For a random pair $(X, Y)$, the correlation $r(X, Y)$ is defined as

$$ r(X, Y) ~ = ~ E\Big{(} \big{(}\frac{X-\mu_X}{\sigma_X}\big{)}\big{(}\frac{Y-\mu_Y}{\sigma_Y}\big{)} \Big{)} ~ = ~ E(X^*Y^*) $$

where $X^*$ and $Y^*$ are respectively $X$ and $Y$ measured in standard units.

Thus by definition, correlation is a pure number and has no units.

You have seen several properties of correlation in Data 8. Some are obvious, such as $r(X, Y) = r(Y, X)$. Some require proof.

In this brief section we will prove one principal property, which is that correlation is a number between $-1$ and $1$. You will prove a few other properties in exercises. In the next section we will specify the sense in which correlation measures clustering about a straight line.

### Lower Bound

As a preliminary, recall that

$$ E(X^*) = 0, ~~~ Var(X^*) = 1 = E({X^*}^2) $$

So also $E(Y^*) = 0$ and $E({Y^*}^2) = 1$.

We know the expected squares, and what we need is a bound on the expected product $E(X^*Y^*)$. A result that connects the squares and the product of two numbers is $(a+b)^2 = a^2 + 2ab + b^2$.

So let's find $E\big{(}(X^* + Y^*)^2\big{)}$ and see what that gives us.

$$ \begin{align*} E\big{(}(X^* + Y^*)^2\big{)} ~ &= ~ E({X^*}^2) + 2E(X^*Y^*) + E({Y^*}^2) \\ &= ~ 2 + 2E(X^*Y^*) \end{align*} $$

Since $E\big{(}(X^* + Y^*)^2\big{)} \ge 0$, we have

$$ 2 + 2E(X^*Y^*) \ge 0 $$

which is the same as

$$ E(X^*Y^*) \ge -1 $$

### Other Properties

As you know from Data 8, correlation measures linear association. In exercises you will show that if $Y$ is a linear function of $X$ then $r(X, Y)$ is either $1$ or $-1$.

You will also find the relation between $r(X, Y)$ and $r(X, W)$ where $W$ is a linear function of $Y$.

In the next section we will return to regression and formalize the idea that correlation measures clustering about a straight line. Our result will imply that if $r(X, Y)$ is either $1$ or $-1$, then the relation between $X$ and $Y$ must be perfectly linear.