3

The wikipedia page on the log-uniform distribution says:

A positive random variable X is log-uniformly distributed if the logarithm of X is uniform distributed,

$$\ln X \sim \mathcal{U}(\ln(a), \ln(b))$$

Whereas this answer reverses the relationship and says:

$X∼U[a,b]$ where $[a,b]⊂R>0$ then $Y=-\ln X$ is log-uniformly distributed

What am I missing here?

Starlight
  • 2,684

1 Answers1

3

I think the Wikipedia page is right, that is $X$ is log-uniformly distributed iff $\ln(X)$ is uniformly distributed. For mainly three reasons:

  1. I believe there is a very strong consensus that $X$ is log-normally distributed iff $\ln(X)$ is normally distributed. I would find it weird to agree on the reverse convention for the uniform distribution.
  2. I can see very well how it is useful to name the distribution of $X$ such that $\ln(X)$ is uniformly distributed. Because you want to be able to describe a phenomenon that is equally likely to happen on any interval on a log-scale, that is for example $\mathbb P(X\in[10,100])=\mathbb P(X\in[100,1000])$. In that case you would choose for instance $X=10^U$ where $U$ is uniformly distributed.
  3. If $U$ is uniformly distributed on $(0,1)$, then $-\ln(U)$ is exponentially distributed. So if you want to name it something, I would say $\ln(U)$ is neg-exponentially distributed.
Will
  • 8,358