 # Quantile absolute deviation of the Exponential 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 Exponential distribution.

## Preparation

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

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

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

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

## Exponential distribution

We consider the standard exponential distribution $$\Exp(1)$$ given by the CDF $$F(x)=1 - e^{-x}$$ with the median value $$M=\ln 2$$. Since $$F$$ is defined only for $$x > x_{\min} = 0$$, we have to consider two cases: $$M - v_p \leq 0$$ and $$M - v_p > 0$$. The critical value $$v_p^*$$ is defined by $$v_p^* = M - x_{\min} = M = \ln 2$$. Now it is easy to get the value of $$p^*$$ using (1):

$p^* = F(2M - x_{\min}) = F(2M) = 1 - e^{-2\ln 2} = 1 - 0.25 = 0.75.$

Let us consider the first case when $$p \leq p^* = 0.75$$. From (1), we have:

$(1 - e^{-\ln 2 - v_p}) - (1 + e^{-\ln 2 + v_p}) = p,$

which is the same as

$e^{v_p} - e^{-v_p} = 2p.$

By multiplying both sides of the equation by $$e^{v_p}$$, we get:

$(e^{v_p})^2 - 2p \cdot (e^{v_p}) - 1 = 0.$

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

$e^{v_p} = \frac{-b \pm \sqrt{D}}{2a} = \frac{2p \pm \sqrt{4p^2 + 4}}{2} = p \pm \sqrt{p^2+1}.$

Since $$e^{v_p}$$ is always positive, only the plus is applicable for $$\pm$$. Taking the natural logarithm from both parts, we get the result:

$v_p = \ln(p + \sqrt{p^2+1}).$

Now let us consider the second case when $$p > p^* = 0.75$$. Equation (1) has the following form:

$(1 - e^{-\ln 2 - v_p}) = p,$

which is the same as

$e^{-\ln 2 - v_p} = 1 - p.$

Taking the natural logarithm from both parts, we can easily express $$v_p$$:

$v_p = -\ln 2 - \ln (1 - p).$

Thus, if $$X \sim \Exp(1)$$,

$\lim_{n \to \infty} \E[\QAD(X, p)] = \begin{cases} \ln(p + \sqrt{p^2+1}), & \textrm{if}\; p \leq 0.75,\\ -\ln 2 - \ln (1 - p), & \textrm{if}\; p > 0.75. \end{cases}$

Here is the corresponding plot: