r*****o
发帖数: 7 | 1 f(x) is a probability distribution function
Suppose f(x) > 0 when x is not equal to + infinity or - infinity.
Is it possible that the integration of 1/f(x) over (-t, t) is a finite value
as t goes to infinity?
If it is possible, could you please give me an example? Thank you. | B****n
发帖数: 11290 | 2 impossible. You can use the cauchy schwarz inequality to prove.
g1=sqrt(f) g2=1/sqrt(f)
inf=(\int(g1*g2))^2<=\int(g1^2)*\int(g2)^2
【在 r*****o 的大作中提到】
: f(x) is a probability distribution function
: Suppose f(x) > 0 when x is not equal to + infinity or - infinity.
: Is it possible that the integration of 1/f(x) over (-t, t) is a finite value
: as t goes to infinity?
: If it is possible, could you please give me an example? Thank you.
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