most recent 30 from http://mathoverflow.net 2013-05-22T04:23:43Z http://mathoverflow.net/feeds/user/9566 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/40233/why-does-this-inequality-hold unknown (google) 2010-09-28T00:30:35Z 2010-09-28T08:57:09Z

<p>Hi people. Can you help me realize why this is true? I can tell you that $P_i$ and $P_j$ are probabilities, i.e. $0 \leq P_i, P_j \leq 1$.</p> <p>$\displaystyle \sum_{i=1}^\infty \sum_{j=1}^\infty ijP_iP_j \leq \sum_{i=1}^\infty \sum_{j=1}^\infty j^2P_jP_i$.</p>

http://mathoverflow.net/questions/40233/why-does-this-inequality-hold/40285#40285 2010-09-29T07:22:58Z 2010-09-29T07:22:58Z

This is not quite the same. It would be, if you just had the $j^2P_j$ term on the right hand side, but $P_i$ is being multiplied onto there, which is in fact $Pr(X \neq 0)$ in this case, so the RHS is $E[X^2] \cdot Pr(X \neq 0)$.

http://mathoverflow.net/questions/40233/why-does-this-inequality-hold 2010-09-28T01:05:35Z 2010-09-28T01:05:35Z

Will, sorry, I am not very confident with these kinds of infinite sums, so I did not know that. I might add the sum of all $P_i$ should be 1, since it is a probability distribution.

http://mathoverflow.net/questions/40233/why-does-this-inequality-hold 2010-09-28T01:02:00Z 2010-09-28T01:02:00Z

This inequality is one I have derived from an inequality which I want to prove. Can I not cancel out the $P_jP_i$ on both sides, and consider just the sums over ij and j^2 respectively?