Does anyone know what is the name of this formula?
$$M_i = \displaystyle\frac{0.6745(x_i - \hat{x})}{\mathrm{MAD}}$$
where $\textrm{MAD}$ is the median absolute deviation and $\hat{x}$ is the median of $x$.
Does it appear in some scientific publication? I also wonder where the constant comes from (0.6745 is roughly 29/43). I am using it for outlier detection.
Suppose $x$ follows a standard normal distribution.
The $\mathbf{MAD}$ will converge to the median of the half normal distribution, which is the 75% percentile of a normal distribution, and $\mathbf{N}(0.75) \simeq 0.6745$
Since you are multiplying by $(x-\hat{x})$, this means that, for any normal distribution, your formula will converge to 1 for a large enough sample size.
The formula was given by Iglewicz and Hoaglin$^1$ (reference below).
Let the mad for a vector $x$ of $n$ observations be defined as $m(x) = \text{median}(|x- \text{median}(x)|)$. If $x$ is normally distributed, it can be shown that $$ \lim_{n\rightarrow \infty}E(m(x)) = \sigma\Phi^{-1}(0.75) $$ where $\Phi^{-1}(0.75) \approx 0.6745$ is the $0.75^\text{th}$ quantile of the standard normal distribution and is used for consistency. That is, so that $m(x)/0.6745$ is a consistent estimator of the standard deviation $\sigma$.
If you can't assume normality, you can use the 0.75$^\text{th}$ quantile of any other distribution that is symmetric about some value (not necessarly the mean) standardised to have mean 0 and standard deviation 1. Typically a t-distribution is used if fat-tail are assumed.
Iglewicz and Hoaglin suggest using $\pm3.5$ as cut-off value but this a matter of choice ($\pm3$ is also often used).
$^1$ Boris Iglewicz and David Hoaglin (1993), "Volume 16: How to Detect and Handle Outliers", The ASQC Basic References in Quality Control: Statistical Techniques, Edward F. Mykytka, Ph.D., Editor.
