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?

  • 4
    $\begingroup$ $\overline{XY}=\overline{X}\,\overline{Y}$ is true for independent variables $X,Y$ but not in general. $\endgroup$
    – anon
    Jan 5, 2020 at 6:29
  • $\begingroup$ Which type of average are you referring to? There are many including geometric mean, arithmetic mean and others. $\endgroup$
    – Karl
    Jan 5, 2020 at 8:54
  • $\begingroup$ 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 $b$? $\endgroup$
    – almagest
    Jan 5, 2020 at 9:19

2 Answers 2


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.

Read the other answer for more details on the above phenomena.

  • $\begingroup$ I think that taking $\bar{ab}$ as the average of the terms $a_ib_j$ is an unusual use of terminology. $\endgroup$
    – almagest
    Jan 5, 2020 at 8:49
  • $\begingroup$ Yes, but that is what I think was meant. $\endgroup$ Jan 5, 2020 at 9:08

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