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Let $C\subset \mathbb R^n$ be convex and symmetric about the origin. I am trying to prove that $\gamma(C) \geq \gamma(C+x)$ for any $x\in \mathbb R^n$, where $\gamma$ is the standard Gaussian measure. I tried using the following, which can be deduced from the Prekopa Leindler inequality:

$$\gamma(tA + (1-t)B) \geq t\gamma(A) + (1-t)\gamma(B)$$

for convex sets $A$ and $B$. However, for any of the various choices of $A$ and $B$ I can think of, this does not yield the desired inequality for $C$. Any suggestions for how to precede?

EDIT: The inequality I stated above is wrong; the right side should have the geometric mean. Here is a proof of the correct form of the inequality, using Prekopa Leindler. Let $f(x) = \chi_{tA + (1-t)B} (x) e^{-|x|^2/2}$, $g(x) = \chi_{A} (x) e^{-|x|^2/2}$, $h(x) = \chi_{B} (x) e^{-|x|^2/2}$, where $\chi$ denotes an indicator function. We claim that $f,g,h$ satisfy the hypothesis of the Prekopa Leindler inequality, i.e. $f(tx + (1-t)y) \geq g(x)^t h(y)^{1-t}$. This is obvious if $x\notin A$ $y\notin B$. If $x\in A$ and $y\in B$, then our claim reduces to showing

$$ e^{-|tx + (1-t)y|^2/2} \geq e^{-t|x|^2 - (1-t)|y|^2}$$

which is immediate from convexity of $x\mapsto |x|^2$. Thus Prekopa Leindler gives

$$\int f(x)\, dx \geq \left(\int g(x)\, dx\right)^t \left(\int h(x)\,dx\right)^{1-t}$$

which is equivalent to

$$\gamma(tA + (1-t)B ) \geq \gamma(A)^t \gamma(B)^{1-t} .$$

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Can you explain how your inequality for $\gamma$ follows from Prekopa-Leindler? My understanding is that one gets a geometric mean of $\gamma(A)$ and $\gamma(B)$ (I also commented on did's answer below). –  user53153 Jan 2 '13 at 1:12
    
I have added an explanation. –  user15464 Jan 2 '13 at 2:17
    
Thanks for the addition. In some speculative way, we can guess results about Gaussian measure by looking at results for Lebesgue measure and letting $n\to \infty$. The Lebesgue measure version of the Brunn-Minkowski inequality expresses the concavity of ${\rm volume}^{1/n}$ with respect to Minkowski addition. In the limit $n\to \infty$ we should expect concavity of the logarithm. // To make an even stronger case, think of Brunn-Minkowski as an inequality for generalized mean with $p= 1/n$. When $p=0$ we have the geometric mean. –  user53153 Jan 2 '13 at 5:27
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1 Answer

up vote 3 down vote accepted

Brunn-Minkowski inequality for gaussian measures states that, for $t$ in $(0,1)$ and $A$ and $B$ both convex, $\gamma(tA+(1-t)B)\geqslant\gamma(A)^t\gamma(B)^{1-t}$. For $t=\frac12$, $A=C+x$ and $B=C-x$, the fact that $C=\frac12(A+B)$ by convexity of $C$, and the fact that $A=B$ by symmetry of $C$ indeed yield $\gamma(C)\geqslant\gamma(C+x)$.

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Does the Brunn-Minkowski inequality work this way for $\gamma$? I thought it had the geometric mean instead of arithmetic on the right hand side. –  user53153 Jan 1 '13 at 23:34
    
@PavelM You are right, I modified the answer. Thanks. –  Did Jan 2 '13 at 0:01
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