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?
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?
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
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.
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.