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$\theta_i (i=1,\ldots,N)$ are real numbers and we have $$ \sum_{i=1}^N \theta_i = 1 $$ For any $i\neq j$, $$ \sum_{w\in W} \frac{f_i(w)}{\sum_{k=1}^N \theta_k f_k(w)} = \sum_{w\in W} \frac{f_j(w)}{\sum_{k=1}^N \theta_k f_k(w)} $$

Here $W$ is a set, and for any $i$, $f_i()$ is a function that maps an element in $W$ to a scalar.

I guess there should be a closed form solution for $\theta_i (i=1,\ldots,N)$ (in terms of $f_i$ and $W$) but couldn't figure it out. Thank you!


The equations above are what I get by applying the Karush–Kuhn–Tucker conditions to the following optimization problem:

Maximize $$ \prod_{w\in W} \sum_{k=1}^N \theta_k f_k(w) $$ subject to $$ \sum_{i=1}^N \theta_i = 1\\ \forall i, \theta_i\geq 0 $$

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Closed form solution for what, in terms of what? $f$ in terms of the $\theta_i$? Also, what kinds of things are $w$ and $\theta_i$? Real numbers perhaps? –  Zev Chonoles Jun 22 '12 at 23:22
I've clarified the question. Thanks for your comments. –  took Jun 22 '12 at 23:39
If you multiply through by $\sum_{k=1}^N \theta_k f_k(w)$ you will get an equation that doesn't involve $\theta_k$. That is, you can pick any $\theta_k$ as long as the resulting term is non-zero. –  copper.hat Jun 22 '12 at 23:49
@copper.hat, that sum depends on $w$, so I don't see how you are going to "multiply through" by it. –  Gerry Myerson Jun 22 '12 at 23:56
Can you do $N=2$ for example? –  GEdgar Jun 23 '12 at 0:00

1 Answer 1

I looked at $N=2$. To simplify notation, I took $\theta_1=a$, $\theta_2=b$, $W=\{{r,s\}}$, $f_1(r)=u$, $f_1(s)=v$, $f_2(r)=w$, $f_2(s)=x$, and I got $$a={ux+vw-2wx\over2(u-w)(x-v)}$$ with something similar for $b$.

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Thanks. Your solution is correct. But when the size of $W$ increases, the closed form solution becomes much more complicated even with $N=2$. I guess a general closed form solution does exist but would be too complicated to write down. –  took Jun 23 '12 at 0:40

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