MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

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.

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.

Since all closed superlevel sets are bounded, they are compact and attain their maximum.

share|cite|improve this question

closed as off topic by George Lowther, Qiaochu Yuan, Chris Godsil, Douglas Zare, Anthony Quas May 29 '12 at 4:52

Questions on MathOverflow are expected to relate to research level mathematics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question.

I think that if $f$ is a continuous density with compact support $\sum p_i \lambda_i f(\frac x {\lambda_i} + v_i$ is a continuous and has no global maximum, choosing $\sum p_i = 1, \lambda_i \rightarrow \infty, \v_i \rightarrow \infty$, – mike May 28 '12 at 23:40
The statement is false. L(v) need not be bounded, and the neighbourhood of an unbounded set in R^n need not have infinite measure. (Try constructing an open neighbourhood of the rationals in R with finite measure). – George Lowther May 28 '12 at 23:57
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. – Dimas Abreu Dutra May 29 '12 at 3:49
No, the modified proposition is still false. Smooth the previous counterexample and then apply $\arctan$ and rescale. – Douglas Zare May 29 '12 at 4:03
@Dimas: You recover your result if it is assumed that the probability density is uniformly continuous, although that is a much stronger condition. – George Lowther May 31 '12 at 20:16
up vote 5 down vote accepted

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

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.