Median absolute deviation (MAD) and SD of different distributions

Question

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?

Answer 1

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.

Answer 2

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 $\sigma$.