# Derivative of a 'weighted average' of decreasing fractions

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) $0<A_{i}<1, 0<B_{i}<1$.

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.

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No, AV is not necessarily increasing in $N$ under the stated assumptions. Proof by counterexample:

Consider the case where there are only two fractions which are equally weighted:

$$AV\left( N\right) =\frac{0.5A_{1}\left( N\right) +0.5A_{2}\left( N\right) }{0.5B_{1}\left( N\right) +0.5B_{2}\left( N\right) }=\frac {A_{1}\left( N\right) +A_{2}\left( N\right) }{B_{1}\left( N\right) +B_{2}\left( N\right) }$$

We now want to know if $AV\left( N^{\prime}\right) >AV\left( N\right)$ (where $N^{\prime}>N$).

Consider the following values for $A$ and $B$:

$$A_{1}\left( N\right) =\frac{5}{8},A_{2}\left( N\right) =\frac{1}{2}$$ $$A_{1}\left( N^{\prime}\right) =\frac{5}{16},A_{2}\left( N^{\prime}\right) =\frac{2}{5}$$ $$B_{1}\left( N\right) =\frac{1}{2},B_{2}\left( N\right) =\frac{5}{8}$$ $$B_{1}\left( N^{\prime}\right) =\frac{1}{4},B_{2}\left( N^{\prime}\right) =\frac{1}{2}$$

Which satisfies assumptions (b)-(c) strictly, and (a) weakly.We then have that:

$$AV\left( N\right) =\frac{\frac{5}{8}+\frac{1}{2}}{\frac{1}{2}+\frac{5}{8} }=1$$ $$AV\left( N^{\prime}\right) =\frac{\frac{5}{16}+\frac{2}{5}}{\frac{1} {4}+\frac{1}{2}}=\frac{19}{20}<1$$

So that $AV\left( N\right)$ is decreasing in $N$.

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