15
$\begingroup$

For normally distributed data, the standard deviation $\sigma$ and the median absolute deviation $\text{MAD}$ are related by:

$\sigma=\Phi^{-1}(3/4)\cdot \text{MAD}\approx1.4826\cdot\text{MAD},$

where $\Phi()$ is the cumulative distribution function for the standard normal distribution.

Is there any similar relation for other distributions?

$\endgroup$
  • $\begingroup$ Which distribution did you have in mind? $\endgroup$ – gung Jan 2 '15 at 3:58
  • $\begingroup$ No specific distribution. I just come across some strange sets of data, and I would like to know if there is a possible range of values of the constant... $\endgroup$ – vic Jan 2 '15 at 4:12
  • $\begingroup$ Yes, for many distributions -- but the numbers are different. $\endgroup$ – Glen_b Jan 2 '15 at 13:00
  • 2
    $\begingroup$ If you want to know the possible range of values that could convert MAD to SD, why not ask that in the question? $\endgroup$ – Glen_b Jan 2 '15 at 13:08
  • 2
    $\begingroup$ Please explain what the "MAD" is: it has more than one conventional meaning! (And both of them give the same values for Normal distributions.) $\endgroup$ – whuber Jan 2 '15 at 16:12
8
$\begingroup$

To address the question in comments:

I would like to know if there is a possible range of values of the constant

(I assume the question is intended to be about the median deviation from median.)

  1. The ratio of SD to MAD can be made arbitrarily large.

    Take some distribution with a given ratio of SD to MAD. Hold the middle $50\%+\epsilon$ of the distribution fixed (which means MAD is unchanged). Move the tails out further. SD increases. Keep moving it beyond any given finite bound.

  2. The ratio of SD to MAD can easily be made as near to $\sqrt{\frac{1}{2}}$ as desired by (for example) putting $25\%+\epsilon$ at $\pm 1$ and $50\%-2\epsilon$ at 0.

    I think that would be as small as it goes.

enter image description here

$\endgroup$
  • 2
    $\begingroup$ Is your interpretation of "MAD" the median absolute deviation from the mean or from the median (which is often used and is explicitly the interpretation in Xi'an's answer)? $\endgroup$ – whuber Jan 2 '15 at 16:11
  • 3
    $\begingroup$ @whuber - it's important to be clear, thanks. I'm interpreting it as from the median, as Xi'an has. (Did I make an error somewhere?) $\endgroup$ – Glen_b Jan 2 '15 at 22:32
  • 2
    $\begingroup$ I didn't see any error--it just wasn't clear either in the question or your answer which interpretation was intended (although with some analysis a reader could figure out which one you are using). I recall seeing a question about the deviation-from-mean interpretation only a few weeks ago. $\endgroup$ – whuber Jan 2 '15 at 22:47
4
$\begingroup$

For any given distribution with density $f(x;\theta)$, the median absolute deviation is given by $\text{MAD}_\theta=G^{-1}_\theta(1/2)$ where $G_\theta$ is the cdf of $|X-\text{MED}_\theta|$ and $\text{MED}_\theta=F^{-1}_\theta(1/2)$ where $F_\theta$ is the cdf of $X$.

  1. In cases when $\theta=\sigma$, i.e., when the standard deviation is the only parameter, $\text{MAD}_\theta$ is therefore a deterministic function of $\sigma$.
  2. In cases when $\theta=(\mu,\sigma)$ and $\mu$ is a location parameter, i.e., when $$f(x;\theta)=g(\{x-\mu\}/\sigma)/\sigma$$ Then the distribution of $|X-\text{MED}_\theta|$ is the same as the distribution of $|\{X-\mu\}-\{\text{MED}_\theta-\mu\}|$, and hence is independent from $\mu$. Therefore $G_\theta$ only depends on $\sigma$ and $\text{MAD}_\theta$ is again a deterministic function of $\sigma$.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.