If a prior integrates to a finite constant that is not 1, is it still considered proper? Is a prior only improper if it integrates to an nonfinite value?
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Since in the case of $p (\theta) \propto f (\theta) $ with $\int_{\theta \in \Theta} f (\theta) d\theta = c$ integrating to some constant you just need to normalize the density to $f (\theta)/c $, that defines a proper prior.
Björn
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2That is true, but also note that you do not even need to take care of that as the c cancels out in the integrals when calculating the posterior. – Erik Oct 07 '15 at 06:42
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The classical definition of an improper prior in Bayesian statistics is one of a measure $\text{d}\pi$ with infinite mass $$\int_\Theta \text{d}\pi(\theta)=+\infty$$ See, e.g., Hartigan's Bayes Theory, which formalises quite nicely the use of improper priors. Any measure $\text{d}\pi$ with finite mass can be normalised into a probability measure with mass $1$.
See also this related Cross validated entry on improper priors.
