# Quantile absolute deviation of the Exponential distribution

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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: