Growth rates are at the heart of ergodicity economics, and economic news are full of them, too — “GDP grew by 3% last year,” something like that. Sometimes we also hear “national debt grew by $1,271,000,000,000 over the last year” (which is dimensionally different from 3% per year). So since growth rates come in very different forms: what are they, really?

1 Growth rates are key

As we continue to re-develop economic theory by asking the ergodicity question, things are becoming simpler and clearer, and growth rates have emerged as key mathematical objects. Computing the right growth rate in a setting without uncertainty, it turns out, produces discounting. People who optimize the right growth rate behave as if they were discounting payments exponentially, or hyperbolically, or whichever way the dynamic dictates.

In the setting of decisions under uncertainty, optimizing appropriate growth rates can be mapped to Laplace’s expected utility theory, (which is worked out in Peters&Gell-Mann(2016) and inspired the Copenhagen experiment). The growth rate contains a function that we can identify with the utility function in Laplace’s theory (not in Bernoulli’s expected utility theory, which is inconsistent).

In other words: ergodicity economics unifies different branches of decision theory (including intertemporal discounting and expected utility theory) into one concept: growth rate maximization.

Hence the question:

How do we determine the appropriate growth rate for a given dynamic?

We all know at least two types of growth rates, probably more. Below, we’ll develop what a growth rate is, really, by going through two well-known examples, spotting similarities, and then generalizing.

Our first example of a growth rate is the simple rate of change.

(1) g_a=\frac{\Delta x}{\Delta t},

and our second example will be the exponential growth rate

(2) g_e=\frac{\Delta \ln x}{\Delta t}.

2 The growth rate for linear (additive) growth

We use Eq.(1) when something grows linearly, according to

(3) x(t)=x(0)+\gamma t.

The rate of change, Eq.1, is then a good growth rate whose value is \gamma. But how do we know that? What is it about the dynamic, Eq.(3), that makes Eq.(1) the “right” growth rate? Or put differently: why not state the exponential growth rate, Eq.(2), when someone asks how fast x is growing?

Answer: for the additive dynamic, Eq.(3), the growth rate g_a in Eq.(1) has a special and very useful property: it is independent of time — no matter at what t I measure Eq.(1), I always get g_a=\gamma. Because of this time-independence, evaluating Eq.(1) is a useful way to say how fast the process is.

For an additive dynamic, the growth rate is constant in time, in our case its value is \gamma. Whether I measure it a bit earlier or later — I get the same result.

Actually, let’s write the dynamic, Eq.(3), as a differential equation.

(4) dx = \gamma dt

or equivalently, because \gamma depends neither on t nor on x,

(5) dx = d (\gamma t)

This second way of writing tells us that the growth rate \gamma is really a sort of clock speed. There’s no difference between rescaling t and rescaling \gamma (by the same factor). Intriguing.

We make a mental note: the growth rate is a clock speed.

Let’s dig a little deeper here. What kind of clock speed are we talking about? What’s a clock speed anyway?

Or: what’s a clock? A clock is a process that we believe does something repeatedly at regular intervals. It lets us measure time by counting the repetitions. By convention, after 9,192,631,770 cycles of the radiation produced by the transition between two levels of the caesium 133 atom we say “one second has elapsed.” That’s just something we’ve agreed on. But any other thing that does something regularly would work as a clock – the Earth spinning around its axis etc.

When we say “the growth rate of the process is \gamma,” we mean that it advances \gamma units on the process-scale (here x) in one standard time unit (in finance we often choose one year as the unit, Earth going round the Sun). So it’s a conversion factor between the time scales of a standard clock and the process clock.

Of course, a clock is no good if it systematically speeds up or slows down. For processes other than additive growth we have to be quite careful before we can use them as clocks, i.e. before we can state their growth rates.

3 The growth rate for exponential (multiplicative) growth

Now what about the exponential growth rate, Eq.(2)? Let’s use what we’ve just learned and derive the process for which Eq.(2) is a good growth rate. We expect to find exponential growth.

