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 | &lt; 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?

Question

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      |    &amp;lt;      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?

flasheadita237m · Accepted Answer

Step 1Uniform 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 2This is a cdf which is uniformly continuous and for which       ∑          n        F  (      γ    n    ) does not converge for any   0  &amp;lt;  γ  &amp;lt;  1.For additional assumptions, it suffices for F to be   α-Hölder continuous at zero for some   α  &amp;gt;  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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