Why the Pitman estimator is given by the sample mean of X and Y?

Question

Let $(X,Y)$ be bivariate normally distributed with $E[X] = E[Y] = \theta$, $Var[X] = Var[Y] = 1$ and $cov[X, Y] = \rho, |\rho| < 1$, where $\rho$ and $\theta$ are unknown. Find the minimum risk location equivariant estimator (the Pitman estimator) for $\theta$ with respect to squared error loss $L(\theta, d)=(d−\theta)^2$.

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I am confused why the Pitman estimator is given by the sample mean of X and Y (i.e. $\hat{\theta}=\frac{1}{2}(X+Y)$). I compute that $$ R(\theta, \hat{\theta})=E[(\hat{\theta}-\theta)^2]=\frac{1}{2}(1+\rho) $$

So this is the risk of the estimator, which is minimized when ρ is minimized. Since |ρ| < 1, the minimum value of ρ is -1, so the minimum risk is 1/2(1 - 1) = 0.

But how to find that the sample mean is the MREE?

Answer 1

The general formula for the best equivariant estimator is $$\delta(x)=\frac{\int_{-\infty}^{\infty} \theta f(x_1-\theta,\dots,x_n-\theta)\,\text d\theta}{\int_{-\infty}f(x_1-\theta,\dots,x_n-\theta)\,\text d\theta}$$ which formally equates with the posterior expectation under the improper (Lebesgue) prior $\pi(\theta)=c$, an arbitrary constant. The posterior distribution of $\theta$ given $(X,Y)=(x,y)$ is \begin{align*} \pi(\theta|x,y) &\propto f(x,y|\theta)\\ &\propto \exp\frac{-\left\{(x-\theta)^2+(y-\theta)^2-2\rho(x-\theta)(y-\theta)\right\}}{2(1-\rho^2)}\\ &\propto \exp-\frac{2\theta^2-2\theta(x+y)-2\rho\theta^2+2\rho\theta(x+y)}{2(1-\rho^2)}\\ &=\exp\left\{-\theta^2(1-\rho)+2(1-\rho)\theta(x+y)\right\}\big/(1-\rho^2)\\ &\propto \exp-\frac{(1-\rho)\left\{\theta-(x+y)\right\}^2}{(1-\rho^2)}\\ &= \exp-\frac{\left\{\theta-(x+y)\right\}^2}{1+\rho}\\ \end{align*} which shows that the posterior distribution on $\theta$ has expectation $$(x+y)\big/2$$ which is indeed the Pitman estimator.

About the comment that the minimum risk is equal to $0$, I want to stress that this is not directly relevant. Each value of $\rho$ defines a different model, with a corresponding risk. That the value of the risk is zero for $\rho=-1$ is incidental, if correct since $Y=2\theta-X$ with probability one in that case.