# 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.

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.