Thursday, 24 November 2011

pr.probability - A probabilistic inequality

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)(10p224p+15). I then plugged 2p(p4)(10p224p+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).

No comments:

Post a Comment