I come across the beta distribution quite frequently when solving exercises for my statistics class. However, I have not been able to fully grasp how to work with it.

Exponential family form is:

$$ f(x)=h(x)c(\theta)\Big(\sum W_i(\theta)t_i(x)\Big) $$

Beta distribution is defined as:

$$ f(x) = \frac{x^{\alpha-1}(1-x)^{\beta-1}}{B(\alpha,\beta)}$$

Where $\alpha$, $\beta$ or both may be unknown.

Also when arranging the Beta distribution in the form of exponential family:

$$ f(x)= \frac{1}{B(\alpha,\beta)} e^{(\alpha-1)ln(x)(\beta-1)ln(x)}$$

Hence, $h(x)=I_{x\in(0,1)}(x)$ (If you see that $h(x) = 1$, that is a cue to use an indicator function that ranges through the support of $x$

What does this mean?





How are these determined?

One of the exercises specified that the distribution is $B(\alpha,1)$ where $\alpha$ is unknown $\alpha>0$. Then the distribution becomes:

$$ f(x)= \alpha x^{\alpha -1}, 0<x<1$$

How can this be derived?

I have been reading about the Beta distribution, but i cannot find anything that is easy to understand for someone with minimal mathematical background.

  • $\begingroup$ You know what the form of an exponential family is. But you do not seem to know what the functions involved in the pdf are. I suggest you look into your textbook and get this clarified. Once this is clear, it would mean that $(\ln x,\ln (1-x))$ is a jointly sufficient statistic for $(\alpha,\beta)$ when both parameters are unknown. For the special case, $\ln x$ would be sufficient for $\alpha$ by a similar logic. $\endgroup$ – StubbornAtom Aug 12 '18 at 18:12
  • $\begingroup$ And for the special case, you mean $\alpha$ is unknown since we are estimating $\alpha$. Also, you are not being careful while properly writing the density. You have to match the density with the form of the exponential family structure. Note that $\theta=(\alpha,\beta)$ is a vector when both parameters are unknown. $\endgroup$ – StubbornAtom Aug 12 '18 at 18:18
  • $\begingroup$ @StubbornAtom you are right, $\alpha$ is unknown in the special case. Regardless, why is $f(x)=\alpha x^{\alpha-1}?$. $\endgroup$ – user1607 Aug 12 '18 at 22:44
  • $\begingroup$ $$f(x)=\frac{x^{\alpha-1}(1-x)^{1-1}}{B(\alpha,1)}=\frac{\Gamma(\alpha+1)x^{\alpha-1}}{\Gamma(\alpha)\Gamma(1)}=\frac{\alpha\Gamma(\alpha)x^{\alpha-1}}{\Gamma(\alpha)}$$ $\endgroup$ – StubbornAtom Aug 13 '18 at 3:37

The Beta distribution is useful for modeling random variables between $0$ and $1$

If $X \sim Exp(\lambda)$ is an exponentially distributed random variable then the mass function of

$$f(x;\lambda ) =\begin{align}\begin{cases} \lambda e^{-\lambda x} & \textrm{ for } x \geq 0 \\ \\ 0 & \textrm{ for } x < 0 \end{cases} \end{align}$$

Now if we add two exponentially distributed random variables $X=X_{1}+X_{2}$ we get

$$f_{X}(x) = \int_{-\infty}^{\infty} f_{X_{1}}(x-x_{2})f_{X_{2}}(x_{2}) dx_{2} $$ $$ f_{X}(x) = \int_{0}^{x} \lambda e^{- \lambda (x-x_{2})}\lambda e^{-\lambda x_{2} } dx_{2}$$ $ f_{X}(x) = \int_{0}^{x} \lambda^{2} e^{-\lambda x} dx_{2}$ $ \lambda^{2} x e^{-\lambda x}$ $f_{X}(x) =\begin{align}\begin{cases} \lambda^{2} x e^{-\lambda x} & \textrm{ for } x \geq 0 \\ \\ 0 & \textrm{ for } x < 0 \end{cases} \end{align}$

Now the gamma density as you said is

$ f(x,\alpha, \beta) = \frac{\beta^{\alpha}}{\Gamma(\alpha)}x^{\alpha-1}e^{-\beta x}$

So if we have $ X \sim Gamma(1,\lambda)$ we get an exponential distribution clearly.

$ f(x,1,\lambda) = \frac{\lambda}{\Gamma(1)}x^{0}e^{-\lambda x} = \lambda e^{-\lambda x}$

now the gamma function is simply

$$\Gamma(x) = \int_{0}^{\infty} u^{x-1} e^{-u} du$$

The connection with beta distribution is that the beta density is given as

$$f(x;\alpha, \beta) = \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}x^{\alpha-1} (1-x)^{\beta-1}$$ $$=\frac{1}{B(\alpha,\beta)}x^{\alpha-1} (1-x)^{\beta-1}$$ note here $$= \frac{x^{\alpha-1}(1-x)^{\beta-1}}{\int_{0}^{1}u^{\alpha}(1-u)^{\beta-1} du}$$ $B(\alpha, \beta) $is that integral on the bottom. essentially they are methods of talking about continuous probability distributions in time and the parameters of their shape.

Fairly clearly when we insert $\beta =1$ we will get an exponential density again

$$=\frac{1}{B(\alpha,1)}x^{\alpha-1} (1-x)^{0} = \frac{1}{B(\alpha,1)}x^{\alpha-1} = \frac{\Gamma(1+\alpha)}{\Gamma(1)\Gamma(\alpha)}x^{\alpha-1}$$ Note that $\Gamma(1) = 1 $, $ \Gamma(1+\alpha) = \alpha!$ , $\Gamma(\alpha) = (\alpha-1)! $ then we have

$$\frac{\Gamma(1+\alpha)}{\Gamma(1)\Gamma(\alpha)}x^{\alpha-1}= \frac{\alpha!}{(\alpha-1)!}x^{\alpha-1} = \frac{\alpha * (\alpha-1)!}{(\alpha-1)!}x^{\alpha-1} = \alpha x^{\alpha-1} $$

Also note now..that if $a=b=1$ then our density is

$$f(x;1, 1) = \frac{\Gamma(1+1)}{\Gamma(1)\Gamma(1)}x^{1-1} (1-x)^{1-1} =1$$

which is a uniform density

  • $\begingroup$ edited, it was a misunderstanding on my part. $\endgroup$ – user3417 Aug 13 '18 at 10:20

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