I'm having some trouble showing the following statement (which intuitively seems to hold):
Suppose I have a series of fractions indexed by $i$ , each of them a function of $N:f_{i}\left( N\right) =\frac{A_{i}\left( N\right) }{B_{i}\left( N\right) }$. Assume that:
- (a) $f_{i}\left( N\right) $ is increasing in $N$
- (b) $B_{i}\left( N\right) $ is decreasing in $N$
- (c) $0
.
Now consider the following 'weighted average':
$AV\left( N\right) =\frac{\sum\limits_{i}p_{i}A_{i}\left( N\right) }{\sum\limits_{i}p_{i}B_{i}\left( N\right) }$ where $\sum p_{i}=1$.
Q: Are the above conditions sufficient to guarantee that $AV\left( N\right) $ is increasing in $N$ as well?
Below you will find what I have been able to show. Any help or insights would be greatly appreciated!
(a) implies that $\frac{d A_{i}}{d N}B_{i}-A_{i}\frac{d B_{i}}{d N}>0$
Now note that $\frac{dAV\left( N\right) }{dN}>0$ if
$\left( \sum\limits_{i}p_{i}\frac{dA_{i}\left( N\right) }{dN}\right) \left( \sum\limits_{j}p_{j}B_{j}\left( N\right) \right) -\left( \sum\limits_{i}p_{i}A_{i}\left( N\right) \right) \left( \sum \limits_{j}p_{j}\frac{dB_{j}\left( N\right) }{dN}\right) >0$
$\sum\limits_{i}\sum\limits_{j}\left( p_{i}p_{j}\left[ \frac{dA_{i}\left( N\right) }{dN}B_{j}\left( N\right) -A_{i}\left( N\right) \frac {dB_{j}\left( N\right) }{dN}\right] \right) >0$
$\sum\limits_{i}\left( \left( p_{i}\right) ^{2}\left[ \frac{dA_{i}\left( N\right) }{dN}B_{i}\left( N\right) -A_{i}\left( N\right) \frac {dB_{i}\left( N\right) }{dN}\right] \right)$
$+\sum\limits_{i}% \sum\limits_{j\neq i}\left( p_{i}p_{j}\left[ \frac{dA_{i}\left( N\right) }{dN}B_{j}\left( N\right) -A_{i}\left( N\right) \frac{dB_{j}\left( N\right) }{dN}\right] \right) >0$
$\sum\limits_{i}\left( \left( p_{i}\right) ^{2}\left[ \frac{dA_{i}\left( N\right) }{dN}B_{i}\left( N\right) -A_{i}\left( N\right) \frac {dB_{i}\left( N\right) }{dN}\right] \right)$
$+\sum\limits_{i}% \sum\limits_{j\neq i}\left[ \left( p_{i}p_{j}\right) \frac{dA_{i}\left( N\right) }{dN}B_{j}\left( N\right) \right] -\left( p_{i}p_{j}\right) \sum\limits_{i}\sum\limits_{j\neq i}\left[ A_{i}\left( N\right) \frac{dB_{j}\left( N\right) }{dN}\right] >0$
The first sum is positive (follows from (a)) and the third sum is negative (follows from (b)). But the second sum does is not necessarily positive.