# Quantile absolute deviation of the Pareto distribution

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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: