# Median absolute deviation (MAD) and SD of different distributions

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?

• Which distribution did you have in mind? Jan 2, 2015 at 3:58
• 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...
– vic
Jan 2, 2015 at 4:12
• Yes, for many distributions -- but the numbers are different. Jan 2, 2015 at 13:00
• If you want to know the possible range of values that could convert MAD to SD, why not ask that in the question? Jan 2, 2015 at 13:08
• Please explain what the "MAD" is: it has more than one conventional meaning! (And both of them give the same values for Normal distributions.)
– whuber
Jan 2, 2015 at 16:12

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.

• 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)?
– whuber
Jan 2, 2015 at 16:11
• @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?) Jan 2, 2015 at 22:32
• 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.
– whuber
Jan 2, 2015 at 22:47

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 $\sigma$.