most recent 30 from http://mathoverflow.net 2013-05-24T03:16:42Z http://mathoverflow.net/feeds/user/24041 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 Dimas 2012-07-19T01:06:31Z 2012-07-19T01:06:31Z

@George: Yes, I confused uniform continuity with continuity. That condition is indeed much stronger but usually met by &quot;garden variety&quot; densities, and good enough for my application. Anyways, do you know if this is a known result? It seems like a very important requirement for anyone doing maximum a posteriori estimation.

http://mathoverflow.net/questions/98228/is-it-known-that-every-pdf-continuous-in-all-rn-has-a-maximum Dimas 2012-05-29T03:49:11Z 2012-05-29T03:49:11Z

Never mind, sorry for the fallacy... An $\epsilon$-neighbourhood of an unbounded set has infinite Lebesgue measure for any finite $\epsilon$, but as the counterexample showed it not necessarily is containded by the lower superlevel set. The case I'm working is simpler though, I actually know my function is bounded above, it is differentiable and its gradient is continuous. Does it make sense saying it attains the maximum? ps.: George, I'm a fan of your blog.