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:

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