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Let $X_1,\cdots,X_n$ iid with density $p_\theta(x) = \theta x^{\theta -1} 1_{[0,1]}(x)$ and $\pi(\theta) = e^{-\theta}$ as prior on the parameterspace $\Theta$. I have to calculate the posterior measure on $\Theta$ and the Bayes estimator of $\theta$. I did the following $$ p_\theta(X)\pi(\theta) = e^{-\theta}\prod_{i=1}^n \theta x_i^{\theta -1} $$

But I dont see what distribution this might be.

Please help :)

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The posterior density $g_x$ is such that $g_x(\theta)$ is proportional to $p_x(\theta)\pi(\theta)$ with $x=(x_i)_{1\leqslant i\leqslant n}$. Writing each $x_i^{\theta-1}$ as $x_i^{-1}\mathrm e^{\theta\log x_i}$, one gets $$ g_x(\theta)=c(x)\,\theta^n\,\exp(-a(x)\theta),\qquad a(x)=1-\sum_{i=1}^n\log(x_i). $$ This is the gamma distribution with parameters $(n+1,a(x))$ hence $$ c(x)=\frac{a(x)^{n+1}}{n!}. $$ Irrespectively of the value of $c(x)$, the density $g_x$ is maximal at $n/a(x)$. On the other hand, the minimum mean square estimator of $\theta$, equal to the mean of the posterior distribution $g_x$, is $$ \hat\theta(x)=\int\theta g_x(\theta)\mathrm d\theta=c(x)\,\frac{(n+1)!}{a(x)^{n+2}}=\frac{n+1}{a(x)}. $$

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  • $\begingroup$ Thanks. I thought that the estimator is given by the epected value of $g$ given the data $X$ thus by $\mathbb E(g(\theta)| X = (x_1,\cdots,x_n)) = \frac { n + 1}{a(x)}$ ? $\endgroup$ – user42761 Apr 1 '13 at 10:17
  • $\begingroup$ See Edit. $ $ $ $ $\endgroup$ – Did Apr 1 '13 at 10:27

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