# 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!

• What techniques are you familiar with for establishing minimaxity? Apr 5, 2021 at 1:08
• One theorem that I have used majorly is the constant risk function theorem. That was the approach that I would use if a question like this is given to me but I'm not sure how to deploy it on this one. @user257566 Apr 5, 2021 at 1:17
• Which textbook are you using for support? Apr 5, 2021 at 6:22
• Mathematical statistics basic ideas and selected topics, vol I, Bickel and Docksum. @Xi'an Apr 5, 2021 at 14:18

$$X\sim Poisson(\lambda), \pi(\lambda)=Gamma(1,1/k)$$. 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.