Chebychev-like inequality based on the median absolute deviation (about the median) Is it possible to bound the fractions of observations that lie outside $m \pm k \lambda$ where $m$ is the median of the data and $\lambda$ is the median absolute deviation (about the median)? Basically, a robust version of Chebychev's inequality? 
$$
   P(|X - m| > k\lambda) < f(k)
$$
I've seen reference to similar inequality for the mean absolute deviation about the median and the median absolute deviation about the mean, both of which have a $1/k$ relationship but which either are centered around the mean or which use the mean absolute deviation for the scale [Pham-Gia and Hung]. I'm wondering if these can be extended to the more robust estimates used in the median absolute deviation.
 A: If $\lambda$ in the inequality is median absolute deviation, it can easily be zero for some distribution and you'll never get bound $f(k)$ that converges to zero. One simple example.
$$P(X=-1) = 0.1, P(X=0) = 0.8, P(X=1) = 0.1$$ Obviously median absolute deviation $\lambda=0$, since we have $P(|X|\le 0) \ge 0.5$ and $P(|X|\ge 0)\ge0.5$. So you get $$P(|X-m|>k\lambda)=P(|X|>0)=0.2$$ for any $k$. 
Therefore in order to get the general result you desire, you should restrict the behavior of the distribution of $X$ somehow. You can easily see that $X$ being continuous random variable is not enough because you can always have a continuous approximation of the above example.
A: It does make sense when you think about the mean absolute deviation about the median instead of the median absolute deviation about the median.
Consider the survival function $S(y)  = P(Y\geq y)$ of the transformed variable $Y = \frac{|X-m|}{\lambda}$.
In the image below is an example for the Laplace distribution and the Normal distribution.

This survival function $S(y)$ has two properties

*

*$S(y)$ must be monotonic decreasing function.

*The integral of the function is equal to the mean absolute deviation about the median and must integrate to one since we have a standardized variable $\int_0^\infty S(y) dy = 1$
For a given value $S(k)$ we have that the integral has a lower limit (because of the monotonic decreasing property)
$$1=\int_0^\infty S(y) dy \geq S(k) \cdot k$$
This we can convert in a limit for $S(k)$
$$S(k) = P\left(\frac{|X-m|}{\lambda}\geq k\right) \leq \frac{1}{k}$$
which you can convert into
$$P({|X-m|}\geq k\lambda) \leq \frac{1}{k}$$
This type of generalization is also described here for higher moments: https://en.m.wikipedia.org/wiki/Chebyshev%27s_inequality#Higher_moments
