# Average of products VS. product of averages

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?

• Your example proves that it’s not generally true, but you may want a way to make it seem intuitively wrong. One thought: Suppose the data was such that for every product, one (just one) of the $M$ factors was zero, but it wasn’t always the same factor. Then all the prices would be zero, but none of the factor averages would be zero, so the product of the averages wouldn’t be zero. I don’t see any other sort of “check” on the averages, since the prices depend on which values of one being averaged go with which values of another. Commented Feb 14, 2014 at 4:31

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.
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.