# How to prove 'For a symmetric distribution with zero mean, the population MAD is the 75th percentile of the distribution'

For a symmetric distribution with zero mean, the population MAD is the 75th percentile of the distribution

• For a symmetric distribution with mean zero, the median is also zero. The MAD is thus the median of the absolute deviation from 0. Look at all the positive values of $X$. What is the number $\theta$ such that half the positive values are larger than $\theta$ and half the positive are smaller? Clearly the 75th percentile. By symmetry, half the negative values of $X$ are smaller than the 25th percentile and half are larger. Exercise: what's symmetry got to do with the matter? Why won't it work for $f_X(x)=\begin{cases}1,&-0.5<x<0,\\0.5e^{-x},&x>0,\end{cases}~~~~????$ – Dilip Sarwate Feb 17 '18 at 4:54
(Proof basically copied from comments) For a symmetric distribution with mean zero, the median is also 0. Then, it is enough to look at the positive part of the distribution. We want a number $\theta$ such that the probability a positive observation is larger than $\theta$ is 1/2, and the probability it is lower than $\theta$ is also one half. But that is clearly the third quartile $Q_3$. By symmetry we get the corresponding result for the negative part of the distribution as $Q_1$, the first quartile (whose absolute value is $Q_3$). That should establish that the MAD, median absolute deviation from median is $Q_3$.
• One needs to interpret this argument carefully: the MAD will be a version of Q3 but it might differ from a conventional Q3. Consider the dataset $(-2,-1,0,0,0,0,1,2)$ for instance: although its Q3 is usually taken to be $0,$ the MAD is the median of $(2,1,0,0,0,0,1,2),$ which conventionally is $1/2.$ Thus, the two R implementations as function(x) quantile(x,3/4) and function(x) median(abs(x-median(x))) may return different values. – whuber Oct 2 '19 at 16:56