 # Quantile absolute deviation of the Normal distribution

Update: this blog post is a part of research that aimed to build a new measure of statistical dispersion called quantile absolute deviation. A preprint with final results is available on arXiv: arXiv:2208.13459 [stat.ME]. Some information in this blog post can be obsolete: please, use the preprint as the primary reference.

In this post, we derive the exact equation for the quantile absolute deviation around the median of the Normal distribution.

## Preparation

We consider the quantile absolute deviation around the median defined as follows:

$\newcommand{\E}{\mathbb{E}} \newcommand{\PR}{\mathbb{P}} \newcommand{\Q}{\operatorname{Q}} \newcommand{\QAD}{\operatorname{QAD}} \newcommand{\median}{\operatorname{median}} \QAD(X, p) = \Q(|X - \median(X)|, p),$

where $$\Q$$ is a quantile estimator.

We are looking for the asymptotic value of $$\QAD(X, p)$$. For simplification, we denote it by $$v_p$$:

$v_p = \lim_{n \to \infty} \E[\Q(|X-M|, p)],$

where $$M$$ is the true median of the distribution.

By the definition of quantiles, this can be rewritten as:

$\PR(|X_1 - M| < v_p) = p,$

which is the same as

$\PR(-v_p < X_1 - M < v_p) = p.$

Hence,

$\PR(M - v_p < X_1 < M + v_p) = p.$

If $$F$$ is the CDF of the considered distribution, the above equality can be rewritten as

$F(M + v_p) - F(M - v_p) = p. \tag{1}$

## Normal distribution

We consider the standard normal distribution $$\mathcal{N}(0, 1)$$ given by the CDF $$F(x)=\Phi(x)$$ with the median value $$M=0$$. From (1), we have:

$\Phi(v_p) - \Phi(-v_p) = p.$

Using $$\Phi(-v_p) = 1 - \Phi(v_p)$$, we get:

$\Phi(v_p) = \frac{p+1}{2},$

which is the same as

$v_p = \Phi^{-1} \Big( \frac{p+1}{2} \Big).$

Thus, if $$X \sim \mathcal{N}(0, 1)$$,

$\lim_{n \to \infty} \E[\QAD(X, p)] = \Phi^{-1} \Big( \frac{p+1}{2} \Big).$

Here is the corresponding plot: