Open Access
1996 A Proof of a Conjecture of Bobkov and Houdré
S. Kwapien, M. Pycia, W. Schachermayer
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Electron. Commun. Probab. 1: 7-10 (1996). DOI: 10.1214/ECP.v1-972


S. G. Bobkov and C. Houdré recently posed the following question on the Internet (Problem posed in Stochastic Analysis Digest no. 15 (9/15/1995)): Let $X,Y$ be symmetric i.i.d. random variables such that $$P(|X+Y|/2 \geq t) \leq P(|X| \geq t),$$ for each $t>0$. Does it follow that $X$ has finite second moment (which then easily implies that $X$ is Gaussian)? In this note we give an affirmative answer to this problem and present a proof. Using a dierent method K. Oleszkiewicz has found another proof of this conjecture, as well as further related results.


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S. Kwapien. M. Pycia. W. Schachermayer. "A Proof of a Conjecture of Bobkov and Houdré." Electron. Commun. Probab. 1 7 - 10, 1996.


Accepted: 26 February 1996; Published: 1996
First available in Project Euclid: 25 January 2016

zbMATH: 0854.60014
MathSciNet: MR1386289
Digital Object Identifier: 10.1214/ECP.v1-972

Primary: 60E05
Secondary: 60E15

Keywords: Gaussian distribution

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