean is clearly what we want to use to calculate long-term growth rates of populations when the growth rate varies among generations. Why? Because the long-term growth rate of a population is the product of generation-to-generation growth rates, not the sum.

The first important thing this example illustrates is that when dealing with variation in rates, the usual arithmetic mean isn't nearly as important as the geometric mean. The long-term growth fate of a population is determined by the geometric mean of $1+R_t$, not by the arithmetic mean.² If you want a formula to compare the two, here it is:

$$\begin{eqnarray*} \bar x_{a.m.} &=& (1/K)\sum_i x_i \\ \bar x_{g.m.} &=& \left(... ...t)^{(1/K)} \\ &=& \exp\left((1/K)\sum_i \ln x_i\right) \quad . \end{eqnarray*}$$

The second important thing this example illustrates is that the geometric mean is alway less than the arithmetic mean.³ As a result, the long-term growth rate of a population may be negative even if the arithmetic mean growth rate is positive. In other words, a population may decline over the long term even if if it tends, on average, to increase in size from one year to the next.

It can be shown that the long-term growth rate of a population will be negative, i.e, the population will tend to decline, whenever the variance in growth rate is more than about twice its mean.⁴ To be more precise, if

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

and $\bar R$ is the average $R_t$ and $s^2_R$ is the variance of $R_t$ then the long-term growth rate of a population will be negative if

$$\bar R < \frac{s^2_R}{2} \quad .$$