# a simple probability inequality

For independent Rademacher random variables $\epsilon_i, i=1,2, \cdots, n$, i.e. $P(\epsilon_i=-1)=P(\epsilon_i=1)=\frac{1}{2}$, do we have

$$max_{0\le a_i\le b, i=1,2,\cdots,n}P(|\sum_{i=1}^n\epsilon_ia_i^2|>x)\le max_{0\le a_i\le b, i=1,2,\cdots,n}P(|\sum_{i=1}^n\epsilon_ia_i|>x), \forall x>0,$$ where $0\le b<1$.

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## 1 Answer

No. Let $a_1 = 1, a_2 = 1/2, x=2/3$.

$P(|\epsilon_1 + 1/4 \epsilon_2| > 2/3) = 1$

$P(|\epsilon_1 + 1/2 \epsilon_2| > 2/3) = 1/2$

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I modified the problem. Thanks. –  honglangwang Dec 10 '12 at 1:50
Are you sure that's what you mean? It's clearly an equation since coordinate-wise squaring preserves the unit cube. –  Douglas Zare Dec 10 '12 at 3:18
I think I got it. Thanks. –  honglangwang Dec 10 '12 at 20:03