# Math Help - beta distribution problem

1. ## beta distribution problem

Hi, I am stuck at this proof problem. Please help! Thanks.

Let X1,,,,,Xn be a random sample from a geometric distribution with parameter θ, i.e.,
f(xi|θ) =θ(1-θ)^xi. Show that the beta distribution

(for ) is conjugate for θ .

2. Originally Posted by kqed123
Hi, I am stuck at this proof problem. Please help! Thanks.

Let X1; : : : ;Xn be a random sample from a geometric distribution with parameter θ, i.e.,
f(xi|θ ) = θ(1-θ )^xi . Show that the beta distribution
f(θ  ;α,β ) =Γ(α+ β)/(Γ(α)Γ(β))*(θ^(α-1))*((1-θ)^( β-1))

(for θ Œ [0; 1]) is conjugate for θ .
Please either learn to use our LaTeX system or spell out the non-ASCII charaters you are using because we can't all read them.

CB

3. $f_1(x|\theta)=\theta(1-\theta)^x$

$f_2(\theta)={\Gamma(\alpha+\beta)\over \Gamma(\alpha)\Gamma(\beta)}\theta^{\alpha-1}(1-\theta)^{\beta-1}$

so the joint density is the product...

$f_3(x,\theta)={\Gamma(\alpha+\beta)\over \Gamma(\alpha)\Gamma(\beta)}\theta^{\alpha}(1-\theta)^{\beta+x-1}$

next obtain the marginal of X....

$f_4(x)={\Gamma(\alpha+\beta)\over \Gamma(\alpha)\Gamma(\beta)}\int_0^1\theta^{\alpha }(1-\theta)^{\beta+x-1}d\theta$

making this look like a Beta, we have...

$=\left({\Gamma(\alpha+\beta)\over \Gamma(\alpha)\Gamma(\beta)}\right)\left({\Gamma(\ alpha+1)\Gamma(\beta+x)\over \Gamma(\alpha+\beta+x+1)}\right)$

Instead of reducing, just divide...

$f_5(\theta|x)={f_3(x,\theta)\over f_4(x)}$

$=\left({\Gamma(\alpha+\beta+x+1)}\over\Gamma(\alph a+1)\Gamma(\beta+x)\right)\theta^{\alpha}(1-\theta)^{\beta+x-1}$

which is a $\beta(\alpha+1,\beta+x)$ distribution.

THIS was way too much typing for me.