# How does one derive the Fisher-consistency correction factor for Rousseeuw and Croux's $S_n$ (and $Q_n$)?

In "Alternatives to the Median Absolute Deviation" (Rousseeuw and Croux, J. Amer. Statistical Assoc, 88(424), 1993, pp.1273–1283) and a few other papers from the same authors published in 1992–1993, Rousseeuw and Croux introduce their robust scale metrics $S_n$ and $Q_n$. They offer correction factors for finite sample size and for Fisher consistency for the normal distribution.

In particular, they find that the correction factor $c$ for Fisher consistency for the normal distribution satisfies $$\Phi\left(\Phi^{-1}(3/4)+c^{-1}\right)-\Phi\left(\Phi^{-1}(3/4)-c^{-1}\right) = \frac{1}{2}.\tag{1}$$ In words, $c$ is such that the interval centered at $\Phi^{-1}(3/4)$ and with width $2 c^{-1}$ covers exactly 50% of the probability mass of the normal distribution. On the other hand, more or less by definition of $S_n$, we have (for any distribution) $$c^{-1} = \text{med}_X (\text{med}_Y \ \left\vert X - Y \right\vert),\tag{2}$$ where $\text{med}_Z$ stands for the median over some random variable $Z$.

In simulation experiments, I found $(1)$ to be true, but I don't see how the argument works, and the paper doesn't offer any help. Does anyone see the trick?

My hope is that the argument generalizes to other distributions sometimes. Indeed it seems (from experiments) that for $X \sim \text{Exp}(\lambda)$, we have $$F\left(F^{-1}(r)+c^{-1}\right)-F\left(F^{-1}(r)-c^{-1}\right) = \frac{1}{2}$$ with $r \approx 0.6$ (but for the beta distribution there is no such formula). Is it known if this is true exactly for any $r$, and if so, what is that $r$? More generally, are there other ways than equation $(2)$ to obtain a formula (possibly implicit, such as $(1)$) for $c$ for non-normal distributions?

• An observation: Let $Z_1, Z_2,\ldots,Z_n$ be iid $\mathcal N(0,1)$. Note that $\mathbb P(|Z| \leq z) = 2 \Phi(z) - 1$, so the median $m$ of the distribution of $|Z|$ solves $\mathbb P(|Z| \leq m) = 1/2$, hence $m = \Phi^{-1}(3/4)$. Since the normal distribution is continuous, the sample quantiles converge everywhere, implying $\hat{m}_n = \mathrm{median}\{Z_1,\ldots,Z_n\} \to \Phi^{-1}(3/4)$ in probability. In particular, this means that if $X_1,\ldots,X_n$ are iid $\mathcal N(0,\sigma^2)$, then $\mathrm{median}\{X_1,\ldots,X_n\} / \Phi^{-1}(3/4)$ is consistent for $\sigma$. – cardinal Mar 7 '12 at 23:55
• it's really late here, but i would think there is an assumption here that F has a center-outward decreasing density (i.e. in lay term that F is unimodal). – user603 Mar 8 '12 at 0:46
• Of course, I dropped the modulus signs above inside the median. It should have read $\hat m_n = \mathrm{median}\{|Z_1|,\ldots,|Z_n|\}$ and $\mathrm{median}\{|X_1|,\ldots,|X_n|\}$. – cardinal Mar 8 '12 at 19:09