# Quantile absolute deviation of the Pareto 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 Pareto(1,1) distribution.

## Preparation

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

$\newcommand{\E}{\mathbb{E}} \newcommand{\PR}{\mathbb{P}} \newcommand{\Q}{\operatorname{Q}} \newcommand{\QAD}{\operatorname{QAD}} \newcommand{\median}{\operatorname{median}} \newcommand{\Exp}{\operatorname{Exp}} \newcommand{\Pareto}{\operatorname{Pareto}(1, 1)} \QAD(X, p) = \Q(|X - \median(X)|, p),$

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

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

## Pareto distribution

We consider the $$\Pareto$$ distribution given by $$F(x)=1-1/x$$ with the median value $$M=2$$. Since $$F$$ is defined only for $$x \geq x_{\min} = 1$$, we have to consider two cases: $$M - v_p \leq 1$$ and $$M - v_p > 1$$. The critical value $$v_p^*$$ is defined by $$v_p^* = M - x_{\min} = 1$$. Now it is easy to get the value of $$p^*$$ using Equation (1):

$p^* = F(2M - x_{\min}) = F(3) = 2/3.$

Let us consider the first case when $$p \leq p^* = 2/3$$. Equation (1) has the following form:

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

which is the same as

$\frac{1}{2 - v_p} - \frac{1}{2 + v_p} = p.$

By multiplying both sides of this equation on $$(2 - v_p)(2 + v_p)$$, we get:

$(2 + v_p) - (2 - v_p) = p (2 - v_p)(2 + v_p),$

which is the same as

$\frac{2}{p} v_p = 4 - v_p^2.$

Hence,

$v_p^2 + \frac{2}{p} v_p - 4 = 0.$

This is a quadratic equation for $$v_p$$ with coefficients $$a=1$$, $$b=2/p$$, $$c=-4$$. The discriminant is given by $$D = b^2 - 4ac = 4/p^2 + 16$$. The solution of the quadratic equation is

$v_p = \frac{-2/p \pm \sqrt{4/p^2+16}}{2} = \frac{-1}{p} \pm \sqrt{\frac{1}{p^2} + 4}.$

Since $$v_p$$ is always positive, only the plus is applicable for $$\pm$$.

Now let us consider the second case when $$p > p^* = 2/3$$. Equation (1) has the following form:

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

which is the same as

$(2+v_p)(1 - p) = 1.$

Hence,

$2 - 2p + v_p - p v_p = 1.$

From this, we can express $$v_p$$:

$v_p = \frac{2p - 1}{1 - p}.$

Thus, if $$X \sim \Pareto$$,

$\lim_{n \to \infty} \E[\QAD(X, p)] = \begin{cases} \frac{-1}{p} + \sqrt{\frac{1}{p^2} + 4}, & \textrm{if}\; p \leq 2/3,\\ \frac{2p - 1}{1 - p}, & \textrm{if}\; p > 2/3. \end{cases}$

Here is the corresponding plot: