Unknown Y has a galenshore (a,$\theta$) distribution if the density is
f(y|a,$\theta$)= $\frac{2}{\gamma(a)}\theta^{2a}y^{2a-1}e^{-\theta^2y^2}$
Assume a is known and $y_1…y_n$ independent observations from a galensh
Suppose a galenshore($\theta,\beta$) prior is assumed for Param $\theta$, show that the posterior distribution of $\theta$ given $y_1…y_n$ Is a galenshore $(na+\alpha,\sqrt{\beta^2+\sum_{i=1}^{n}Y_i^2})$
I understand that posterior is likelihood times prior, have them all but I have been stuck on this for so long now I have no idea how is simplifies to that posterior.
Best Answer
First, you have an error in your question; you mean that the prior desntiy of $\theta$ is Galenshore with hyperparameters $\alpha$ and $\beta$, not $\theta$ and $\beta$.
Next, ignore all factors that are not functions of the posterior parameter $\theta$. That is to say, look only at the kernel: $$f(\boldsymbol y \mid \theta) = f(y_1, \dots, y_n \mid \theta) = \prod_{i=1}^n \frac{2}{\Gamma(a)} \theta^{2a} y_i^{2a-1} e^{-\theta^2 y_i^2} \propto (\theta^{2a})^n e^{-\theta^2 \sum y_i^2}.$$ We don't care about $2/\Gamma(a)$ or $y_i^{2a-1}$ because these factors are not functions of $\theta$. Similarly, $$f(\theta \mid \alpha, \beta) = \frac{2}{\Gamma(\alpha)} \beta^{2\alpha} \theta^{2\alpha-1} e^{-\beta^2 \theta^2} \propto \theta^{2\alpha - 1} e^{-\beta^2 \theta^2}.$$ Consequently, $$\begin{align*} f(\theta \mid \boldsymbol y) &\propto f(\boldsymbol y \mid \theta)f(\theta \mid \alpha,\beta) \\ &\propto (\theta^{2a})^n e^{-\theta^2 \sum y_i^2} \theta^{2\alpha - 1} e^{-\beta^2 \theta^2} \\ &= \theta^{2(na+\alpha)-1} e^{-\theta^2 (\beta^2 + \sum y_i^2)} \end{align*}$$ which is the kernel of a density from the same parametric family with posterior parameters $na + \alpha$ and $\sqrt{\beta^2 + \sum_{i=1}^n y_i^2}$.