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Consider the sum of 2 independent random variables $P$ and $Q$ with unknown distributions.

It is known that the variance of the sum of the variables is the sum of the variances.

Is there any relation that allows to estimate what happens to the Mean Absolute Deviation (MAD) of the sum, knowing the $P$ e $Q$ MADs? For example, is there some possibility of the $MAD / \sigma$ grows, if the $MAD / \sigma$ for $P$ and $Q$ are similar? In some simulations that I've made it seems that this ratio never increases.

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You have not stated whether you are taking the mean of absolute deviations from the mean or the median, but this counter-example to your conjecture works for both. Suppose $P$ takes the value $0$ with probability $\frac12$ and the values $+1$ or $-1$ each with probability $\frac14$, and that $Q$ is independent of $P$ though with the same distribution

Then the mean (and median) of $P$ (and $Q$) is $0$, the variance is $\frac12$, the standard deviation $\frac1{\sqrt{2}}$, and the mean absolute deviation is $\frac12$. The mean absolute deviation divided by the standard deviation is $\frac1{\sqrt{2}} \approx 0.707$

Here $P+Q$ takes the value $0$ with probability $\frac38$, the values $+1$ or $-1$ each with probability $\frac14$, and the values $+2$ or $-2$ each with probability $\frac18$. So the mean (and median) of $P+Q$ is $0$, the variance is $1$, the standard deviation $1$, and the mean absolute deviation is $\frac34$. The mean absolute deviation divided by the standard deviation is $\frac34 = 0.75$, so higher than before, contrary to your conjecture

I would not be surprised if the mean absolute deviation of $P+Q$ is always less than or equal to the sum of the mean absolute deviations of $P$ and of $Q$, with equality in some special cases (e.g. one of the mean absolute deviations being $0$, or $P$ and $Q$ having some form of dependence) and being able to get arbitrarily close to equality in independent cases with positive mean absolute deviations, particularly when $P$ and $Q$ are highly leptokurtic

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  • $\begingroup$ It's a great counterexample, but it's a discrete and not smooth distribution. It may perhaps have a class of continuous distribution functions where the MAD / $\sigma $ ratio would always fall with a sum of independent P and Q distributions. $\endgroup$ – Paulo Buchsbaum Apr 23 '19 at 19:09
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    $\begingroup$ @PauloBuchsbaum - take the sum of two i.i.d. Laplace distributed random variables and you will get pretty much the same increase for the ratio with $P+Q$. Take two iid normal distributions and the ratio for $P+Q$ will be unchanged. Take two iid uniform distributions and the ratio for $P+Q$ will fall. So there is no general rule, but it seems to be related somewhat to kurtosis $\endgroup$ – Henry Apr 23 '19 at 23:47
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Comment; Not that is obvious to me.

One of the disadvantages of dealing with absolute values is that they do not share the 'nice' mathematical properties of squares.

Here is a simple simulation, illustrating that the sum of MADs is not MAD of sum.

set.seed(421);  m = 10^6
x = rexp(m, 1);  y = runif(m,0,1);  s = x+y
v.x = var(x);  v.y = var(y);  v.s = var(s)
v.x;  v.y;  v.x+v.y;  v.s
[1] 1.00127
[1] 0.08335595
[1] 1.084626    ## essentially
[1] 1.084834    ##    equal

mad.x = mean(abs(x-mean(x)));  mad.y = mean(abs(y-mean(y)))
mad.s = mean(abs(s-mean(s)))
mad.x;  mad.y;  mad.x+mad.y;  mad.s
[1] 0.7360768
[1] 0.2499935
[1] 0.9860703   ## clearly
[1] 0.7671409   ##    unequal
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  • $\begingroup$ I know that! I use MAD in my heuristics not for statistical reasons. From sd (P) = d, sd (Q) = f d, sd (P + Q) = sqrt (1 + f ^ 2) d, MAD (P) / SD (P) Q) ~ u, I'm trying to prove MAD (P + Q) / SD (P + Q) <= u $\endgroup$ – Paulo Buchsbaum Apr 21 '19 at 20:25

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