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Let $X$ and $Y$ be continuous random variables both having some density, not identically distributed but independent.

Imagine I'm interested in the quantile $q_{X+Y}(\alpha)$ for some $\alpha \in (0,1)$.

Does it then hold that

$$q_{X+Y}(\alpha)=\int\limits_{-\infty} ^{\infty} f_X(x)(x+q_{Y} (\alpha)) dx$$

So the idea of this expression would be the following:

Go over all possible values of $X$. If $X$ has some fixed value $x$, then the quantile of $X+Y$ is just the quantile of $Y$ plus some constant $x$.

But I might be very wrong in my intuition

Ivan
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  • Try constructing a counterexample. Note that integral is equal to $EX$ plus $q_Y(\alpha)$. – Jarle Tufto May 27 '21 at 09:40
  • @Jarle Tufto Oh damn, then my expression looks very wrong - while I look for a counter example: Would there be any similar expression that is actually correct? – Ivan May 27 '21 at 09:45
  • Okay, already if the two distributions are normally distributed it seems to be violated – Ivan May 27 '21 at 10:08
  • Did you delete your comment? Anyway, thanks for sharing this similar question. Now I know that quantiles seems to suck as you can't really do a lot with it unless you know the very specific distribution... :-) – Ivan May 27 '21 at 11:28
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    Yes, there is also a good answer to a similar question at https://stats.stackexchange.com/questions/134976/compute-quantile-of-sum-of-distributions-from-particular-quantiles – Jarle Tufto May 27 '21 at 11:30
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    Since the answer is "no," this isn't a useful way to pursue the ideas behind your question. What do you *really* want to know? How to find quantiles of a sum of independent random variables? There are general formulas for that. Or would you have two particular variables in mind? If so, why not ask directly about them? – whuber May 27 '21 at 15:38

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