Unbiased estimator for the smaller of two random variables Suppose $X \sim \mathcal{N}(\mu_x, \sigma^2_x)$ and $Y \sim \mathcal{N}(\mu_y, \sigma^2_y)$
I am interested in $z = \min(\mu_x, \mu_y)$.  Is there an unbiased estimator for $z$?
The simple estimator of $\min(\bar{x}, \bar{y})$ where $\bar{x}$ and $\bar{y}$ are sample means of $X$ and $Y$, for example, is biased (though consistent).  It tends to undershoot $z$.
I can't think of an unbiased estimator for $z$.  Does one exist?
Thanks for any help.
 A: This is just a couple of comments not an answer (don't have enough rep. point).
(1). There is an explicit formula for the bias of the simple estimator $min(\bar{x},\bar{y})$ here: 
Clark, C. E. 1961, Mar-Apr. The greatest of a ﬁnite set of random variables.
Operations Research 9 (2): 145–162.
Not sure how this helps though
(2). This is just intuition, but I think such an estimator doesn't exist. If there is such an estimator, it should also be unbiased when $\mu_x=\mu_y=\mu$. Thus any 'downgrading' which makes the estimator less than say the weighted average of the two sample means make the estimator biased for this case.
A: You are right that an unbiased estimator doesn't exist. The problem is that the parameter of interest is not a smooth function of the underlying data distribution due to non-differentiability at $\mu_x=\mu_y$.
The proof is as follows. Let $T(X,Y)$ be an unbiased estimator. Then $E_{\mu_x,\mu_y}[T(X,Y)]=\min\{\mu_x,\mu_y\}$. The left-hand side is differentiable everywhere with respect to $\mu_x$ and $\mu_y$ (differentiate under the integral sign). However, the right-hand side is not differentiable at $\mu_x=\mu_y$, which leads to a contradiction.
Hirano and Porter have a general proof in a forthcoming Econometrica paper (see their Proposition 1). Here is the working paper version.
A: There is an estimator for the minimum (or the maximum) of a set of numbers given a sample. See Laurens de Haan, "Estimation of the minimum of a function using order statistics," JASM, 76(374), June 1981, 467-469.
A: I'd be fairly sure an unbiased estimator does not exist.  But unbiased estimators don't exist for most quantities, and unbiasedness is not a particularly desirable property in the first place.  Why do you want one here?
