-1
$\begingroup$

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. $0 \leq P_i, P_j \leq 1$.

$\displaystyle \sum_{i=1}^\infty \sum_{j=1}^\infty ijP_iP_j \leq \sum_{i=1}^\infty \sum_{j=1}^\infty j^2P_jP_i$.

$\endgroup$
10
  • 2
    $\begingroup$ If you don't know how to prove it yourself, but are sure it is true, what is the reference? $\endgroup$
    – Will Jagy
    Sep 28, 2010 at 0:47
  • 1
    $\begingroup$ Will, even if neither side converges, the sums still have a value in [0,+∞], so it makes sense to ask if one is ≤ the other. $\endgroup$
    – user5810
    Sep 28, 2010 at 1:11
  • 1
    $\begingroup$ In that case you should edit your question to reflect the hypothesis, and in general give more motivation. $\endgroup$
    – Will Jagy
    Sep 28, 2010 at 1:11
  • 5
    $\begingroup$ YOU MOST DEFINITELY CANNOT CANCEL OUT THE $P_jP_i$ ON BOTH SIDES! (takes deep breath) $\endgroup$
    – Yemon Choi
    Sep 28, 2010 at 2:03
  • 1
    $\begingroup$ "Can I not cancel out the $P_jP_i$ on both sides, and consider just the sums over $ij$ and $j^2$ respectively?" This makes me wonder if this is indeed an exercise/homework; it seems odd to get to this inequality and not know how to prove it. Voting to close. $\endgroup$
    – Yemon Choi
    Sep 28, 2010 at 2:15

2 Answers 2

3
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ I was going to write out a slightly different spin on your argument, based on the fact that when all terms in a double series are positive one has recourse to Fubini's theorem (for series rather than integrals) regardless of any assumptions of sums being finite. But on 2nd thoughts it doesn't seem worth it; your answer is fine. $\endgroup$
    – Yemon Choi
    Sep 28, 2010 at 2:13
0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Are you sure there is such a theorem? Because it would imply $x(1-x)\leq 0$ for all $x\in[0,1]$... $\endgroup$ Sep 28, 2010 at 9:09

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