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Since priors seem to defined in terms of what they are proportional to, rather than what they are equal to, I'm a little confused on how to tell whether or not a prior is improper. I think I have a general understanding of the concept, but I'm not sure how to apply it to a specific example, and every explanation I can find online seems pretty "hand-wavy" to me.

I've calculated Jeffrey's prior for the variance $\theta$ of a normal distribution, where the mean is known to be 0, and found that:

$\pi_{J}(\theta) \space \propto \space \frac{1}{\theta\sqrt{2}} $

Now, will calculating $\int_{0}^{\infty} \frac{1}{\theta\sqrt{2}} \space d\theta$ and checking whether or not it's finite tell me whether or not $\pi_{J}(\theta)$ is improper? Or is there something more I must check?

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  • $\begingroup$ note: I removed an extra minus sign in your integral and wrote the integral from $0$ to $+\infty$, rather than from $-\infty$ to $+\infty$. $\endgroup$
    – Xi'an
    Dec 5 '16 at 14:03
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You're pretty much right. A prior is proper if and only if it's a probability density function, which means it has to integrate to 1—or if you only know a function that is proportional to the true prior, then that function has to have a defined, finite integral. So if the integral is infinite or undefined, you have an improper prior.

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  • $\begingroup$ Besides checking the function is measurable of course. $\endgroup$
    – Xi'an
    Aug 27 '19 at 7:33
  • $\begingroup$ @Xi'an Isn't Lebesgue-integrability sufficient? I mean, if a function has a Lebesgue integral of 1 but it has a non-measurable preimage of a measurable set, isn't it still a proper prior? $\endgroup$ Aug 27 '19 at 13:01

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