Aggregation of distributions: Independent vs. perfectly rank correlated Suppose $X_1, X_2,\ldots,X_n$ are random variables distributed according to p.d.f's $p_{X_1},p_{X_2},\ldots,p_{X_n}$ (most of which are extremely right-skewed). We randomly sample $S$ values from each distribution. 
Consider two ways of aggregating these distributions:
a. Assuming that $X_1, X_2,\ldots,X_n$ are all independent of each other, in which case, we can randomly sum up values from each distribution and arrive at an aggregated distribution represented by $p_\text{indep-aggr}$. Let the 90th percentile be denoted by $P90_\text{indep-aggr}$.
b. Assuming that $X_1, X_2,\ldots,X_n$ are perfectly rank correlated, in which case, we can sum up the percentile values for each distribution to arrive at the respective percentile value of the resulting aggregated distribution $p_\text{dep-aggr}$. Let the 90th percentile be denoted by $P90_\text{dep-aggr}$.
Here are my questions:


*

*Should $P90_\text{dep-aggr}$ be greater than $P90_\text{indep-aggr}$? Can this be proved?

*Should the interquartile range be greater for $p_\text{dep-aggr}$ than for $p_\text{indep-aggr}$

*How about variance? Should $\sigma^2_{p_\text{dep-aggr}} > \sigma^2_{p_\text{indep-aggr}}$?

 A: I looked at your question since I found it interesting. I think the answer to question 1) is "no". Here is a conter-exemple :
Take $n=2$ and consider $X_1$ and $X_2$ following a Bernoulli distribution $\mathcal{B}(0.9)$. 
Then, if $X_1$ and $X_2$ are independent : $P90_{indep\_aggr} = 1$
If $X_1$ and $X_2$ are "perfectly rank correlated", $P90_{dep\_aggr} = 0$.
In fact, the notion of a vector of random variable being perfectly rank correlated is often referred as being comonotonic. For instance I found this article : https://lirias.kuleuven.be/bitstream/123456789/118650/1/OR_0119.pdf
Apparently, the right way to understand the domination of the comonotonic sum $S_{dep} = X_1 + ... + X_n$ upon the iid sum $S_{indep} = X_1 + ... + X_n$ at the tail of the distribution is to think in terms of convex order. 
(i.e., in the article it is shown that for all non-decreasing and convex function $f$, $E[f(S_{indep})] \le E[f(S_{dep})]$)
In your case, you were wondering if, for $d = P90_{dep\_aggr}$ :
$P(S_{indep} > d) = E[1_{S_{indep} > d}] \le 0.1$ (if yes we would have $P90_{indep\_aggr} \le d$)
But the function $x \mapsto 1_{x > d}$ is not convex hence the theorem doesn't apply !!
