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suppose $X_1,X_2,...,X_n$ be a random sample of distribution with Probability density function $f_\theta(x)= \displaystyle \frac{2x}{\theta^2}e^{\displaystyle\frac{-x^2}{\theta^2}}, x>0, \theta>0$. under loss function $L(\delta, \theta)= (\displaystyle\frac{\delta}{\theta}-1)^2$how can I calculate minimax estimator for $\theta$

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    $\begingroup$ I might be wrong, but I don't think there is a general way to find minimax estimators analytically. If you know the minimax estimator, you can verify it though. Another approach is guessing a parametric family of "least-favorable" distributions, then optimizing the parameters until the Bayes estimator is also minimax. Sorry I can't be of much help. You may get a better result asking on the Theoretical Computer Science stack exchange. $\endgroup$ – AaronDefazio Mar 31 '17 at 4:32
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In this problem, as in many, one standard approach to minimaxity is to use least favourable priors, often equal to Jeffreys' priors. Note that $$S=\sum_{i=1}^n X_i^2$$ is a sufficient statistic, distributed as a Gamma $G(n,1/\theta^2)$ and $\theta^2$ is a scale parameter for $S$. I would thus seek the reference prior $\pi(\theta)=1/\theta$ (or is it $1/\theta^2$?) and check whether or not the associated Bayes estimator has a fixed risk. Since this is presumably the invariant prior in Pitman best invariant estimator, the risk, if finite, must be constant!

A (possibly wrong) resolution is found in this paper by W.A. Jasim, establishing that $S/2n+c$ is the minimax estimator with no specification of $c$.

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