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

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.

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