Minimax estimator for geometric distribution I'm trying to solve this problem:

Let $X$ be a single sample from Geo($p$) where $p ∈ (0, 1)$. Find a minimax estimator for $p$ under the loss $L(p, δ(x)) = (p−δ(x))^2/
p(1−p)$ .

I'm trying to put different priors for $p$ and find one with constant risk. Because we know Bayesian estimator with constant risk is minimax. But all I got gives me sth like $\delta(X) = 1/X$ and it doesn't have constant risk. Is there any hint?

The weighted quadratic risk of an estimator $\delta$ is given by
$$R(p,\delta)=\sum_{x=0}^\infty (p-\delta(x))^2\times (1-p)^{x-1}$$
Since the first term of this sum is $(p-\delta(0))^2\times (1-p)^{-1}$, it diverges to infinity as $p$ goes to $1$, unless $\delta(0)=1$. In order to secure a finite minimax risk, the minimax estimator $\delta^\star$ must satisfy $$\delta(0)=1\tag{1}\,.$$ This implies that $R(1,\delta^\star)=0$.
Similarly, to achieve a finite risk at $p=0$,
$$R(0,\delta)=(1-0)^{2-1}+\sum_{x=1}^\infty \delta(x)^2<\infty\tag{2}$$
the series
$$\sum_{x=1}^\infty \delta(x)^2$$
must converge. If $\delta^\star(x)=0$ for $x>0$, this is obviously the case, with a risk function equal to
$$R(p,\delta^\star)=(1-p)+\sum_{x=1}^\infty p^2 (1-p)^{x-1}=1-p+p\underbrace{\sum_{x=0}^\infty p (1-p)^{x}}_{=1}=1$$
Therefore $\delta^\star$ has a constant risk function. And since (2) is greater than $1$, $\delta^\star$ is minimax.
