# Why is this average rule true?

Let $$\overline{a}$$ represent the average of the quantity $$a$$.

I have seen this law being used many times: $$\overline{ab}=\overline{a}\times\overline{b}$$

Can anyone share a proof on why this is true?

• $\overline{XY}=\overline{X}\,\overline{Y}$ is true for independent variables $X,Y$ but not in general.
– anon
Jan 5, 2020 at 6:29
• Which type of average are you referring to? There are many including geometric mean, arithmetic mean and others.
– Karl
Jan 5, 2020 at 8:54
• What exactly do you mean by the average of $ab$? Are you looking at a set of data? If so, do you have the same number of values for $a$ as for $b$? Jan 5, 2020 at 9:19

Let's represent the values of the quantities you mention by $$a_1,\ldots,a_n$$ and $$b_1,\ldots,b_n$$ so that $$\overline a=\frac{a_1+\cdots+a_n}{n},\qquad \overline b=\frac{b_1+\cdots+b_n}{n},$$ and similarly $$\overline{ab}=\frac{a_1b_1+\cdots+a_nb_n}{n}.$$ Your question is asking whether (or under which conditions) $$\frac{a_1b_1+\cdots+a_nb_n}{n}=\frac{a_1+\cdots+a_n}{n}\times\frac{b_1+\cdots+b_n}{n},$$ or equivalently $$n(a_1b_1+\cdots+a_nb_n)=(a_1+\cdots+a_n)(b_1+\cdots+b_n).$$ This is just an equation involving the $$a_i,b_i$$ values and it could be either true or false depending on their values. On the right side there are $$n^2$$ terms, one for each possible $$(i,j)$$ combination $$a_ib_j$$. The left side consists of the $$n$$ diagonal terms $$a_ib_i$$, but each is multiplied by $$n$$ to make up for the discrepancy.

One thing to notice is that if $$a_1=a_2=\cdots=a_n$$, then the equation is true. (And likewise for $$b_1=\cdots=b_n$$.) But there are many more solutions, which is expected since there are $$2n$$ unknowns and only $$1$$ equation, so there are $$2n-1$$ degrees of freedom.

Okay, so just looking at the equation itself doesn't tell us too much - we need to put it in context by bringing in probability (this was already suggested in the comments before I wrote this answer). The average is a special case of a more general quantity - the expected value. Expected values are defined for certain random variables. In the case when the values $$a_1,\ldots,a_n$$ are distinct and the random variable takes each of these values with equal probability $$1/n$$, then the expected value is the average. In symbols, we can call the random variable $$A$$, satisfying $$\mathbb P(A=a_i)=\frac{1}{n},\qquad \textrm{for every i=1,\ldots,n},$$ and likewise for $$B$$. Then the statement that "average equals expectation" translates into $$\overline a=\mathbb E[A],\qquad \overline b=\mathbb E[B],$$ where $$\mathbb E$$ is the shorthand for "taking expectation of a random variable".

So your question boils down to asking when $$\mathbb E[AB]=\mathbb E[A]\mathbb E[B].$$ This equation holds if and only if $$A$$ and $$B$$ are uncorrelated. In fact, this equation is so important that it is the basis for the concept of covariance, defined as $$\textrm{covariance}(A,B)=\mathbb E[AB]-\mathbb E[A]\mathbb E[B],$$ i.e. the left side minus the right side of the previous equation. Saying that $$A$$ and $$B$$ are uncorrelated is the same thing as saying that their covariance is zero.

The most common way for random variables to be uncorrelated is if they are independent, which is a more advanced concept. (But not too much more...)

Let $$a$$ be the average of $$a_i, i = 1 \to n$$ and $$b$$ be the average of $$b_j, j = 1 \to m$$. Then, $$na = \sum_{i=1}^n a_i$$ and $$mb = \sum_{j=1}^m b_j$$.

Simply multiplying both sides, $$abmn = \sum_{i=1}^n \sum_{j=1}^m a_ib_j$$. Taking the $$mn$$ to the other side, $$\boxed{ ab = \frac{\sum_{i=1}^n \sum_{j=1}^m a_ib_j}{mn}}$$

The right hand side, however, is the sum of all products $$a_ib_j$$ divided by the number of such elements $$a_ib_j$$. Therefore, the RHS is infact the average of the $$a_ib_j$$, while the left hand side is the product of the individual averages, as desired.

• I think that taking $\bar{ab}$ as the average of the terms $a_ib_j$ is an unusual use of terminology. Jan 5, 2020 at 8:49