Note that in this particular case, the deterministic trajectory is predicted to be the same as the average of the stochastic trajectories. If we take the expectation of Eq. (7.2), we have

E{X (t + 1)} = E{(1 + I)X (t)} + E{Z (t)} = (1 + I)E{X (t)} (7.3)

which is the same as Eq. (7.1), so that the deterministic dynamics characterize what the population does ''on average.'' This identification of the average of the stochastic trajectories with the deterministic trajectory only holds, however, because the underlying dynamics are linear. Were they nonlinear, so that instead of (1 + I)X(t), we had a term g(X(t)) on the right hand side of Eq. (7.2), then the averaging as in Eq. (7.3) would not work, since in general E{g(X)} = g(E{X}).

The deterministic trajectory shown in Figure 7.1 accords with our experience with exponential growth. Since the growth parameter is small, the trajectory grows exponentially in time, but at a slow rate. How about the stochastic trajectories? Well, some of them are close to the deterministic one, but others deviate considerably from the deterministic one, in both directions. Note that the largest value of X(t) in the simulated trajectories is about 23 and that the smallest value is about — 10. If this were a model of a population, for example, we might say that the population is extinct if it falls below zero, in which case one of the ten trajectories leads to extinction. Note that the trajectories are just a little bit bumpy, because of the relatively small value of the variance (try this out for yourself by simulating your own version of Eq. (7.2) with different choices of I and a ).

The transition from Eq. (7.1) to Eq. (7.2), in which we made the dynamics stochastic rather than deterministic, is a key piece of the art of modeling. We might have done it in a different manner. For example, suppose that we assume that the growth rate is composed of a deterministic term and a random term, so that we write X(t + 1) = (1 + I(t))X(t), where I(t) = I + Z(t), and understand I to be the mean growth rate and Z(t) to be the perturbation in time of that growth rate. Now, instead of Eq. (7.2), our stochastic dynamics will be

Note the difference between Eq. (7.4) and Eq. (7.2). In Eq. (7.4), the stochastic perturbation is proportional to population size. This slight modification, however, changes in a qualitative nature the sample paths (Figure 7.2). We can now have very large changes in the trajectory, because the stochastic component, Z(t), is amplified by the current value of the state, X(t).

Which is the ''right'' way to convert from deterministic to stochastic dynamics - Eq. (7.2) or Eq. (7.4)? The answer is ''it depends.'' It depends

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