## Electronic Communications in Probability

### A Proof of a Conjecture of Bobkov and Houdré

#### Abstract

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.

#### Article information

Source
Electron. Commun. Probab., Volume 1 (1996), paper no. 2, 7-10.

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

https://projecteuclid.org/euclid.ecp/1453756493

Digital Object Identifier
doi:10.1214/ECP.v1-972

Mathematical Reviews number (MathSciNet)
MR1386289

Zentralblatt MATH identifier
0854.60014

Subjects
Primary: 60E05: Distributions: general theory
Secondary: 60E15: Inequalities; stochastic orderings

Keywords
Gaussian distribution

Rights