# Difference between power law distribution and exponential decay

This is probably a silly one, I've read in Wikipedia about power law and exponential decay. I really don't see any difference between them. For example, if I have a histogram or a plot that looks like the one in the Power law article, which is the same as the one for $e^{-x}$, how should I refer to it?

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$$y = x^{(\text{constant})}$$

$$y = (\text{constant})^x$$

That's the difference.

As for "looking the same", let's not be silly. Both are positive and go asymptotically to $0$, but with, for example $y=(1/2)^x$, the value of $y$ actually cuts in half every time $x$ increases by $1$, whereas, with $y = x^{-2}$, notice what happens as $x$ increases from $1\text{ million}$ to $1\text{ million}+1$. The amount by which $y$ gets multiplied is barely less than $1$, and if you put "billion" in place of "million", then it's even closer to $1$. With the exponential function, it always gets multiplied by $1/2$ no matter how big $x$ gets.

Also, notice that with the exponential probability distribution, you have the property of memorylessness.

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is it correct to say that exponential decays goes to 0 faster than power law? –  user19821 Jun 29 '12 at 7:40
@user19821 : Yes. You can see why that happens if you figure out what happens to $x^{-2}$ as $x$ increases from $1\text{ million}$ to $1\text{ million}+1$. It doesn't diminish anywhere near as much as to half of what it was. –  Michael Hardy Jun 29 '12 at 17:02
"let's not be silly" - let's not be pretentious. –  Kevin H. Lin Nov 13 '13 at 0:20

very different. A power law just says that some variable is a power of the other. For example, in physics

$$y=3x^2$$

is a power law between $y$ and $x$ where the power is $2$ (the coefficient doesn't matter).

$$y=x^2+x$$

is not. It must be one term of the form $cx^n$.

Exponential decay, on the other hand, is a similar idea, but formed around $Ce^{-kt}$ instead, for some constants $c$ and $k$.

The image in the wikipedia page on the power law is probably something like $\frac 1 x$, not an exponential decay curve.

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