benatudq

2022-10-22

Convergence of geometric series of probabilities.

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 $\sum _{n=1}^{\mathrm{\infty}}P(|X|<{\gamma}^{n})=\sum _{n=1}^{\mathrm{\infty}}F({\gamma}^{n})-F(-{\gamma}^{n})$ 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?

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 $\sum _{n=1}^{\mathrm{\infty}}P(|X|<{\gamma}^{n})=\sum _{n=1}^{\mathrm{\infty}}F({\gamma}^{n})-F(-{\gamma}^{n})$ 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?

flasheadita237m

Beginner2022-10-23Added 17 answers

Step 1

Uniform continuity is insufficient. Let $F(x)=\{\begin{array}{ll}0& x\le -{e}^{-2}\\ 1/2-1/\mathrm{ln}(-1/x)& -{e}^{-2}\le x\le 0\\ 1/2+1/\mathrm{ln}(1/x)& 0\le x\le {e}^{-2}\\ 1& x\ge {e}^{-2}\end{array}$.

Step 2

This is a cdf which is uniformly continuous and for which $\sum _{n}F({\gamma}^{n})$ does not converge for any $0<\gamma <1$.

For additional assumptions, it suffices for F to be $\alpha $-Hölder continuous at zero for some $\alpha >0$. So if $F(x)=1/2+\text{sign}(x)\sqrt{|x|}$ in a neighborhood of 0, then your sum converges, even though F is not Lipschitz.

Uniform continuity is insufficient. Let $F(x)=\{\begin{array}{ll}0& x\le -{e}^{-2}\\ 1/2-1/\mathrm{ln}(-1/x)& -{e}^{-2}\le x\le 0\\ 1/2+1/\mathrm{ln}(1/x)& 0\le x\le {e}^{-2}\\ 1& x\ge {e}^{-2}\end{array}$.

Step 2

This is a cdf which is uniformly continuous and for which $\sum _{n}F({\gamma}^{n})$ does not converge for any $0<\gamma <1$.

For additional assumptions, it suffices for F to be $\alpha $-Hölder continuous at zero for some $\alpha >0$. So if $F(x)=1/2+\text{sign}(x)\sqrt{|x|}$ in a neighborhood of 0, then your sum converges, even though F is not Lipschitz.

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