# Unbiased median absolute deviation for n=2

Update: this blog post is a part of research that aimed to build an unbiased median absolute deviation estimator based on various quantile estimators. A preprint with final results is available on arXiv: arXiv:2207.12005 [stat.ME]. Some information in this blog post can be obsolete: please, use the preprint as the primary reference.

I already covered the topic of the unbiased median deviation based on the traditional sample median, the Harrell-Davis quantile estimator, and the trimmed Harrell-Davis quantile estimator. In all the posts, the values of bias-correction factors were evaluated using the Monte-Carlo simulation. In this post, we calculate the exact value of the bias-correction factor for two-element samples.

Let $$X = \{ X_1, X_2 \}$$ be a sample of two random variables from the standard normal distribution $$\mathcal{N}(0, 1^2)$$. Regardless of the chosen median estimator, the median is unequivocally determined:

$\operatorname{median}(X) = \dfrac{X_1 + X_2}{2}.$

Now let’s calculate the median absolute deviation $$\operatorname{MAD}$$:

$\begin{split} \operatorname{MAD}(X) & = \operatorname{median}(|X - \operatorname{median}(X)|) = \\ & = \operatorname{median}(\{ \, |X_1 - (X_1 + X_2)/2|\,,\, |X_2 - (X_1 + X_2)/2|\, \}) = \\ & = \operatorname{median}(\{ \, |(X_1 - X_2)/2|\,,\, |(X_2 - X_1)/2| \,\}) = \\ & = |X_1 - X_2|/2. \end{split}$

Since $$X_1, X_2 \sim \mathcal{N}(0, 1^2)$$ which is symmetrical, $$|X_1 - X_2|/2$$ is distributed the same was as $$|X_1 + X_2|/2$$. Let’s denote the sum of two standard normal distributions as $$Z = X_1 + X_2$$. It gives us another normal distribution with modified variance:

$Z \sim \mathcal{N}(0, \sqrt{2}^2).$

Since we take the absolute value of $$Z$$, we get the half-normal distribution. The expected value of a half-normal distribution formed from the normal distribution $$\mathcal{N}(0, \sigma^2)$$ is $$\sigma \sqrt{2/\pi}$$. Thus,

$\mathbb{E}(|Z|) = \sqrt{2} \sqrt{2/\pi} = 2/\sqrt{\pi}.$

Finally, we have:

$\begin{split} \mathbb{E}(\operatorname{MAD}(X)) & = \mathbb{E}\Bigg( \frac{|X_1 - X_2|}{2} \Bigg) = \mathbb{E}\Bigg( \frac{|X_1 + X_2|}{2} \Bigg) = \\ & = \mathbb{E}\Bigg( \frac{|Z|}{2} \Bigg) = \frac{2/\sqrt{\pi}}{2} = \frac{1}{\sqrt{\pi}}. \end{split}$

The bias-correction factor $$C_2$$ for $$\operatorname{MAD}(X)$$ is the reciprocal value of its expected value:

$C_2 = \frac{1}{\mathbb{E}(\operatorname{MAD}(X))} = \sqrt{\pi} \approx 1.77245385090552.$