Exponential Decay

Let's leave the simple disease transmission model for a while and think about a simple survival process.

In particular, suppose that at the outset, there is some large number N(0) of organisms in a population. We're going to assume that each one has a risk of dying during one time interval. Equivalently, we could say that each has a certain probability of living through the interval; let's call that probability s.

If there are N(0) organisms, and each one has a probability s of surviving the interval, then the expected value of the number of survivors is going to be N(0) × s. Really, we'd like to think of N as being discrete; it is really a count of the number of living organisms in the system we're looking at, and it should be confined to integer values. But N(0) × s could be anything; it does not have to be an integer. For instance, if N(0) is one million, and s is 1/3, then N(0) × s is about 333333.333.

It's common in situations like this to make the approximation that when the population is fairly large, we can just treat the number of individuals as continuous, and not worry about the fact that it is really a discrete count. For instannce, in the example above, 333333.333 is not very far away from the nearest integer 333333, so there is not much error arising from the approximation.

So let's do it. Let's suppose that N(0) is the number of organisms at the outset, and N(1) is the number after one time step. We're going to just say that N(1)=N(0) × s. What we've done is to say that the number of organisms after one time step is equal to the expected value. We are neglecting the probabilistic aspect, which is reasonable when N(0) is large (we'll discuss this is much more detail later). And we're also treating the number of organisms as a continuous variable that can take on any values, and not just integer counts.

What this all boils down to is that the number of survivors after one time step is equal to the number we started with, times the chance of survival over one time step!

If we say that the survival probability is the same for every time step, then we could calculate the number that survive to the beginning of the next time step: N(2)=N(1) × s. In fact, we can write another difference equation that applies to any time t:
N(t+1) = N(t) × s

Actually this is the same equation we had for exponential growth. We already know the solution:
N(t)=N(0) × s^t
Let's forget about the possibility that s=1 for a moment. Since s is between 0 and 1, the bigger t is, the smaller s^t is.

Eventually the population will get so small that we can't neglect chance effects anymore. Let's forget this and just look at the equation N(t)=N(0) × s^t. I want to convince you that N(t) approaches zero as t grows without bound. Just because something is getting smaller and smaller does not mean it is approaching zero. But here it does. The way to see this is to pick some distance away from zero, and show that eventually N(t) is always less than that distance from zero. In other words, no matter how small a size we pick, eventually the population gets and stays smaller than that. So let's prove it. We'll call our tiny size n; let's find out how long it takes for N(t) to get down to n. We'll solve for t in the equation
n=N(0) × s^t.
This gives us
t logs = log(n/N(0))
Divide both sides by log s, and then round this up to the next largest integer (since t is after all discrete!), and call this answer t₀. Any time t is larger than t₀, we know N(t) is smaller than n. Since we could pick n as tiny as we like (as close to zero as we like), this satisfies the mathematician's definition of what it means for something to approach zero.

Simple exercise: suppose that the initial number N(0) is one million, and that the survival probability over one year (taking the time step as one year) is 0.97. In other words, the death rate is three percent per year. How many are alive after one year? How many after two years? How long does it take for there to be only one thousand expected to be living?

When we examined exponential growth, we were multiplying by a, which was greater than one, and the values kept getting bigger and bigger over time. Here, we're multiplying by s, which is between 0 and 1, and the values keep getting smaller and smaller over time. If either a or s is equal to one, then the population keeps the same value every time unit that passes.

Really, we have the difference equation
N(t+1)=N(t) × a
with some initial condition N(0). If N(0)=0, this gives us some dull results: the population is always zero at every time. (As Andy Gutierrez says, "no cats, no kittens"). We have a single a, and we've seen that the behavior of the equation is different depending on the value of a. If a is greater than one, then we have exponential growth, and the population gets bigger and bigger. If a exactly equals one, the population stays exactly the same always, and if a is between 0 and 1, then we have exponential decay, and the population gets smaller and smaller, closer and closer to zero as time gets larger.

One final note: when we're thinking about exponential growth, we can also make the assumption that the population size is continuous, rather than a discrete count, and we may consider a as an average number of new organisms at the next time step for every organism at this time step and neglect the randomness as before. We'll talk more about this later.

So in this section, we learned that one may use continuous variables to describe population numbers, that constant per-capita death fractions lead to exponential decay, and we touched on the idea that for large populations, chance effects may sometimes be neglected. In the next section we will take our first look at continuous time.