Quantile absolute deviation of the Uniform distribution
In this post, we derive the exact equation for the quantile absolute deviation around the median of the Uniform distribution.
Preparation
We consider the quantile absolute deviation around the median defined as follows:
\[\newcommand{\E}{\mathbb{E}} \newcommand{\PR}{\mathbb{P}} \newcommand{\Q}{\operatorname{Q}} \newcommand{\QAD}{\operatorname{QAD}} \newcommand{\median}{\operatorname{median}} \QAD(X, p) = \Q(|X - \median(X)|, p), \]
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
\[F(M + v_p) - F(M - v_p) = p. \tag{1} \]
Uniform distribution
We consider the standard uniform distribution \(\mathcal{U}(0, 1)\) given by the CDF \(F(x)=x\) on \([0;1]\) with the median value \(M=0.5\). From(1), we have:
\[(0.5 + v_p) - (0.5 - v_p) = p, \]
which is the same as
\[v_p = p / 2. \]
Thus, if \(X \sim \mathcal{U}(0, 1)\),
\[\lim_{n \to \infty} \E[\QAD(X, p)] = \frac{p}{2}. \]
Here is the corresponding plot: