Unbiased median absolute deviation for n=2

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:

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