I have a problem at work where prices of things are determined by multiplying together a series of factors. For example assume each price is made up of three factors, $A, B, C$, so that $\text{Price} = A \cdot B \cdot C$.
For example:
Product 1, $\text{price} = 200 \cdot 0.5 \cdot 1.50 = 150$.
Product 2, $\text{price} = 100 \cdot 0.25 \cdot 1.20 = 30$.
Product 3, $\text{price} = 50 \cdot 0.75 \cdot 1.8 = 67.5$.
My boss wants me to create a report showing the average A, average B, and average C, and then as a check would like to see that
$$\text{avg}(A) \cdot \text{avg}(B) \cdot \text{avg}(C) = \text{avg}(\text{price}).$$
So in the above example, average price is
$$\frac{150+30 + 67.5}{3} = 82.5,$$
but
$$ \begin{align} \text{avg}(A) & = \frac{200+100+50}{3} = 116.67, \\ \text{avg}(B) & = \frac{.5+.25+.75}{3} = 0.50, \\ \text{avg}(C) & = \frac{1.5+1.2+1.8}{3} = 1.5, \end{align} $$
and then $\text{avg}(A) \cdot \text{avg}(B) \cdot \text{avg}(C) = 116.67 \cdot 0.50 \cdot 1.5 = 87.50$, which is off from $82.50$.
I'm pretty sure from thinking about this that it will not generally be true that the product of averages equals the average of the products. Is there an easy to understand proof that this is true? Say for $N$ products and $M$ factors in each product?
Is there some other creative way of taking averages that will make this idea work out? E.g. geometric averages or something?
To prove that your boss' idea won't work, you only need one example. I would use $A=1,5,1000, B=1000,1,5, C=5,1000,1$. Then all the prices are $5000$ and all the averages are $335\frac 13$ so the product of the averages is $335\frac13^3\approx 37,707,712$. How you present this without getting fired is not a mathematical problem.
What would work is (product of the A's)*(product of the B's)*(product of the C's)=(product of the prices) Depending on how you implement this in the software, it may not be much of a check, but it should satisfy your boss. If you want to take the geometric mean, that will just raise this relation to the $\frac 1n$ power, where $n$ is the number of products. It works the same.