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 .
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 .
$\displaystyle f_1(x|\theta)=\theta(1-\theta)^x$
$\displaystyle f_2(\theta)={\Gamma(\alpha+\beta)\over \Gamma(\alpha)\Gamma(\beta)}\theta^{\alpha-1}(1-\theta)^{\beta-1}$
so the joint density is the product...
$\displaystyle f_3(x,\theta)={\Gamma(\alpha+\beta)\over \Gamma(\alpha)\Gamma(\beta)}\theta^{\alpha}(1-\theta)^{\beta+x-1}$
next obtain the marginal of X....
$\displaystyle f_4(x)={\Gamma(\alpha+\beta)\over \Gamma(\alpha)\Gamma(\beta)}\int_0^1\theta^{\alpha }(1-\theta)^{\beta+x-1}d\theta$
making this look like a Beta, we have...
$\displaystyle =\left({\Gamma(\alpha+\beta)\over \Gamma(\alpha)\Gamma(\beta)}\right)\left({\Gamma(\ alpha+1)\Gamma(\beta+x)\over \Gamma(\alpha+\beta+x+1)}\right)$
Instead of reducing, just divide...
$\displaystyle f_5(\theta|x)={f_3(x,\theta)\over f_4(x)}$
$\displaystyle =\left({\Gamma(\alpha+\beta+x+1)}\over\Gamma(\alph a+1)\Gamma(\beta+x)\right)\theta^{\alpha}(1-\theta)^{\beta+x-1}$
which is a $\displaystyle \beta(\alpha+1,\beta+x)$ distribution.
THIS was way too much typing for me.