If $Z=X+Y$ ($X$ and $Y$ being random samples), what is the relationships between the respective order statistics? Suppose $X$ and $Y$ are two random samples (not necessarily iid, but one can make this assumption) and that $Z=X+Y$.
If one computes the order statistics of $X$ and $Y$, what can be said about the relative order statistic of $Z$?
To be more clear, let $\tilde{X}_{0.99}$ and $\tilde{Y}_{0.99}$ be the 0.99th quantiles of $X$ and $Y$, respectively, does it exist a relationship $f(\cdot)$ with $\tilde{Z}_{0.99}$ (i.e., the 0.99th quantile of $Z$) such that $\tilde{Z}_{0.99}=f(\tilde{X}_{0.99},\tilde{Y}_{0.99})$?
Sorry for the possibly ill-posed question... I'm not a statistician.
 A: You really cannot say very much about $\tilde{Z}_{0.99}$ compared with $\tilde{X}_{0.99}$ and $\tilde{Y}_{0.99}$ without knowing more about the rest of the distributions, even if $X$ and $Y$ are independent. 
For most distributions you will find  $\tilde{Z}_{0.99}  < \tilde{X}_{0.99} + \tilde{Y}_{0.99}$: as an illustration, if $X$ and $Y$ have standard normal distributions then $\tilde{X}_{0.99} = \tilde{Y}_{0.99} \approx 2.326$ so their sum is about $4.653$ but $\tilde{Z}_{0.99} \approx 3.290$.
However it is easy enough to find a counterexample: for example if $X=0$ with probability $0.992$ and $X=1$ otherwise, and $Y$ similarly, then $\tilde{X}_{0.99} = \tilde{Y}_{0.99} = 0$ so their sum is $0$ but $\tilde{Z}_{0.99} =1$.   
In fact it is possible to find a distribution where $\tilde{X}_{0.99}$ and $\tilde{Y}_{0.99}$ take any given value and $\tilde{Z}_{0.99}$ takes any given value.
A: This was a while back but if you can say that $X$ and $Y$ are approximately Normal then: 
$$
Z_{.99}^2 \sim X_{.99}^2 + Y_{.99}^2
$$
So for the example above, $3_{.29}^2 = 2_{.326}^2 + 2_{.326}^2$.
I'm sure you can prove it.
If the distribution is not zero based then you need to subtract the means (medians perhaps?)
So perhaps: 
$$
(Z_{.99}-Z_{.5})^2 \sim (X_{.99}-X_{.5})^2 + (Y_{.99}-Y_{.5})^2
$$
We use it quite a lot.
Would love to see some proof though  :-).
