An unbiased estimate of the median

Question

Suppose we have a random variable $X$ supported on $[0,1]$ from which we can draw samples. How can we come up with an unbiased estimate of the median of $X$?

We can, of course, generate some samples and take the sample median, but I understand this will not in general be unbiased.

Note: this question is related, but not identical, to my last question, in which case $X$ could only be sampled approximately.

Answer 1

Such an estimator does not exist.

The intuition is that the median can stay fixed while we freely shift probability density around on both sides of it, so that any estimator whose average value is the median for one distribution will have a different average for the altered distribution, making it biased. The following exposition gives a little more rigor to this intuition.

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We focus on distributions $F$ having unique medians $m$, so that by definition $F(m) \ge 1/2$ and $F(x) \lt 1/2$ for all $x \lt m$. Fix a sample size $n \ge 1$ and suppose that $t: [0,1]^n \to [0,1]$ estimates $m$. (It will suffice that $t$ only be bounded, but usually one doesn't seriously consider estimators that produce obviously impossible values.) We make no assumptions about $t$; it does not even have to be continuous anywhere.

The meaning of $t$ being unbiased (for this fixed sample size) is that

$$E_F[t(X_1, \ldots, X_n)] = m$$

for any iid sample with $X_i \sim F$. An "unbiased estimator" $t$ is one with this property for all such $F$.

Suppose an unbiased estimator exists. We will derive a contradiction by applying it to a particularly simple set of distributions. Consider distributions $F = F_{x,y,m, \varepsilon}$ having these properties:

1. $0 \le x \lt y \le 1$;

2. $0 \lt \varepsilon \lt (y-x)/4$;

3. $x + \varepsilon \lt m \lt y - \varepsilon$;

4. $\Pr(X = x) = \Pr(X = y) = (1-\varepsilon)/2$;

5. $\Pr(m-\varepsilon \le X \le m+\varepsilon) = \varepsilon$; and

6. $F$ is uniform on $[m-\varepsilon, m+\varepsilon]$.

These distributions place probability $(1-\varepsilon)/2$ at each of $x$ and $y$ and a tiny amount of probability symmetrically placed around $m$ between $x$ and $y$. This makes $m$ the unique median of $F$. (If you are concerned that this is not a continuous distribution, then convolve it with a very narrow Gaussian and truncate the result to $[0,1]$: the argument will not change.)

Now, for any putative median estimator $t$, an easy estimate shows that $E[t(X_1, X_2, \ldots, X_n)]$ is strictly within $\varepsilon$ of the average of the $2^n$ values $t(x_1, x_2, \ldots, x_n)$ where the $x_i$ vary over all possible combinations of $x$ and $y$. However, we can vary $m$ between $x + \varepsilon$ and $y - \varepsilon$, a change of at least $\varepsilon$ (by virtue of conditions 2 and 3). Thus there exists an $m$, and whence a corresponding distribution $F_{x,y,m,\varepsilon}$, for which this expectation does not equal the median, QED.

Answer 2

Finding an unbiased estimator without having a parametric model would be difficult! But you could use bootstrapping, and use that to correct the empirical median to get an approximately unbiased estimator.

Answer 3

I believe quantile regression will give you a consistent estimator of the median. Given the model $Y = \alpha + u$. And you want to estimate $\text{med}(y) = \text{med}(\alpha + u) = \alpha + \text{med}(u)$ since $\alpha$ is a constant. All you need is the $\text{med}(u) = 0$ which should be true so long as you have independent draws. However, as far as unbiasedness, I don't know. Medians are difficult.