Scale invariance of power law distribution.

Question

On wikipedia i have find this statement:

...it is scale invariant, and the only continuous distribution that fits this (scale invariance) is one whose logarithm is uniformly distributed.

how can be proven?

Answer 1

This is the argument as it's sketched by Julian Havil in Gamma: Exploring Euler's Constant; some fiddly details (e.g., involving domains of definition) are ignored.

Let $X$ be a scale-invariant continuous random variable with probability density function $f(x)$ and cumulative density function $F(x)$. Fix a lower bound $u$ on $X$. Scale invariance means that for any $a>0$, $$P(u<X<x)=P\left(u<\frac1aX<x\right)=P(au<X<ax)\;,$$ and hence $F(ax)-F(au)=F(x)-F(u)$. Differentiating with respect to $x$, we see that $af(ax)=f(x)$, so $$f(ax)=\frac1af(x)\;.\tag{1}$$

Now fix a base $b$ for logarithms and let $Y=\log_b X$, with pdf $g(y)$ and cdf $G(y)$. Then $$G(y)=P(Y\le y)=P(\log_bX\le y)=P(X\le b^y)=F(b^y)=F(x)\;,$$ and hence

$$\begin{align*} g(y)&=\frac{d}{dy}G(y)=\frac{d}{dy}F(x)\\\\ &=\frac{d}{dx}F(x)\frac{dx}{dy}=f(x)\frac{dx}{dy}\\\\ &=xf(x)\ln b\;.\tag{2} \end{align*}$$

Substitution of $ax$ for $x$ and using $(1)$ and $(2)$ yields

$$\begin{align*}g(\log_b ax)&=axf(ax)\ln b\\ &=ax\cdot\frac1af(x)\ln b\\\\ &=xf(x)\ln b\\\\ &=g(\log_b x)\;, \end{align*}$$

which can be rewritten as $g(\log_b x+\log_b a)=g(\log_b x)$. But $a$ was an arbitrary positive scaling factor, so $g(y+c)=g(y)$ for all $c\in\Bbb R$, and therefore $g$ must be a constant function.

This shows that if $X$ is continuous and scale-invariant, then $\log_bX$ is uniformly distributed.

Answer 2

If you start by assuming that $X\sim\operatorname{Unif}[0,1]$ and then examine the leading digit of $Y=b^X$ base $b$, I think you will be able to derive the law. Let $B$ be the leading digit base $b$ (B for Benford). Then $B = \lfloor Y\rfloor$ and $$P\left(B=k\right)=\log_b(k+1)-\log_b k=\log_b\left(1+\frac1k\right)$$ follows from observing that $$k\le b^X\lt k+1\qquad\iff\qquad\log_bk\le X\lt \log_b(k+1)$$ and $X$ is uniform.