0
$\begingroup$

Show that the family of beta distributions where parameters $α$ and $β$ are unknown is an exponential family.

I know that the beta distribution is $f(x; \alpha, \beta)={1\over B(\alpha, \beta)}x^{\alpha -1}(1-x)^{\beta -1}$ and that $B(\alpha, \beta)=\int_0 ^1x^{\alpha - 1}(1-x)^{\beta -1}dx={\Gamma(\alpha)\Gamma(\beta)\over \Gamma(\alpha +\beta)}$.

So this becomes $f(x; \alpha, \beta)={\Gamma(\alpha+\beta)\over\Gamma(\alpha)\Gamma(\beta)}x^{\alpha -1}(1-x)^{\beta -1}={\Gamma(\alpha+\beta)\over\Gamma(\alpha)\Gamma(\beta)}{x^\alpha \over x}{(1-x)^{\beta} \over (1-x)}=[x(1-x)]^{-1}[e^{(\alpha \ln{x}+\beta \ln{1-x})}+\ln{\Gamma (\alpha+\beta)}-\ln{\Gamma(\alpha)}-\ln{(\beta)}]$

I'm not sure how to show it is an exponential family. Any help is greatly appreciated.

$\endgroup$
1
$\begingroup$

According to Wikipedia, a distribution is in the exponential family if its PDF can be expressed as $$ f(x;\vec{\theta})=h(x)g(\vec{\theta})\exp\big(\vec{\eta}(\vec{\theta})^\text{T}\vec{T}(x)\big). $$ (The functions $\eta$ and $T$ are vector-valued). Exponentiating (as you did), we have $$ x^{\alpha-1}=\exp\big((\alpha-1)\log{x}\big) $$ and $$ (1-x)^{\beta-1}=\exp\big((\beta-1)\log(1-x)\big). $$ Hence $$ f(x;\alpha,\beta)=\frac{1}{B(\alpha,\beta)}\exp\big((\alpha-1)\log{x}+(\beta-1)\log(1-x)\big). $$ From this you can pick out the functions $h$, $g$, $\vec{\eta}$ and $\vec{T}$.

$\endgroup$
  • $\begingroup$ How would I choose $h$ and $g$? I see the other two inside the exp expression. $\endgroup$ – ddswsd Jan 26 at 4:25
  • $\begingroup$ Forget $h$ for a second--can you see $g$? For the case of the Beta distribution, $\vec{\theta}=(\alpha,\beta)$. $\endgroup$ – David M. Jan 26 at 4:26
  • $\begingroup$ Let g be ${\Gamma(\alpha+\beta)\over\Gamma(\alpha)\Gamma(\beta)}$? $\endgroup$ – ddswsd Jan 26 at 4:29
  • $\begingroup$ That's the idea $\endgroup$ – David M. Jan 26 at 4:30

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.