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?


5 Answers 5


$$ \begin{array}{rl} \text{power law:} & y = x^{(\text{constant})}\\ \text{exponential:} & y = (\text{constant})^x \end{array} $$

That's the difference.

As for "looking the same", they're pretty different: 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.

  • 1
    $\begingroup$ is it correct to say that exponential decays goes to 0 faster than power law? $\endgroup$
    – user19821
    Commented Jun 29, 2012 at 7:40
  • $\begingroup$ @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. $\endgroup$ Commented Jun 29, 2012 at 17:02
  • 4
    $\begingroup$ This answer doesn't even explain which is which. $\endgroup$
    – MountainX
    Commented Mar 12, 2014 at 4:53
  • 1
    $\begingroup$ Note that in general, there may be other constants in each expression, e.g. $y = C x^n + b$ and $y = C^{kx+b}+d$, though these may be set to zero for certain problems. It's always important to know the data when fitting it. $\endgroup$
    – jvriesem
    Commented May 29, 2018 at 22:43
  • $\begingroup$ @jvriesem No, that is not actually the case. Looking at your expressions, $b=0$ in a power law. You can of course have an expression where $b\ne0$, but ain't power law. It can though, exhibit power law behavior in some region (power law tail). $\endgroup$
    – myradio
    Commented Mar 29, 2021 at 13:36

How is a power law different from an exponential? (I'm putting this answer here mainly for my own future reference. Hopefully someone else may find it useful.)

Power Law function
(notice the exponent, $k,$ is a constant) $$ y = x^k $$

Exponential function
(notice the exponent is a variable) $$ y = a^x $$

Technical definition of Power Law:

A power law is any polynomial relationship that exhibits the property of scale invariance.

Scale invariance (from Wikipedia)

One attribute of power laws is their scale invariance. Given a relation $f(x) = ax^k,$ scaling the argument $x$ by a constant factor $c$ causes only a proportionate scaling of the function itself. That is,

$$ f(c x) = a(c x)^k = c^k f(x) \propto f(x) $$

That is, scaling by a constant $c$ simply multiplies the original power-law relation by the constant $c^k.$

Thus, it follows that all power laws with a particular scaling exponent are equivalent up to constant factors, since each is simply a scaled version of the others. This behavior is what produces the linear relationship when logarithms are taken of both $f(x)$ and $x,$ and the straight-line on the log-log plot is often called the signature of a power law.

  • 2
    $\begingroup$ I edited out the part that is not an answer. This is not a blog or a bulletin board. $\endgroup$ Commented Jun 16, 2014 at 23:08
  • 1
    $\begingroup$ However I found very useful if, aside of the answer, there si more information that answerer felt useful to add. Maybe instead of editing out marking it is as "for future reference" ? $\endgroup$
    – user305883
    Commented Oct 27, 2017 at 22:27
  • 2
    $\begingroup$ I would like to add back the additional information that was removed, if the majority don't object. $\endgroup$
    – MountainX
    Commented Mar 22, 2020 at 22:27

If there is anybody landing on this from Nassim Nicholas Taleb's The Black Swan the issue at stake is how doubling a random variable affects the probability in power law distributions as opposed to a normal or Gaussian distribution.

In the case of continuous random variables, the pdf illustrates the difference:

A power law distribution has a pdf of the form

$$f_X(x) =C x^{-\alpha},$$

where $C$ is a constant, and $\alpha$ is the law's exponent.

The effect of doubling $x$ will remain constant across the domain: For example, the ratio in the pdf between people who make $\$50,000$ per year to those that make $\$25,000$ will be the same as the ratio between people that make $\$10,000,000$ to those that make $\$5,000,000:$


an attribute called scale invariance.

This is in contrast to the rapid decay in the normal distribution, which in the standardized form has the following pdf:

$$f_X(x)=\frac{1}{\sqrt{2\pi}}\exp\left( -\frac{x^2}{2}\right)$$

Doubling the value of $x,$ amounts to raising to the third power the exponential (un-normalized) part of the pdf:

$$\begin{align} \frac{f_X(2x)}{f_X(x)}&=\exp\left(-\frac{(2x)^2}{2} +\frac{{}x^2}{2}\right)\\[2ex] &=\exp\left( -\frac{3x^2}{2}\right)\\[2ex] &=\left(\exp\left(-\frac{x^2}{2} \right)\right)^{3}\\[2ex] &=\frac{1}{\left(\mathrm e^{x^2/2}\right)^3} \end{align}$$

which can be visually plotted as:

enter image description here

Therefore, at asymptotic values the relative probability between an extreme event and that of its value doubled will rapidly tend to zero, whereas in a power law it will remain stubbornly constant.


If you have a limited data set, one way to tell the difference is to put the data into spreadsheet software capable of exponential and power regressions and see which gives the better correlation coefficient. Presumably the coefficient is calculated by comparing the least squares errors of the semi-log and log-log plots. A bit more on that...

Let's call an exponential law one like $y = Ca^x$ and a power function one like $y = Cx^p$. If we take the logarithm of both sides of an exponential function, we get $$ \log y = \log C + x \log a. $$ That is, the collection of ordered pairs $(x, \log y)$ (the semi-log plot) should be roughly linear for exponential data.

On the other hand, for a power function we get $$ \log y = \log C + p \log x, $$ so the collection of ordered pairs $(\log x, \log y)$ (the log-log plot) should be roughly linear for power law data.

Determining which of these two plots is more line-like can tell whether exponential or power laws best model the original data.


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


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


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .