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!

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    $\begingroup$ What techniques are you familiar with for establishing minimaxity? $\endgroup$ – user257566 Apr 5 at 1:08
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    $\begingroup$ 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 $\endgroup$ – statwoman Apr 5 at 1:17
  • $\begingroup$ Which textbook are you using for support? $\endgroup$ – Xi'an Apr 5 at 6:22
  • $\begingroup$ Mathematical statistics basic ideas and selected topics, vol I, Bickel and Docksum. @Xi'an $\endgroup$ – statwoman Apr 5 at 14:18

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.


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