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Sum of percentiles compared to percentile of the sum
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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.
percentile
asked Apr 10 '15 at 17:13
ShNShN
1
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add a comment |
$begingroup$
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.
percentile
asked Apr 10 '15 at 17:13
ShNShN
1
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Did you find an answer to your question?
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– Confounded
Jan 26 '17 at 19:10
add a comment |
$begingroup$
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.
percentile
asked Apr 10 '15 at 17:13
ShNShN
1
$endgroup$
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.
percentile
percentile
asked Apr 10 '15 at 17:13
ShNShN
1
asked Apr 10 '15 at 17:13
ShNShN
1
asked Apr 10 '15 at 17:13
ShNShN
1
asked Apr 10 '15 at 17:13
ShNShN
1
asked Apr 10 '15 at 17:13
ShNShN
1
1
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Did you find an answer to your question?
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– Confounded
Jan 26 '17 at 19:10
add a comment |
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Did you find an answer to your question?
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– Confounded
Jan 26 '17 at 19:10
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Did you find an answer to your question?
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– Confounded
Jan 26 '17 at 19:10
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Did you find an answer to your question?
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– Confounded
Jan 26 '17 at 19:10
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1 Answer
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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.
answered Apr 10 '15 at 17:25
user2566092user2566092
21.5k1947
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$begingroup$
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.
answered Apr 10 '15 at 17:25
user2566092user2566092
21.5k1947
$endgroup$
add a comment |
$begingroup$
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.
answered Apr 10 '15 at 17:25
user2566092user2566092
21.5k1947
$endgroup$
add a comment |
$begingroup$
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.
answered Apr 10 '15 at 17:25
user2566092user2566092
21.5k1947
$endgroup$
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.
answered Apr 10 '15 at 17:25
user2566092user2566092
21.5k1947
answered Apr 10 '15 at 17:25
user2566092user2566092
21.5k1947
answered Apr 10 '15 at 17:25
user2566092user2566092
21.5k1947
answered Apr 10 '15 at 17:25
user2566092user2566092
21.5k1947
21.5k1947
add a comment |
add a comment |
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– Confounded
Jan 26 '17 at 19:10