Consider a likelihood $x|\theta \sim N(\theta,\sigma^2)$. Now if there is no prior information an uniform prior can be chosen, that is: $p(\theta)=c$.

As such: $p(\theta|x)=\frac{1}{(2\pi\sigma^2)^{0.5}}e^{-\frac{(x-\theta)^2}{2\sigma^2}}$

All of this is fine, apart from the fact that normal distribution of course ranges over infinite values. As such $p(\theta)=c$ should not be possible, as $\theta\sim\text{Unif}(-\infty,\infty)$ is not a valid distribution.

How should the uniform prior be thought about in cases such as this?

  • 3
    $\begingroup$ It's not quite clear what kind of information you're seeking there (it would help to try to clarify what you want to know), but see en.wikipedia.org/wiki/Prior_probability#Improper_priors ... there are a number of posts on site which mention this particular improper prior or improper priors in general $\endgroup$ – Glen_b May 7 '18 at 0:55
  • $\begingroup$ An improper prior is not a probability density but a positive measure. Notions like uniformity and expectations and simulations do not make sense for measures. $\endgroup$ – Xi'an May 7 '18 at 11:55

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