3
$\begingroup$

Fix positive numbers $x_1,\dots,x_n$ with $\sum_{i=1}^nx_i=1$. For any positive numbers $a_1\leq\dots\leq a_n$, we define $$Y=a_1x_1+\dots+a_nx_n,$$ and $y_i=\frac{a_ix_i}{Y}$for all $1\leq i\leq n$.

What are all the possible sequences $(y_1,\dots,y_n)$, i.e., sequences that appear as $(y_1,\dots,y_n)$ for some $a_1\leq\dots\leq a_n$? Are there some succinct sufficient or necessary conditions?

For example, one necessary condition is $y_1\leq x_1$. This is because $$y_1=\frac{a_1x_1}{a_1x_1+\dots+a_nx_n}\leq\frac{a_1x_1}{a_1x_1+\dots+a_1x_n}=x_1.$$ Similarly it is necessary that $y_n\geq x_n$. Are these two together already sufficient?

$\endgroup$
  • $\begingroup$ I don't have any precise set of conditions (honestly, I don't think it is possible to express them simply), but the two conditions you mentioned are not sufficient. Consider the sequence $x_1, \dots, x_4 = 1/4, \dots, 1/4$, and the coefficients $a_i = i$. The sequence $(y_i)$ satisfies your two conditions, but so does the sequence ($y_1, y_3, y_2, y_4$). $\endgroup$ – Mariuslp Aug 24 '16 at 15:54
  • $\begingroup$ @Mariuslp I take your point that the two conditions that I mentioned are not sufficient. So for $(x_1,\dots,x_4)=(1/4,\dots,1/4)$, the sequence $(y_i)$ that can appear are exactly those with $y_1\le y_2\le y_3\le y_4$, right? This is a simple condition for this particular case. $\endgroup$ – pi66 Aug 24 '16 at 16:44
  • $\begingroup$ If all the $x_i$ are equal to some $x$, then yes, because your $y_i$ are defined by $y_i = K\cdot a_i$, (with $K = \frac xY \geq 0$) and the sequences that satisfy your conditions are the ones such that the $a_i$ are increasing with $i$, which is (in this case) equivalent to the $y_i$ increasing with $i$ $\endgroup$ – Mariuslp Aug 24 '16 at 22:57
  • $\begingroup$ It's necessary and sufficient that $y_i/x_i \le y_j/x_j$ for all $i < j$. I'm not sure what other kind of condition you would want... $\endgroup$ – arghbleargh Aug 26 '16 at 0:00
  • $\begingroup$ @arghbleargh : There are two more conditions needed. $\endgroup$ – Michael Aug 28 '16 at 14:50
1
+50
$\begingroup$

Fix a vector $(x_1, ..., x_n)$ with positive entries. Then $(y_1, ..., y_n)$ is a vector that satisfies the requirements with respect to $(x_1, ..., x_n)$ if and only if it satisfies the following three conditions:

1) $y_i>0$ for all $i \in \{1, ..., n\}$.

2) $\frac{y_i}{x_i} \leq \frac{y_{i+1}}{x_{i+1}}$ for all $i \in \{1, ..., n-1\}$.

3) $\sum_{i=1}^n y_i = 1$.

The proof of necessity is simple. The proof of sufficiency is just a little more work, but not hard and you can likely do that yourself.

These three conditions are "minimal" in the sense that no two of the conditions implies the third.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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