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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?

2 Answers 2

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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 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.

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    @emanuele: I'd have to work through the algebra, but it doesn't look like it.2012-05-06
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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.