Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$\theta_i (i=1,\ldots,N)$ 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!

Update

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 $$ \sum_{i=1}^N \theta_i = 1\\ \forall i, \theta_i\geq 0 $$

share|improve this question
    
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
1  
Can you do $N=2$ for example? –  GEdgar Jun 23 '12 at 0:00
show 1 more comment

1 Answer

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 $b$.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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