I think it's true.
My reasoning goes like this (and I kept making mistakes with the algebra/arithmetic so check carefully)
P(S6lt3delta)leP(hboxexactly4Xihboxslessthandelta)+P(hboxexactly5Xihboxslessthandelta)+P(hboxall6Xihboxslessthandelta). So if we let p=P(X1ltdelta), then the right hand part of this inequality is
(p4)(10p2−24p+15). I then plugged 2p−(p4)(10p2−24p+15) into R and got a function that looks always non-negetive between 0 and 1 (shouldn't be too bad to show this by derivatives but I'm lazy and have computer power).
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