We require that Eq.(2) yield a constant, let’s call that \gamma again, irrespective of when we measure it.

(6) g_e=\frac{\Delta \ln x}{\Delta t}=\gamma,


(7) \Delta \ln x = \gamma\Delta t,

or indeed, in differential form, and revealing that again the growth rate is a clock speed: \gamma plays the same role as t,

(8) d\ln x = d(\gamma t).

This differential equation can be directly integrated and has the solution

(9) \ln x(t)- \ln x(0)=\gamma t.

We solve for the dynamic x(t) by writing the log difference as a fraction

(10) \ln \left[\frac{x(t)}{x(0)}\right] = \gamma t ,

and exponentiating

(11) x(t)=x(0)\exp(\gamma t)

As expected, we find that the exponential growth rate, Eq.(2), is the appropriate growth rate (meaning time-independent) for exponential growth.

In terms of clocks, what just happened is this: we insisted that Eq.(2) be a good definition of a clock speed. That requires it to be constant, meaning that the process has to advance on the logarithmic scale, specified in Eq.(2), by the same amount in every time interval (measured on the standard clock, of course — Earth or caesium).

For multiplicative (exponential) dynamics, the growth rate g_e:=\frac{\Delta \ln x}{\Delta t} is independent of time. It’s not the slope of x itself that’s constant but that of a non-linear function of (the logarithm).

4 The growth rate for general growth

Now let’s raise the stakes and assume less, namely only that x(t) grows according to a dynamic that can be written down as a separable differential equation. We could be even more general, but this is bad enough.

How do we define a growth rate now?

Well, we insist that the thing we’re measuring will be a time-independent rescaling of time, as before. We enforce this by writing down the dynamic in differential form, containing the growth rate as a time rescaling factor. Then we’ll work backwards and solve for g:

(12) dx=f(x) d(g t)

(for linear growth f(x) would just be f(x)=1, and for exponential growth it would be f(x)=x, but we’re leaving it general). We separate variables in Eq.(12) and integrate the differential equation

(13) \int_{x(t)}^{x(t+\Delta t)}\frac{1}{f(x)}dx= g \Delta t,

and we’ve got what we want, namely the functional form of g:

(14) g= \frac{ \int_{x(t)}^{x(t+\Delta t)}\frac{1}{f(x)}dx}{\Delta t}.

To make the equation a bit simpler, let’s give the definite integral a name, the letter u, so that

(15) \Delta u=\int_{x(t)}^{x(t+\Delta t)}\frac{1}{f(x)}dx.

Then we have

(16) g= \frac{ \Delta u[x(t)]}{\Delta t}


(17) \Delta u[x(t)]=g\Delta t.

More generally, there may be some transformation u[x(t)] whose increments change at a constant rate, i.e. at a rate that doesn’t depend on when I measure it. That rate of change is then the appropriate growth rate (allowing us to use the process as a clock).

5 The growth rate from the inverse of the dynamic

There’s a trick to find the growth rate for a given dynamic, without having to solve integrals.

Imagine the growth rate has the numerical value \gamma=1. To keep things clear we’ll introduce a subscript 1 for this rate-1 process:

(18) dx_1=f(x_1)dt.

The time that passes, in standard time units, between two levels of x_1 is then just \Delta u, i.e. we have \Delta u[x_1(t)]=\Delta t. That’s achieved if u(x_1) is the inverse of the dynamic, u(x_1)=x_1^{(-1)}[x_1(t)]=t.

We measure the growth rate by using the actual process as a clock (not the rate-1 process). We take the actual process, generated with whatever the value of the growth rate actually is, dx=f(x) d(\gamma t), we measure it at two different points in time, x(t) and x(t+\Delta t) (where time is defined by our standard clock, like that ^{133}{\text{Cs}} atom), invert it according to x_1^{(-1)}, and compare how much time has elapsed on the time scale of the process (which contains \gamma) to how much time has elapsed on our standard clock.

The result is the growth rate.

(19) g:=\frac{x_1^{(-1)}[x(t+\Delta t)]-x_1^{(-1)}[x(t)]}{\Delta t}=\gamma,

and we conclude that u(x)=x_1^{(-1)}[x(t)].

The required non-linear transformation is the inverse of the rate-1 process. Nice!

That makes total sense, of course: x(t) grows in some wild way, and we just want to know its clock speed \gamma. To find that, we have to get rid of the wild growth, i.e. we have to invert the growth — namely we have to undo how time would be transformed in the rate-1 case x_1(t).

Let’s quickly check this for the additive and multiplicative dynamics, and then try out a different growth to see that everything works out.

5.1 Growth rate as an inverse for additive dynamics

For additive dynamics, Eq.(3), we have the linear function x(t)=\gamma t. The inverse of the rate-1 process is x^{(-1)}[x_1(t)]=x: we’re just inverting the identity transformation. So we expect u(x) to be the identity function, which it is: comparing Eq.(16) to Eq.(1), we have u(x)=x. We’ve dropped x(0) here because it makes no differences to \Delta u.

5.2 Growth rate as an inverse for multiplicative dynamics

For multiplicative dynamics — as ever — it’s more interesting. The inverse of the rate-1 process for Eq.11 is u(x)=\ln \frac{x}{x(0)}. Again, it fits: Eq.(17) and Eq.(2) match if u(x)=\ln x (in the growth rate computation x(0) cancels out).

5.3 Growth rate as an inverse for square dynamics

Now a case that’s neither exponential (multiplicative) nor linear (additive).

(20) x(t)=(\gamma t)^2.

It’s a trivial example, but it shows the mechanics. Without any differential equations, we’ll just find the growth rate from the inverse function of the rate-1 process x_1(t). That’s

(21) u(x)=x_1^{(-1)}[x(t)]=x^{1/2}.

According to Eq.(16), g:=\frac{\Delta u(x)}{\Delta t},

which is

(22) \frac{x(t+\Delta t)^{1/2}-x(t)^{1/2}}{\Delta t}

=\frac{ [\gamma^2(t+\Delta t)^2]^{1/2}-[\gamma^2(t)^2]^{1/2}}{\Delta t}


(Nerdy aside: Section 5 shows that the restriction to dynamics of the form dx=f(x)dt in Section 4 amounts to assuming that x(t) has a differentiable inverse.)

6 Why the letter u, and what the hell is going on?

Beautiful. It all works out. But doesn’t this remind you of something? Of course, a null model of human behavior must be that people maximize the growth rate of their wealth. That means they do different things, depending on the dynamics. Let’s fix \Delta t for a moment. Under additive dynamics they’ll then optimize \Delta x, under multiplicative dynamics they’ll optimize \Delta \ln x, and under general dynamics \Delta u(x).

So people optimize the change in a generally non-linear function of wealth… that’s utility theory, and that’s why we called the non-linear transformation u. Turns out, this has less to do with your idiosyncratic psychology and more to do with the dynamic to which your wealth is subjected.

I’ll leave the extension of this treatment to a random environment as an exercise. Hint: in a deterministic environment, growth rates are constant in time. In a random environment they are ergodic (that’s why at the London Mathematical Laboratory we don’t say “utility function” but “ergodicity mapping”).

You can read more about this in Chapter 2 of our lecture notes (which we’re constantly revising), or in this paper with Alex Adamou: The time interpretation of expected utility theory. [2019-05-02 addendum: a nice example of a growth process that’s neither linear nor exponential is the body mass of organisms]

I thank Yonatan Berman and Alex Adamou for headache-inducing discussions about inverse functions and the like.

6 thoughts on “What’s a growth rate, really?

      1. 😉 Thanks!

        Regarding 5.1, I am a little bit lost. I’d say that:

        x^{-1}(x) = x / \gamma
        x_1^{-1}(x) = x / \gamma


        x^{-1} (x_1 (t)) = x_1(t) / \gamma = t / \gamma

        Or am I missing something here?


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