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?
[self-study]
tag & read its wiki. $\endgroup$ – gung - Reinstate Monica Nov 4 '16 at 14:05