# Quantile absolute deviation of the Pareto distribution

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: