# The difference of normal means is also minimax?

Let $$X_i \sim N(\xi, \sigma^2)$$ and $$Y_i \sim N(\eta, \tau^2)$$ for known $$\sigma^2$$ and $$\tau^2$$.

I know that $$\bar{X}$$ and $$\bar{Y}$$ are minimax under squared error loss since their variance is fixed, and a sequence of Bayes estimators can be constructed such that their Bayes risk converge to the maximum risk of $$\bar{X}$$ and $$\bar{Y}$$.

I am wondering how I can show $$\delta(X,Y) =\bar{Y}-\bar{X}$$ is also minimax for $$\eta-\xi$$?

Essentially I need to show that for any $$T(X,Y)$$, we have

$$\sup_{\xi,\eta} E(T(X,Y)-(\eta-\xi))^2 \leq \sup_{\eta,\xi} E((\bar{Y}-\bar{X})-(\eta-\xi))^2 = \frac{\sigma^2+\tau^2}{n}$$

The only thing I can think of doing is constructing another sequence of priors whose Bayes risks converge to the RHS. However, this seems kind of tedious now that we're in 2 dimensions. I feel like there's a "trick" here that I should be using?

• It looks to me that for any permutation $\sigma,$ the sequence $Y_i-X_{\sigma(i)}$ is an iid sample from $N(\eta-\xi, \tau^2+\sigma^2)$ under the implicit assumptions of (1) independence of all $X_i$ and $Y_j$ and (2) equinumerous datasets. Apply your assertion (after "I know") to this sequence. – whuber May 29 '19 at 11:31
• I concur with @whuber that the problem is identical to a single Normal problem – Xi'an May 29 '19 at 12:37
• I understand why it would be minimax for a dataset $(X_i-Y_i)$, and I also understand how all permutations $\sigma(i)$ would give an equivalent distribution. But doesn't this assume the minimax estimator is of the form $g(X_i-Y_i)$? What about other linear estimators $g(aX_i-bY_i)$ etc? If that makes sense – Xiaomi May 30 '19 at 10:04
• It makes sense. The concept of sufficient statistic is at the bottom of this: a sufficient statistic for $\eta-\xi$ is $\bar Y-\bar X;$ apart from nonzero multiples of this, no other linear combination is sufficient. The proof lies in examining the likelihood function. – whuber Jun 5 '19 at 21:59