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Let $X_{1},...,X_{n}$ be a random sample of size n from the continuous distribution with pdf:

$f_{X}(x|\alpha,\beta) = \frac{2*\beta^{\alpha}}{\Gamma(\alpha)}*(\frac{1}{x})^{2*\alpha+1}*\exp({\frac{-\beta}{x^{2}}})*I_{(0,\infty)}(x)$

where $\alpha$ > 0 is fixed and $\beta$ > 0. Assume the prior distribution on $\beta$ has the pdf:

$\pi(\beta|\lambda) = \lambda*e^{-\lambda*\beta}$

where $\lambda$ > 0 is fixed. Find the Bayes estimator of $\beta$.

So to find the Bayes estimator of $\beta$ I know I need to find the posterior distribution, which I did, and I got

$\pi(\Theta|X)$ = $e^{\frac{\lambda*\beta*x^{2}-\beta}{x^{2}}*2*\beta^{\alpha}}$

which I am questioning if it is right, especially since I'm not sure my marginal distribution of x is right, either, but I thought that since both $\alpha$ and $\lambda$ were fixed, then I could treat them like constants, but I wasn't sure what I need to replace in the joint to make sure that it integrated to 1 so then it just 'disappeared' because it integrated to 1. Also, to find the Bayes estimator I also know that I need to take the conditional expected value of the posterior distribution, but when I'm not sure if my posterior is correct, I didn't want to start that. Any help would be greatly appreciated!

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Let $\mathcal{\Theta}$ be the parameter space and $L: \Theta \times \Theta \mapsto \mathcal{R}$ be a loss function. Let $X$ be your data. The Bayes Estimator is the one which minimizes your expected posterior loss, that is:

$$\hat{\theta}_{\text{Bayes}} = \arg \inf_{\hat{\theta}}E[L(\hat{\theta},\theta)|X]$$

Your question is incomplete in the sense that the Bayes Estimator depends on the loss function (and you didn't provide any). A very common loss function is the squared loss, $L(\hat{\theta},\theta) = (\hat{\theta}-\theta)^{2}$. For this loss function, it can be shown that $\hat{\theta}_{\text{Bayes}} = E[\theta|X]$, that is the posterior mean of $\theta$.

When we wish to compute the posterior we usually drop constants, and only compute it up to a constant of proportionality:

$$f(\beta|x) \propto f(\beta)f(x|\beta) \propto$$ $$\propto e^{-\lambda \beta} \beta^{\alpha} \exp{(\beta/x^{2})} = \beta^{\alpha} e^{-(\lambda+ x^{-2})\beta} $$

This is the pdf you wrote rearranged in another way. Recall that the part which is a variable here is $\beta$. The constant of proportionality is not 1. Before trying to solve this integral you should ask yourself: Is this distribution known? This is one of the famous distributions!

This is a Gamma Distribution with parameters $\alpha+1$ and $\lambda+x^{-2}$.

the Bayes Estimator (for the squared error loss function) is just it's mean:

The mean of a Gamma$(\alpha+1,\lambda+x^{-2})$ is $\frac{\alpha+1}{\lambda+x^{-2}}$.

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  • $\begingroup$ I think what confused me was the 1/x part of it, because when I tried doing it again, I eventually got to it being gamma, but because of the 1/x, I thought it might be the inverted gamma distribution. But Thanks for the help! $\endgroup$ – user61752 Feb 11 '13 at 6:21

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