RE: Conjugate Prior Distributions

Sipes, Summer ( (no email) )
Tue, 12 Oct 1999 13:22:52 -0500

It's the mixed distribution that has its variance equal to the sum of the
VHM and EPV. (This is most easily seen in the Normal-Normal conjugate prior
case.)

For Beta-Bernoulli, the mixed distribution is a bernoulli, with Variance =
ab/[(a+b)^2], which does equal VHM + EPV.

EPV + VHM = ab + ab
..
(a+b) (a+b+1) (a+b)^2(a+b+1)

= ab(a+b) + ab .
(a+b)^2(a+b+1) (a+b)^2(a+b+1)

= ab(a+b+1) .
(a+b)^2(a+b+1)

= ab .
(a+b)^2

> -----Original Message-----
> From: Michael Ying (STPAULRE) [SMTP:Michael_Ying@STPAULRE.COM]
> Sent: Tuesday, October 12, 1999 12:51 PM
> To: studygroup4b@lists.casact.org
> Subject: Conjugate Prior Distributions
>
> Based on Casualty Study Manual Herzog C 8:
>
> Prior mean for Beta-Binomial = a/(a+b)
> Variance of hypothetical means = ab/(a+b+1)(a+b)^2 = prior variance
> Expected value of process variance = ab/(a+b+1)(a+b)
>
> Does anyone know why the prior variance for Beta-Binomial equals to VHM
> (i.e.
> ab/(a+b+1)(a+b)^2) and not the sum of VHM and EPV (i.e. ab/(a+b)^2)? I
> always
> thought that (total) variance equals VHM + EPV. In Conjugate Prior
> distributions, it doesn't look like this formula applies. (Same for
> Gamma-Poisson: VHM for Gamma-Poisson equals its variance?)
>
> Thanks for any inputs!
>