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Assume $\theta_1,\dots,\theta_n$ are $n$ positive numbers such that $$\theta_1+\dots+\theta_n=1$$ Define $y_{ij}=\frac{\theta_i}{\theta_j}$ for all $i,j \in \{1,\dots,n\}$. Is there a way to transform the above equation in terms of $y_{ij}$.

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    $\begingroup$ The fact that the left-hand side is of degree $1$ (in the $\theta$) while both the right-hand side and $y_{ij}$ are of degree $0$ might make this very difficult. $\endgroup$ – Arthur Sep 18 '15 at 9:15
  • $\begingroup$ May I ask why this was downvoted? $\endgroup$ – dineshdileep Sep 22 '15 at 4:00
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Yes. Use $\theta_i = \theta_j y_{ij}$ to write $\theta_1+\theta_2\dots+\theta_n=1$ as $\theta_j y_{1j}+\theta_j y_{2j}+\dots+ \theta_j y_{nj}= 1$

Collect $\theta_j$ s.t. $\theta_j(y_{1j}+y_{2j}+\dots+y_{nj})= 1$ so $\theta_j = \displaystyle \frac{1}{y_{1j}+y_{2j}+\dots+y_{nj}}=\frac{1}{\displaystyle \sum_{k=1}^{n}y_{kj}}$

$\theta_1+\theta_2\dots+\theta_n=1 = \displaystyle \frac{1}{\displaystyle \sum_{k=1}^{n}y_{k1}} + \frac{1}{\displaystyle \sum_{k=1}^{n}y_{k2}} + \dots+\frac{1}{\displaystyle \sum_{k=1}^{n}y_{kn}} = \displaystyle \sum_{j=1}^{n}\frac{1}{\displaystyle \sum_{k=1}^{n}y_{kj}}$


Edit 2

More simply:

Given $$\theta_1 + \theta_2 + \dots+ \theta_‌​n= 1$$ $$\theta_j > 0$$ $$y_{ij} = \frac{\theta_i}{\theta_j}$$ then $$ \theta_j = \displaystyle \frac{\theta_j}{\theta_1 + \theta_2 + \dots+ \theta_‌​n} = \displaystyle \frac{1}{\frac{\theta_1}{\theta_j}+\frac{\theta_2}{\theta_j}+\dots+\frac{\theta_‌​‌​n}{\theta_j}} = \displaystyle \frac{1}{y_{1j}+y_{2j}+\dots+y_{nj}}$$ Or $$ y_{1j}+y_{2j}+\dots+y_{nj} = \displaystyle \frac{1}{\theta_j}$$ So each $\theta_j$ is reduced to a $y_{kj}$ expression.

So any expression written in $\theta_j$ variables can be transformed into an expression in $y_{kj}$ variables.

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    $\begingroup$ This condition ($\sum\limits_j(1/\sum\limits_k y_{kj}) =1$) is satisfied for ARBITRARY non-zero $\theta_i$: $\sum\limits_j(1/\sum\limits_k y_{kj}) = \sum\limits_j (\sum\limits_k \frac{\theta_k}{\theta_j})^{-1} = \sum\limits_j (\frac{\sum\limits_k \theta_k}{\theta_j})^{-1} = \sum\limits_j \frac{\theta_j}{\sum\limits_k \theta_k} = \frac{\sum\limits_j \theta_j}{\sum\limits_k \theta_k} = 1$ $\endgroup$ – Litho Sep 24 '15 at 7:27
  • $\begingroup$ Yes I agree. That's interesting. The key relationship is: $$\theta_j = \displaystyle \frac{1}{y_{1j}+y_{2j}+\dots+y_{nj}} = \displaystyle \frac{1}{\frac{\theta_1}{\theta_j}+\frac{\theta_2}{\theta_j}+\dots+\frac{\theta_‌​n}{\theta_j}}=\displaystyle \frac{\theta_j}{\theta_1 + \theta_2 + \dots+ \theta_‌​n} = \theta_j$$ This only holds for $\theta_1+\theta_2\dots+\theta_n=1$ $\endgroup$ – arthur Sep 24 '15 at 8:43
  • $\begingroup$ @arthur Isn't the additional conditions , $y_{ij}\geq 0 $ and $y_{ij}y_{ji}=1$ also needed? $\endgroup$ – dineshdileep Sep 25 '15 at 7:30
  • $\begingroup$ @dineshdileep Yes. $\theta_j \gt 0$ (positive numbers) so $y_{ij} \gt 0$. $y_{ij}y_{ji}=1$ follows from the definition of $y_{ij} $ $\endgroup$ – arthur Sep 25 '15 at 8:16
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No. If you multiply all $\theta_i$ by the same positive number $c\neq 1$, then all $y_{ij}$ stay the same, but $\theta_1+\dots+\theta_n$ changes.

In more details: assume that you have a condition or a set of conditions on $y_{ij}=\frac{\theta_i}{\theta_j}$ which is satisfied if and only if $\theta_1+\dots+\theta_n=1$. Let $\overline{\theta}_i$ be positive numbers such that $\overline{\theta}_1+\dots+\overline{\theta}_n=1$. Then, by assumption, the numbers $\overline{y}_{ij}=\frac{\overline{\theta}_i}{\overline{\theta}_j}$ satisfy that condition(s). Now choose a positive number $c\neq 1$ and set $\theta'_i = c\overline{\theta}_i$. Then $y'_{ij} = \frac{\theta'_i}{\theta'_j} = \overline{y}_{ij}$, so the numbers $y'_{ij}$ satisfy the condition(s) as well, but $\theta'_1+\dots+\theta'_n = c(\overline{\theta}_1+\dots+\overline{\theta}_n) = c\neq 1$.

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  • $\begingroup$ but $\sum_{i=1}^{n}\theta_i'\neq 1$?? $\endgroup$ – dineshdileep Sep 24 '15 at 6:40
  • $\begingroup$ @dineshdileep Yes? That's the whole point: you cannot find condition(s) on $y_{ij}$ that would be satisfied ONLY if $\sum\theta_i = 1$. $\endgroup$ – Litho Sep 24 '15 at 7:19
  • $\begingroup$ No, I meant why should $\sum_{i=1}^{n}\theta_i'\,=\,1$?. You didn't pick them like that. So there is nothing contradicting. You just defined $\theta_i'=c\bar{\theta_i}$. That doesn't imply they should sum to 1. $\endgroup$ – dineshdileep Sep 25 '15 at 4:41
  • $\begingroup$ @dineshdileep Yes, I intentionally defined them in such a way that $\sum\theta'_i\neq 1$. Maybe I misunderstand your original question: when you write "Is there a way to transform the above equation in terms of $y_{ij}$", what equation do you mean? That $\sum\theta_i = 1$? $\endgroup$ – Litho Sep 25 '15 at 7:17
  • $\begingroup$ yes. Like the other answer tries to do. $\endgroup$ – dineshdileep Sep 25 '15 at 7:27

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