Let X be an absolutely continuous random variable with strictly positive pdf f and cdf F. Moreover, let gamma in (0,1). I am interested when sum_{n=1}^{infty} P(|X|<gamma^n)=sum_{n=1}^{infty}F(gamma^n)-F(-gamma^n) converges. One option would be to impose that F is K-Lipschitz continuous, since then F(gamma^n)-F(-gamma^n) le 2K gamma^n and thus the series reduces to a geometric one. Is it also sufficient if F is assumed to be uniformly continuous? If not, are there additional assumptions that do ensure this is enough?

benatudq 2022-10-22 Answered
Convergence of geometric series of probabilities.
Let X be an absolutely continuous random variable with strictly positive pdf f and cdf F. Moreover, let γ ( 0 , 1 ). I am interested when n = 1 P ( | X | < γ n ) = n = 1 F ( γ n ) F ( γ n ) converges. One option would be to impose that F is K-Lipschitz continuous, since then F ( γ n ) F ( γ n ) 2 K γ n and thus the series reduces to a geometric one. Is it also sufficient if F is assumed to be uniformly continuous? If not, are there additional assumptions that do ensure this is enough?
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Answers (1)

flasheadita237m
Answered 2022-10-23 Author has 17 answers
Step 1
Uniform continuity is insufficient. Let F ( x ) = { 0 x e 2 1 / 2 1 / ln ( 1 / x ) e 2 x 0 1 / 2 + 1 / ln ( 1 / x ) 0 x e 2 1 x e 2 .
Step 2
This is a cdf which is uniformly continuous and for which n F ( γ n ) does not converge for any 0 < γ < 1.
For additional assumptions, it suffices for F to be α-Hölder continuous at zero for some α > 0. So if F ( x ) = 1 / 2 + sign ( x ) | x | in a neighborhood of 0, then your sum converges, even though F is not Lipschitz.
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