most recent 30 from http://mathoverflow.net 2013-06-19T17:13:44Z http://mathoverflow.net/feeds/question/98228 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/98228/is-it-known-that-every-pdf-continuous-in-all-rn-has-a-maximum Dimas 2012-05-28T23:07:57Z 2012-05-29T00:25:39Z

<p>I'm working with maximum a posteriori estimation and managed to show that every probability density function that is continuous in all $R^n$ always has at least one global maximum. I've search around and asked a few fellow engineers and professors but am not sure if this is widely known. This can actually be extended to any continuous Radon-Nikodym derivative of a finite measure.</p> <p>The proof is simple: let $f$ be the PDF, and be continuous in all $R^n$. If $L(v)$ is the closed superlevel set at $v$, that is: $L(v):= ${$x\in R^n: f(x)\geq v$}, then it must be bounded.</p> <p>That is so because the neighbourhood of any unbounded set in $R^n$ has infinite Lebesgue measure. Due to continuity of $f$, any lower superlevel set of it, for example $L(v/2)$ contains a neighbourhood of $L(v)$. The probability of the superlevel sets is bounded below by $P[L(v)]\geq v \lambda[L(v)]$. This means that if any superlevel set of $f$ were unbounded, then a lower superlevel set would have probability greater than one.</p> <p>Since all closed superlevel sets are bounded, they are compact and attain their maximum.</p>

http://mathoverflow.net/questions/98228/is-it-known-that-every-pdf-continuous-in-all-rn-has-a-maximum/98237#98237 Francois Ziegler 2012-05-29T00:25:39Z 2012-05-29T00:25:39Z

<p>Take $n=1$ and put a triangle with height $2^m$ and width $2^{-2m}$ at each integer $m=0,1,2,\dots$</p>