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? 
 A: 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 $X$. 


*

*In cases when $\theta=\sigma$, i.e., when the standard deviation is
the only parameter, $\text{MAD}_\theta$ is therefore a deterministic
function of $\sigma$.

*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 $\sigma$.

A: 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.)

*

*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.


*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.

