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Is it possible to find an approximate expression of $\frac{\sum_{i=1}^{n} k_i x_i}{\sum_{i=1}^{n} k_i}$ using $\langle k \rangle$, $\langle k^2 \rangle$, $\langle x \rangle$, and $n$? Alternatively if it is possible to express the boundary (biggest and smallest possible value) of $\frac{\sum_{i=1}^{n} k_i x_i}{\sum_{i=1}^{n} k_i}$ using $\langle k \rangle$, $\langle k^2 \rangle$, $\langle x \rangle$ and $n$?

Here: $\langle k \rangle=\frac{\sum_{i=1}^{n} k_i}{n}$, $\langle k^2 \rangle=\frac{\sum_{i=1}^{n} k_i^2}{n}$, $\langle x \rangle=\frac{\sum_{i=1}^{n} x_i}{n}$, $k_i \geq 0$ is an integer, and $0\leq x_i\leq 1$.

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Approximate, yes. Meaningful? Unlikely. Take a long bar and divide it into n varying commensurable lengths, one length for each k_i. The average of the square values gives some information, I saynot enough: take the average of the x's to be 1/2, then color the 1/2 shortest segments one color, and the remaining longest segments another. That represents your variability when you know the average x value. The square average may weakly measure that variability, but it does not say where the desired value (sum of colored lengths) lives. Gerhard "Ask Me About System Design" Paseman, 2012.04.18 – Gerhard Paseman Apr 18 '12 at 19:18
$n\langle x\rangle$ is a (trivial) upper bound, and is attained when the $x_i$ are all equal, regardless of the distribution of the $k_i$. – Gerry Myerson Apr 18 '12 at 23:39
@Gerry, I think you mean $\langle x \rangle$, not $n\langle x \rangle$, too (see my comment to Aaron Meyerowitz's answer). I'm not sure why the OP bothers putting what amounts to an $n\langle k \rangle$ in the denominator of what he wants to express in terms of $\langle k \rangle$ etc. – Barry Cipra Apr 19 '12 at 23:37
@Barry, right you are, of course. – Gerry Myerson Apr 20 '12 at 5:37
up vote 2 down vote accepted

We must assume that $\langle k \rangle \ne 0.$ When $n=1$ we know everything.

Certainly $\langle k^2\rangle \ge\langle k\rangle ^2.$ If $\langle k^2\rangle =\langle k\rangle ^2$ then the $k_i$ are all equal (to $\langle k \rangle$ ) and $\frac{\sum_{i=1}^{n} k_i x_i}{\sum_{i=1}^{n} k_i}=\langle x \rangle$ exactly. The same thing happens if all the $x_i$ are equal although we have no way of know from the given information if that is the case. Also knowing $\sum_{i=1}^{n} x_i^2$ would be helpful. Then $\sum_{i=1}^{n} k_i x_i=C\sqrt{\sum_{i=1}^{n} k_i^2\sum_{i=1}^{n} x_i^2}$ for some $0 \le C \le 1.$

In the special case that $n=3$, and $\langle k^2 \rangle=30$ we can figure that the $k_i$ are $1,2,5$ in some order (given that they are integers). I will ignore these number theoretic features and only assume that the $k_i$ are non-negative reals.

We may assume that $\max_i{x_i}=x_1.$ When $\langle k^2 \rangle \gt \langle k \rangle^2$ (i.e. the $k_i$ are not all equal) all we can say with the given information is that $0 \le \frac{\sum_{i=1}^{n} k_i x_i}{\sum_{i=1}^{n} k_i} \le x_1.$ Equality occurs when $k_1 \gt 0$ but $k_i=0$ for $i \gt 1.$ That only one of the $k_i$ is non-zero can be discovered from $\langle k^2 \rangle =n\langle k \rangle^2$ Now $\max_i{x_i} \le \sum_{i=1}^{n} x_i=n\langle x \rangle$ with equality when $x_i=0$ for $i \ge 2.$

Earlier Careless reading lead me to mention: If the $k_i$ are not all equal, and some of the $x_i$ can be negative, then nothing can be deduced. even if we actually know all of the $k_i.$ Let us assume that $k_2=k_1+\epsilon.$ Assume first that $n=2$. Then given desired values $\langle x \rangle=m$ and $\frac{\sum_{i=1}^{n} k_i x_i}{\sum_{i=1}^{n} k_i}=S$ we must take $x_1=m+\delta$ and $x_2=m-\delta$ for $\delta=\frac{(S-m)(2k_1+\epsilon)}{\epsilon}.$ For larger $n$ I can volunteer the extra information that $x_i=0$ for all $i \gt 2$ and adjust the formulas slightly.

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In your "Certainly" sentence, I think you mean just "$\langle x \rangle$ exactly," not $n\langle x \rangle$. – Barry Cipra Apr 19 '12 at 22:46
No need to worry about the $x_i$ being negative, as the question says $0\le x_i\le1$. – Gerry Myerson Apr 19 '12 at 23:19

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