# The meaning of scale invariance in power law distribution

A function $f(ax)$ that satisfies $$f(ax)=a^\Delta f(x)\,\,\, (\Delta \in R)$$ is said to be scale invariant. The most general function $f(x)$ that satisfies the previous condition is of the form $$f(x)=C x^{\Delta}$$ If we consider the set of distributions, namely the set of all $f(x)$ such that $\int f(x)dx=1$, is it possible to prove that the only distribution that is invariant under scale transformation is $$f(x)=C x^{-1}.$$ This means that in general I can build a generic distribution as a power law that still satisfies the scale invariance properties $f(ax)=a^\Delta f(x)$, but this generic distribution will not be invariant under scale transformations. So, what is the physical meaning of the scale invariance property in a generic power law distribution?

First Edit: For a distribution $f$, if I perform a change of variables, I must preserve the probability, namely $$|f(x)dx|=|g(y)dy|,$$ so that the new distribution $g$ will be given by $f(x)|\frac{dx}{dy}|=g(y)$. If a distribution is invariant under change of variables, I must have $$g(y)=f(y).$$ Replacing $x$ with $ax$ and using the invariance of the distribution $f(x)$ under this transformation, I have $f(x)=C x^{-1}$.

• Something seems off about your second to last sentence: as I understand it, you seem to be saying that in general, a generic power law distribution satisfies the scale invariance property but is not invariant under scale transformations. But those two statements are equivalent - you can't have a distribution that is scale invariant and also not scale invariant. Could you clarify that? Also, what limits should go on your integral $\int f(x)\mathrm{d}x$? May 8, 2012 at 13:33
• What i means is that defining $f(ax)=a^\Delta f(x)$ as scale invariant property is misleading, because the distribution function is not invariant for a generic value of $\Delta$. May 8, 2012 at 13:54
• Why should the exponent $\Delta$ be an integer? May 8, 2012 at 15:25
• The integral of 1/x is divergent at both large and small x, any other powerlaw is either divergent at small x or at large. The question doesn't quite make sense--- what do you want to prove? May 8, 2012 at 16:11
• @emanuele: The boundaries break the scale invariance. Your edit is an answer, you should make it an answer and accept your own answer. Your question could have been much clearer. May 9, 2012 at 4:55

my math isn't advanced enough to give a rigorous answer but my research project is almost exactly this topic (scale-free networks) so hopefully I can give you a sensible PHYSICAL explanation.

So what is the physical meaning of scale-invariance? Well, my understanding is that the distribution in question has no "characteristic length". For instance, take the distribution of adult male heights in US. This would give a nice normal curve with some mean (1st moment of course) and variance (2nd moment). We would expect higher moments to be 0, and the mean in this example has a clear physical interpretation.

Excellent. Now take the example of wealth distribution (which I believe was empirically shown to have power law distribution). Suppose we have a certain number of data points drawn from this population. Now, we can obviously calculate the arithmetic average of this set, but the claim is that such a number does not have a useful meaning in this case. The reason is because as we sample more and more, it becomes increasingly likely that we will sample an income so large as to shift the mean substantially (and the variance, and skewness, and all the higher moments).

So that is my understanding of scale-invariance and power law distribution in general. No matter how many times you sample the "true" distribution, the mean or variance or any of the higher moments won't converge to some particular number because there is always a non-trivial chance of drawing something at the extreme end of the scale. If you like you can also imagine zooming in to a part of some power-law distribution and seeing that the "shape" is exactly the same as when you were zoomed out (like fractals!) Hope that helps!

• you can build distribution without mean and variance that are not scale invariant. See stable distributions on wikipedia. May 9, 2012 at 8:30

I'm still not entirely sure I understand what you're asking, so I'm not sure I'm answering the right question, but consider this: the property of scale invariance is something that applies to functions in general, not just distributions. In essence, scale invariant functions are those which do not require a specific unit for their argument. I've written a blog post explaining this in some detail, but the gist is that with a scale invariant function $f(x)$, you can plug in the value of $x$ in any unit system without changing the character (i.e. the shape) of the function. That's not the case for arbitrary functions; in most cases, you'll need to attach a unit to $x$ because it will be compared with some quantity with units that is built into the definition of the function.

First of all, I don't think the usual terminology is invariant, but rather homogeneous (see Homogeneous function). In particular, we are usually only talking about functions $f$ that, for a fixed $d$, satisfy $$f(ax)=a^df(x),$$ and such functions are said to be homogeneous of degree $d$.

Now, on to the actual question. It seems as if you want to classify all functions $f:\mathbb{R}\rightarrow \mathbb{R}$ homogeneous of degree $-1$ that satisfy $\int f=1$. There are a couple of problems with this. First of all, you don't specify the set of integration. Second of all, of all the natural choices you might mean (e.g., $\int _\mathbb{R}f=1$, $\int _0^1f=1$, $\int _1^\infty f=1$, etc.) all yield trivial answers: there are no such functions. For example: $$\frac{1}{a}\int _0^af(x)dx=\int _0^1f(ax)dx=\frac{1}{a}\int _0^1f(x)dx,$$ so that $$\int _1^af(x)dx=0,$$ which implies that $f$ is $0$ almost everywhere on $(1,\infty )$ (because $a$ is arbitrary). By homogeneity ($f(x)=af(ax)$), this implies that $f$ is $0$ almost everywhere on $(0,\infty )$, and hence it cannot be the case that $\int _0^1f=1$. Similar arguments work for other intervals.

• I think that there is an error on your changing the limits of integration. May 9, 2012 at 8:29