It is known that integral convolution will preserve quasi-convexity for PF2 distribution.

If F(x) is quasi-convex and the density \( \phi \) is PF2, then, \(\int_0 ^\infty F(x-\xi) \phi(\xi) d\xi\) will be quasi-convex.

I have a problem with a 3 dimensional function. \(\int_0 ^\infty F(x-\xi,\xi)\phi(\xi) d\xi \) = quasi-convex? if F is quasi-convex in x ?

I want to know how to prove.


asked 02 Aug '11, 13:28

ksphil's gravatar image

accept rate: 14%

edited 03 Feb '14, 01:30

Read, understand, and master Samuel Karlin "Total Positivity Volume 1" (note that there was never a Volume 2, which would have addressed the multivariate theory had it ever been completed and published; but you can get by with a solid mastery of the univariate theory, as presented in Volume 1, and "grow your own" multivariate extensions). Then the answer will be immediately evident. And inequalities, relating to convexity in any form, will spring forth as a veritable mathematical fountain.


answered 03 Feb '14, 00:42

Mark%20L%20Stone's gravatar image

Mark L Stone
accept rate: 15%

edited 03 Feb '14, 01:05


About 2 and a half years. I am glad I got an answer. I will check out the book from library and will try. Thank you.

(03 Feb '14, 01:33) ksphil
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Asked: 02 Aug '11, 13:28

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Last updated: 03 Feb '14, 01:33

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