# Uniform prior distribution on log scale

Can anyone please suggest me a distribution whose $$log$$ transformation is uniform and it should be a well known distribution? I am not sure if it exists. I know that if we consider a uniform distribution $$x \sim U(a,b)$$ and and take $$y = e^x$$. Then, $$log(y)$$ follows uniform distribution. However, $$y$$ is not a well known distribution. Can you suggest me a well known distribution which is close to this?

• maybe I'm being stupid, but isn't a distribution whose log transform uniform exactly the exponential of the uniform distribution? if so, then there's no answer to your question if you consider your defined $y$ not well-known (which it appears you do). – mathworker21 Nov 16 '19 at 10:40

## 2 Answers

The distribution of your $$y$$ has two names:

1. log-uniform distribution (also spelled as log uniform) which follows properties of the inverse transform
2. Reciprocal distribution because it has a probability density function $$\it f(y) \propto \frac 1y$$

$$y$$ has PDF $$f(y,c,d) = \frac 1{y \ \left(\it ln(d)-\it ln(c)\right)} \ \text for \ y \in [c,d]$$ where $$c=e^a, d=e^b$$

I don't know if it's well known but Google is giving tens of thousands answers.

https://en.wikipedia.org/wiki/Reciprocal_distribution

Let me provide you with a full derivation so that you do not feel you are missing anything.

Let random variable $$X$$ have pdf $$f_X(x)$$. Let $$Y=g(X)$$ where $$g$$ is monotone. Let $$h = g^{-1}$$ be the be inverse function of $$g$$. Then the pdf of $$Y$$ is given by $$f_Y(x)=f_X(h(x))h^\prime(x).$$

You can read here for the above statement.

In your case, you have $$g(X)=ln(X)$$. Hence $$h(x) = e^x$$. You want the distribution of $$Y$$ to be uniform, i.e. $$f_Y(x)= \frac{1_{[a, b]}}{b-a}$$ for some real value $$a < b$$. Plug in h(x), you get $$f_X(e^x)e^x= \frac{1_{[a, b]}}{b-a}.$$

Let $$y=e^x$$. you get $$f_X(y) = \frac{1_{[e^a, e^b]}}{y(b-a)}, y > 0.$$ In the last formula, $$y$$ is just a variable, you can use $$x$$ if you prefer. Then you get $$f_X(x) = \frac{1_{[e^a, e^b]}}{x(b-a)}, x > 0.$$ That is your distribution. You have NO other options.