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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.

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  • $\begingroup$ well the percentiles of a continuous random variable are characterized by the quantile function (the inverse of the CDF). I'm not sure if you will be able to derive a relationship between the quantile function of $Z$ and those of $X,Y$. If it is at all possible, I'd start by noting that the density of $Z$ will be the convolution between the densities of $X$ and $Y$. Using that fact you may be able to get a relationship between the cdf of $Z$ and those of $X$ and $Y$ and perhaps you can go from there. $\endgroup$
    – Macro
    Commented Jul 24, 2011 at 20:56
  • $\begingroup$ The problem here is that I don't know the real underlying distribution since samples $X$ and $Y$ are just observations from a complex system. $\endgroup$
    – seg.fault
    Commented Jul 25, 2011 at 9:39

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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.

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  • $\begingroup$ As stated in the above comment, I don't know the exact theoretical distributions generating $X$ and $Y$. What I can surely say is the support of such distribution(s) is greater than 0 with a known lower bound and possibly no upper bound. Said that, the only distribution I can work with is the empirical one. $\endgroup$
    – seg.fault
    Commented Jul 25, 2011 at 9:50
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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 :-).

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  • $\begingroup$ sorry second 2.326 also squared $\endgroup$
    – Bob Jordan
    Commented Dec 5, 2012 at 4:47
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    $\begingroup$ Fixed it for you. $\endgroup$
    – Glen_b
    Commented Dec 5, 2012 at 4:56

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