TITLE: Average of products VS. product of averages
QUESTION [7 upvotes]: 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?

REPLY [8 votes]: I guess what you want is an insight into this mismatch and to grasp what it depends on etc. Otherwise, a counter example is a sufficient proof.
For the insight into the problem, and to understand the specifics of the relation between avg(A*B) vs. avg(A)*avg(B), check "covariance" in statistics (e.g. https://en.wikipedia.org/wiki/Covariance). 
Let's say $x_1, x_2, ..., x_n$ and $y_1, y_2, ..., y_n$ are two lists of numbers (I am talking about 2 lists, but the idea generalizes to 3 or more, as in your case). Then the covariance of x and y is defined as, $Cov(x_i,y_i)=avg(x_i\cdot y_i)-avg(x_i)\cdot avg(y_i)$ (You will see $E(x)$ instead of $avg(x)$ in math resources, and $x$ will be called a random variable, and $x_i$ are samples). So, the discrepancy you talk about has a name in statistics. It basically shows how much the two sets of numbers are related, vary together (in a linear way). If they are independent or uncorrelated, then the difference is zero and your two calculations give the same result.
Another useful expression relating to covariance is, $$Cov(x_i,y_i)=avg[ (x_i-avg(x_i))\cdot(y_i-avg(y_i) ] = avg(x_i\cdot y_i)-avg(x_i)\cdot avg(y_i)$$You see that, obviously, if the $x_i$ aor $y_i$ are constant and $x_i=avg(x_i)$, there would be no difference in taking the average before product or after.