# What is a good example of a non-informative prior for the uniform distribution?

I recently noticed that for non-informative priors, people usually use something like a uniform prior, which works for many different distributions. However, assuming that your likelihood is nothing more than a uniform $\frac{1}{\theta}$ for a parameter theta, what is a non-informative prior that works here? The first thing coming to mind is the Jeffrey's prior, but that yields $-\frac{1}{\theta^2}$, which when trying to calculate the posterior mean / variance leads to a divergent series when trying to sum across the product of the prior and likelihood. Does anyone have any ideas of what is a good non-informative prior?

• To be clear, this prior is a Pareto(1, 1) distribution. The posterior form is known and has finite moments: Pareto(max(1, data), 1 + sample_size). The fact that the prior does not have a finite first moment shouldn't scare us as long as it is a distribution (e.g., using a Cauchy distribution as a prior). So I'm a little confused as to what the problem is exactly? – user44764 Oct 1 '15 at 0:20
• Are you talking about a Uniform$(0,\theta)$?. (Note that the prior is for the parameter, rather than a distribution.) – Glen_b Oct 1 '15 at 0:52
• @Matt What I mean is that if I want a prior where the expectation and variance exist, what should I choose? – user123276 Oct 1 '15 at 2:10
• Warning: Your likelihood is more than $1/\theta$ in that there is an indicator at play: $\mathbb{I}_{0\le x\le \theta}$. – Xi'an Oct 23 '15 at 14:14
• A similar question with answers: stats.stackexchange.com/questions/69383/… – kjetil b halvorsen Feb 11 '19 at 11:27

I will assume your model is a uniform distribution on the interval $$(0, \theta)$$. So let $$X_1, \dotsc, X_n$$ iid with that distribution, with $$\theta>0$$. Then the likelihood function can be written $$L(\theta) = \theta^{-n} \cdot \mathbb{1}(\theta \ge T)$$ where $$T=\max_{i=1}^n X_i$$. The first idea is the Jeffrey' prior, and your statement of that cannot be right (it is negative!). What I get is $$n/\theta$$ for $$\theta \ge T$$. That may look strange, first, it depends on the data through $$T$$. But in this case that isn't a problem, since the likelihood is zero for $$\theta < T$$, so the prior on that interval is unimportant, it will always be multiplied with zero when calculating the posterior. Second, it is improper, but, as long as $$n\ge 2$$ it leads to a proper posterior (which do have the pareto form). A detailed development is here.
The posterior is $$f(\theta | T) = \frac{n-1}{\theta}\cdot\left(\frac{T}{\theta}\right)^{n-1}$$ for $$\theta \ge T$$.