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Let $X_1,...,X_n$ be independent r.v.'s from the poisson $P(\theta)$ ,$\theta \in (0,\infty)$.

and consider loss function $L(\theta ; \delta ) =\frac{[\theta - \delta(x_1,...,x_n)]^2}{\theta}$.

then risk $R(\theta ; T ) =1/n$ where $T=\frac{X_1+...+X_n}{n}$.

Then $T$ is minimax?

If there is distribution $\lambda$ on $(0,\infty)$ such that $T$ is bayes rule for $\lambda$, then $T$ is minimax because $R(\theta ; T)$.

Could you help me?

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  • $\begingroup$ Is this a question from a course or textbook? If so, please add the [self-study] tag & read its wiki. $\endgroup$ – gung♦ Nov 4 '16 at 14:05
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    $\begingroup$ Hints: (1) what does having a constant risk function tell you about the estimator? (2) What would you guess for a prior distribution on $\theta$ that leads to this estimator $T$? $\endgroup$ – whuber♦ Nov 4 '16 at 17:55
  • $\begingroup$ @whuber (1) I don't know. (2) I think $\theta$ is distributed by negative exponential. $\endgroup$ – Planche Nov 5 '16 at 4:12
  • $\begingroup$ Your answer to (1) indicates you should study the properties of minimax estimators before proceeding. Among other things, they have constant risk functions. $\endgroup$ – whuber♦ Nov 5 '16 at 20:06

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