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 $0\leq x\_i\leq 1$.



Find an approximate expression of a sum of a product using the average of each item
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.


