Showing $X\sim \operatorname{Poi}(\lambda)$ is minimax Assume that $X$ has $\operatorname{Poisson} (\lambda)$ distribution and the loss function is $\ell(\lambda,a)=\frac{(\lambda-a)^2}{\lambda}$. Now, I want to show that $X$ is minimax. A hint that is given is to consider the gamma $\Gamma(k-1,1)$ prior and let $k\rightarrow\infty$. I am pretty confused on how to approach this.
Any help would be appreciated!
 A: Okay I got this one myself.
$X\sim Poisson(\lambda), \pi(\lambda)=Gamma(1,1/k)$. The posterior is obtained as:
$$\pi(\lambda|x)\propto p(x|\lambda)\pi(\lambda)\propto\lambda^xe^{-\lambda}e^{-\lambda/k}\propto\lambda^xe^{-(\lambda(1+1/k))}$$
$$\lambda|x\sim Gamma(X+1,1+1/k)$$
Bayes rule:
$$E[\frac{(\lambda-a)^2}{\lambda}|X]=a^2E(1/\lambda|X)-2a+E(\lambda|X)$$
$$\delta_B(X)=argminE((\lambda-a)^2/\lambda|X)=\frac{1}{E(1/\lambda|X)}$$
$$E(1/\lambda|X)=\frac{(1+1/k)^{X+1}}{\Gamma(X+1)}\int_0^\infty\frac{1}{\lambda}\lambda^Xe^{-\lambda(1+1/k)}d\lambda$$
where it's $\infty$ when $X=0$ and $\frac{1+1/k}{X}$ when $X>0$.
$$\delta_B(X)=\frac{X}{1+1/k}$$
Risk of $\delta_B(X)$:
$$R(\lambda,\delta_B(X))=E(R(\theta,\delta_B))=E(\frac{(\lambda-\frac{x}{1+1/k})^2}{\lambda})=\frac{1+\lambda/k^2}{(1+1/k)^2}$$
Bayes risk:
$$r(\pi,\delta_B)=E(R(\theta,\delta_b))=\frac{1+k/k^2}{(1+1/k)^2}=\frac{1}{1+1/k}$$
Risk of $X$:
$$R(\theta,X)=E(\frac{(\lambda-X)^2}{\lambda})=var(X)/\lambda=1$$
Let $k\rightarrow\infty$:
$$r(\pi,\delta_B)=\frac{1}{1+1/k}\rightarrow 1=R(\theta,X)$$
By Theorem, $\delta=X$ is minimax.
