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Consider random variables X_ij, where i is in {1,...,T} and j is in {1,...,M}. Can we compare these two statements?

Sum{over j}(95th percentile {over i} of X_ij) and 95th percentile {over i}(Sum{over j} of X_ij)

I personally conjecture that 95th percentile of the sum of random variables is less than sum of 95th percentile of the random variables. I am not sure how to go about rigorously proving it though.

ShN
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  • Did you find an answer to your question? – Confounded Jan 26 '17 at 19:10
  • If that can help, the distribution of the quantile $q$ of $n$ drawings of a random variable following $\text{cdf}_X(x)$, follows the Binomial term $\displaystyle\binom n{nq}\text{cdf}^{nq}_X(x)(1-\text{cdf}_X(x))^{np}$, where $p:=1-q$. Of course, you obtain the expectation by integration on $x$. –  Feb 16 '22 at 11:47

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If all your random variables have the same distribution and are perfectly correlated (i.e. they always take the same random value) then the percentile of the sum will be exactly the same as the sum of percentiles.

user2566092
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  • However, if the random variables are independent, the percentiles of the sum of the independent random variables will have a narrow range. – Brian Borchers Jun 25 '22 at 23:19