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.

  • 1
    $\begingroup$ How are you using this for outlier detection? Presumably, you are comparing $M$ to some threshold--and it would be unreasonable to suppose it is equal to $1$. What will matter is the ratio between that threshold and $0.6745$; the actual value of $0.6745$ by itself is not terribly meaningful for this purpose. $\endgroup$
    – whuber
    Nov 13, 2014 at 18:24
  • $\begingroup$ If the absolute value of $M_i$ is larger than three I flag the observation as an outlier. $\endgroup$
    – synonym
    Nov 13, 2014 at 18:29
  • 1
    $\begingroup$ It would be practically the same thing just to compute $M_i^\prime = (x_i-\hat x)/\text{MAD}$ and compare it to $2 \approx 3\times 0.6745$. That might be a little bit simpler to explain and interpret, too. $\endgroup$
    – whuber
    Nov 13, 2014 at 18:36
  • $\begingroup$ B. D. Ripley refers to MAD/0.6745 as the "MAD estimator". See page 2 and 3 of stats.ox.ac.uk/pub/StatMeth/Robust.pdf. For the normal distrubution. MAD is approximately equal to 0.6745 x the standard deviation $\endgroup$ Nov 17, 2014 at 3:34
  • $\begingroup$ Dead link above, now at: web.archive.org/web/20120410072907/http://www.stats.ox.ac.uk/… $\endgroup$
    – Matti Wens
    Nov 4, 2020 at 21:14

2 Answers 2


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.

  • $\begingroup$ Thank you, this was helpful. So I suppose the formula does not have any specific name? $\endgroup$
    – synonym
    Nov 13, 2014 at 18:34
  • $\begingroup$ I don't follow your last sentence. How will $M_i$ 'converge to 1'? $\endgroup$
    – Glen_b
    Nov 14, 2014 at 1:13
  • $\begingroup$ the average of M_i $\endgroup$
    – Arthur B.
    Nov 14, 2014 at 14:57

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.


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