5
$\begingroup$

I know the result of integrating a Gaussian against another Gaussian is still Gaussian, $$\int N(x|\mu_y,\sigma_y)N(y|\mu,\sigma) dy = N(x|\mu',\sigma')\quad.$$

Can I get the same form for Beta distributions? or is there a $p(x|y)$ that makes the result another Beta distribution? like $$\int p(x|y)~\textrm{Beta}(y|a,b) dy = \textrm{Beta}(x| a', b')\quad.$$

Perhaps a Delta function centered at $y$ will do, $p(x|y)=\delta_y(x)$, so that the resulting Beta distribution stays the same, right? But what if I don't want all the mass at one point?

Any suggestions are welcome. Thanks.

EDIT
I'm not asking for a conjugate prior or likelihood function, if you suggest Binomial please at least add some explanation, it is not that obvious (to me).

EDIT
$$p(y|x)=\frac{p(x|y)p(y)}{p(x)}=\frac{p(x|y)p(y)}{\int p(x|y)p(y)dy}$$ AFAIK, we call $p(y)$ a conjugate prior for $p(x|y)$ if $p(y|x)$ stays the same form as $p(y)$. But I want $p(x)$ to be in the same form as $p(y)$ instead. I'm not quite familiar with statistics, are these the same thing?

comments and corrections by Juho Kokkala (thanks :)

So this could be interpreted as 'If we have a Beta prior for parameter $y$, what would be an observation model $p(x|y)p(x|y)$ so that the prior predictive for $x$ is $\textrm{Beta}$ rather than 'what would be an observation model so that the posterior for $y$ is $\textrm{Beta}$. Where the latter would be answered with Binomial, but not the former.

Note that the first claim about Gaussians does not hold if $\mu_y$, $\sigma_y$ are some general functions of $y$ -- $\sigma_y$ must be a constant and $\mu_y$ an affine function of $y$.

$\endgroup$
  • 5
    $\begingroup$ Binomial distribution. $\endgroup$ – Greenparker Mar 21 '16 at 1:46
  • 4
    $\begingroup$ The Binomial is not a conjugate prior for the Beta: rather, the Beta is the conjugate prior for the Binomial. And that, it seems, is precisely what you are asking. One might say that this question is asking for a "conjugate posterior" for the Beta distribution. As such, @greenparker nailed the answer. $\endgroup$ – whuber Mar 21 '16 at 14:37
  • 2
    $\begingroup$ @whuber Um, I don't see how this is asking for that. Note that the first factor of the integrand is supposed to be $p(x \mid y)$ and the outcome is desired to be a Beta density for $x$. So this could be interpreted as 'If we have a Beta prior for parameter $y$, what would be an observation model $p(x|y)$ so that the prior predictive for $x$ is Beta' rather than 'what would be an observation model so that the posterior for $y$ is Beta'. Where the latter would be answered with Binomial, but not the former. $\endgroup$ – Juho Kokkala Mar 21 '16 at 14:59
  • 2
    $\begingroup$ @dontloo Note that the first claim about Gaussians does not hold if $\mu_y,\sigma_y$ are some general functions of $y$ -- $\sigma_y$ must be a constant and $\mu_y$ an affine function of $y$. $\endgroup$ – Juho Kokkala Mar 21 '16 at 15:01
  • 2
    $\begingroup$ One obvious solution is when $f(x|p)=Beta(x|a',b')$, i.e. when $x$ is independent of $p$. $\endgroup$ – Xi'an Apr 24 '16 at 8:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.