# Quantile absolute deviation of the Exponential distribution

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: