# Posts / Quantile absolute deviation of the Normal distribution

In this post, we derive the exact equation for the quantile absolute deviation around the median of the Normal 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

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

We consider the standard normal distribution $\mathcal{N}(0, 1)$ given by the CDF $F(x)=\Phi(x)$ with the median value $M=0$. From (1), we have:

$$ \Phi(v_p) - \Phi(-v_p) = p. $$Using $\Phi(-v_p) = 1 - \Phi(v_p)$, we get:

$$ \Phi(v_p) = \frac{p+1}{2}, $$which is the same as

$$ v_p = \Phi^{-1} \Big( \frac{p+1}{2} \Big). $$Thus, if $X \sim \mathcal{N}(0, 1)$,

$$ \lim_{n \to \infty} \E[\QAD(X, p)] = \Phi^{-1} \Big( \frac{p+1}{2} \Big). $$Here is the corresponding plot:

### Backlinks (1)

- Optimal quantile absolute deviation (2022-08-30)