# Integrating the Lindy Effect

Suppose the following:

1. Your crystallized intelligence is directly proportional to how many useful things you know.
2. Your intelligence increases when you learn things and decreasing as the world changes and the things you know go out-of-date.

How quickly the things you know become irrelevant is directly proportional to how many relevant things you know and therefore proportional to your intelligence $I$ and inversely proportional to the typical lifetime of things you know $L$. Let's use $R$ to denote your rate of learning. Put this together and we get an equation.

$$\frac IL+\frac{dI}{dt}\propto R$$

If we measure intelligence in units of "facts you know" then the proportionality becomes and equality.

$$\frac{dI}{dt}=R-\frac IL$$

The solution to this first order differential equation is an exponential function.

$$I(t)=ce^{-t/L}+RL$$

We must solve for $c$. For convenience, let's declare that your intelligence is $0$ at time $t=0$. Then $c$ must equal $-RL$. This gives us a tidy solution.

$$I(t)=RL\left(1-e^{-t/L}\right)$$ This makes sense intuitively because your intelligence is directly proportional to $R$ and $L$. But wait a minute. $L$ isn't just a coefficient. $L$ is in the exponent too.

## Time $t$ and lifetime $L$

Most human beings reading this article will be between 10 and 100 years old. In other words, $t$ is measured in decades. In other other words, $t$ is on the order of 10 years.

$L$ values, on the other hand, are distributed exponentially across many orders of magnitude.

Order of $L$ in Words Order of $L$ in Years Medium(s)
day 0.001 daily newspaper
week 0.01 Sunday newspaper, political story, sports game outcome
century 100 most spoken languages, classic literature
millennium 1,000 cooking, history
ten thousand years 10,000 human psychology
gigaannum 1,000,000,000 biology
forever math, physics

Most things you can know have a useful lifetime at least one order of magnitude away from the human timescale of decades. In other words, we can assume either $L$ is much greater than $t$ or much less than $t$.

$$I(t)=R\cdot\cases{L &for t \gg L \cr t &for t \ll L \cr}$$

### Suppose $L$ is much less than $t$.

Then your crystallized intelligence equals $I=RL$, a constant. In other words, if $L\ll t$ then how long you have been learning $t$ is irrelevant. $I$ is constant with respect to time. Years and years of studying will not make you smarter.

If you ever stop studying then you will feel yourself getting dumber quickly. But only for a short time. Your crystallized intelligence will restabilize after a period of $t=L$.

### Suppose $L$ is much greater than $t$

Then your crystallized intelligence equals $I=Rt$, a linear function. You get smarter over time. Your crystallized intelligence will be proportional to your age.

## Conclusion

$I=Rt$ grows with respect to time while $I=RL$ stays constant. Eventually, anyone on an $I=Rt$ trajectory will always become smarter than someone on an $I=RL$ trajectory even if the person on the $I=RL$ trajectory has higher $R\neq0$.

The lifetime of things you learn $L$ is far more important than how fast you learn $R$. Over a lifetime of decades, someone who learns a few durable facts slowly will eventually become smarter than someone who learns many transient facts quickly