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

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

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If you don't know how to prove it yourself, but are sure it is true, what is the reference? –  Will Jagy Sep 28 '10 at 0:47
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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. –  Ricky Demer Sep 28 '10 at 1:11
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In that case you should edit your question to reflect the hypothesis, and in general give more motivation. –  Will Jagy Sep 28 '10 at 1:11
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YOU MOST DEFINITELY CANNOT CANCEL OUT THE $P_jP_i$ ON BOTH SIDES! (takes deep breath) –  Yemon Choi Sep 28 '10 at 2:03
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"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. –  Yemon Choi Sep 28 '10 at 2:15
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3 Answers

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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.

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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. –  Yemon Choi Sep 28 '10 at 2:13
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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.

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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 $E[X^2] \cdot Pr(X \neq 0)$. –  user9566 Sep 29 '10 at 7:22
    
I assumed from the original question that since $(P_n)_{n=1}^\infty$ were supposed to come from a probability distribution, $\sum_{n=1} P_n =1$. In other words, this is a RV taking values in the natural numbers. I think your correction is not needed. –  Yemon Choi Sep 29 '10 at 18:10
    
Certainly one needs to make some assumption on $\sum_{n=1}^\infty P_n$, I agree. –  Yemon Choi Sep 29 '10 at 18:12
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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.

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Are you sure there is such a theorem? Because it would imply $x(1-x)\leq 0$ for all $x\in[0,1]$... –  Federico Poloni Sep 28 '10 at 9:09
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