Why does this inequality hold? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T11:54:42Z http://mathoverflow.net/feeds/question/40233 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/40233/why-does-this-inequality-hold 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/40246#40246 Answer by Bob Terrell for Why does this inequality hold? Bob Terrell 2010-09-28T01:35:23Z 2010-09-28T01:35:23Z <p>As Will Jagy said it is not true in general. But assume $S=\sum_{j=1}^\infty j^2 P_j$ converges, and apparently you are assuming $\sum_{i=1}^\infty P_i = 1$. Then the right side converges to $S$. You also know that $i^2+j^2\ge 2ij$ (because $(i-j)^2\ge 0$). Absolute convergence of the right side lets you rearrange it to $\sum_j \sum_i j^2 P_jP_i = \sum_i\sum_j i^2P_iP_j$. So the right side is $\frac{1}{2}\sum_i\sum_j (i^2+j^2)P_iP_j$, which is then greater than or equal to the left.</p> http://mathoverflow.net/questions/40233/why-does-this-inequality-hold/40273#40273 Answer by Péter Komjáth for Why does this inequality hold? Péter Komjáth 2010-09-28T06:49:50Z 2010-09-28T06:49:50Z <p>It suffices to show th finite version of this, i.e,, $\sum_{i,j=1}^n ijP_iP_j\leq \sum_{i,j=1}^n j^2 P_j^2$. There is a theorem (I do not have the reference) that an inequality like this, with polynomials of degree at most 2, holds iff it holds for all choices where each $P_i$ is either 0 or 1. Assume that $P_{a_1},\dots,P_{a_k}$ are 1, the rest are zero. Then the inequality has the form $\sum^k_{i,j}a_ia_j\leq k\sum^k_{j=1}a_j^2$. The LHS is $(\sum a_i)^2$, so by dividing by $k$ we obtain the arithmetic mean - quadratic mean inequality.</p> http://mathoverflow.net/questions/40233/why-does-this-inequality-hold/40285#40285 Answer by Yemon Choi for Why does this inequality hold? Yemon Choi 2010-09-28T08:40:53Z 2010-09-28T08:57:09Z <p>Is it worth me pointing out that the desired inequality is just saying that when X is a discrete random variable taking positive (integer) values, then $(\mathbb{E} X)^2 \leq {\mathbb E} X^2$? Which is true, but a rather basic result in one's study of probability theory, not to mention "just" being the Cauchy-Schwarz inequality.</p>