# Unbiased estimate of sign of mean

Consider the set $\mathcal{P}$ of probability distributions that have a finite first moment and define the function $\operatorname{sgn} :\mathcal{P} \to \mathbb{R}$ as $$\operatorname{sgn}(\mu) = \begin{cases} -1 &\text{ if } \mathbb{E}_{\mu}[x] < 0 \\ +1 &\text{ if } \mathbb{E}_{\mu}[x] \geq 0 \end{cases}$$

Is there an unbiased estimator for this statistic? i.e. some function $S$ operating on $n$ points such that $\mathbb{E}_{\mu}[S(x_1,\dots,x_n)] = \operatorname{sgn}(\mu)$ ?

Or if not in the general case, can I find an estimator for special cases? (Normal distribution, etc)?

I tried looking at the 1 point estimate $$S(x) = \begin{cases} -1 &\text{ if } x < 0 \\ +1 &\text{ if }x \geq 0 \end{cases}$$ and my analysis shows that $\mathbb{E}_{\mu}[S(x)] = 1 - 2\mu(x<0)$, but I don't think this is equivalent to the function $\operatorname{sgn}$.

• It's an intriguing question. Since the sign of the mean is a discontinuous function of the distribution, there won't be an unbiased estimator in general; and for specific distribution families, you would need there to be a discontinuity in the distributions as the mean crosses zero. (That rules out the Normal family, for instance.) A trivial example of the latter would be a family with all non-negative means (such as the Gamma or Lognormal) where the constant estimator $1$ is unbiased! – whuber Sep 20 '18 at 2:16
• @whuber Thanks for your comment.. Any pointers on how I would begin to prove the impossibility of unbiased estimation for such discontinuous statistics? – ttb Sep 23 '18 at 18:56
• Contemplate a distribution with a mean just less than zero. Slightly change the probabilities to makes its mean just greater than zero. How would you determine which of these two distributions governs any particular sample? This thought experiment translates into a more formal argument: the shifting of distributions can be carried out along a continuous path in the space of distributions. The expectation of any statistic, because it depends continuously on the probabilities, must be a continuous function of that path--but that's impossible if its expectation equals the sign of the mean. – whuber Sep 23 '18 at 19:46