most recent 30 from http://mathoverflow.net 2013-05-25T04:09:30Z http://mathoverflow.net/feeds/question/43104 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/43104/expectation-of-rvs-with-poisson-type-decay jat 2010-10-21T22:10:46Z 2010-10-24T18:58:57Z

<p>I need to bound the expectation of a nonnegative random variable that satisfies a Poisson-type tail bound:</p> <p>$\mathbb{P}( X \geq t ) \leq \min( d \cdot (\frac{a}{t} )^{t}, \ 1)$ for $t > 0$ </p> <p>where $a > 0$ and $d \geq 3$. My guess for the mean:</p> <p>$\mathbb{E} X \leq {\rm const} \cdot \max( a,\ \frac{\log d}{\log \log d} )$</p> <p>The reference I checked (Ledoux &amp; Talagrand, 1991) helpfully told me that this calculation is "standard". The argument apparently depends on integration by parts, but I can't figure out the trick.</p>

http://mathoverflow.net/questions/43104/expectation-of-rvs-with-poisson-type-decay/43266#43266 Ori Gurel-Gurevich 2010-10-23T04:55:59Z 2010-10-23T04:55:59Z

<p>It's a bit late now (so maybe I made a trivial mistake), but it seems to me that the statement is actually false. Take some large $d$ and set $a=\log d / \log \log d$. Let $t=Ka=K\log d / \log \log d$ for $K$ to fixed soon. Then the tail bound is $$\mathbb{P}(X\ge t) \le d K^{-K\log d / \log \log d}=exp(\log d - \frac{K \log K \log d}{\log \log d})$$ so if we take $K$ not too large (say $K=\log \log \log d$) then the bound is more than 1 and hence the constant random variable $X=t$ satisfies the constraint but has expectation $$Ka=K\frac{\log d}{\log \log d} >> \max(a, \frac{\log d}{\log \log d}) .$$</p>