# Is it true that $\frac{E[(|X - E[X]|)(|Y - E[Y]|)]}{\sigma_X \sigma_Y} = 1$?

Consider the well-known fact that correlation is bounded between $-1$ and $1$:

$$-1 \le \text{corr}(X,Y) = \frac{E[(X - E[X])(Y - E[Y])]}{\sigma_X \sigma_Y} \le 1.$$

I've been trying to wrap my mind intuitively around why this is so.

Question: Is this because (or is it true that)

$$\frac{E[(|X - E[X]|)(|Y - E[Y]|)]}{\sigma_X \sigma_Y} = 1?$$

(Notice the absolute value signs in the numerator).

Example: I notice that this is true in case $X$ were to take on values $\{1, 3\}$ and $Y$ were to take on values $\{2, 6\}$ in a uniform distribution. That is:

$$\frac{\frac{(1-2) + (3-2)}{2} \cdot \frac{(2-4)+(6-4)}{2}}{1 \cdot 2} = \frac{0 }{2} = 0$$

yet

$$\frac{\frac{|(1-2)| + |(3-2)|}{2} \cdot \frac{|(2-4)|+|(6-4)|}{2}}{1 \cdot 2} = 1.$$

So is it true in general? If so, this would make understanding why correlation is bounded between $-1$ and $1$ quite easy for my mind to wrap around.

EDIT: The claim also seems to work on uniform $\{1,5\}$ and $\{1,7\}$:

$$\frac{\frac{(1-3) + (5-3)}{2} \cdot \frac{(1-4)+(7-4)}{2}}{2 \cdot 3} = \frac{0 \cdot 0}{6} = 0$$

yet

$$\frac{\frac{|(1-3)| + |(5-3)|}{2} \cdot \frac{|(1-4)|+|(7-4)|}{2}}{2 \cdot 3} = \frac{2 \cdot 3}{6} = 1.$$

• Hint: Not the good approach to prove this. Use instead the Cauchy-Schwartz inequality. – dv_bn May 3 '16 at 13:13
• I think you are getting 1 because you are considering only two possible values for the uniforms. Try using a more general uniform variable. – Greenparker May 3 '16 at 13:46

The reason that the corr$(X,Y)$ is between -1 and 1 is due to Cauchy-Schwarz In equality, which says $$Var(Y) \geq \dfrac{Cov(X,Y)^2}{Var(X)}.$$
By your definition $X$ and $Y$ are independent, and so $E\left[|X - E(X)||Y - E(Y)| \right] = E\left[|X - E(X)|\right] E\left[|Y - E(Y)| \right]$
For this to be equal to $\sigma_x\sigma_y$, $\sigma_x = E\left[|X - E(X)|\right]$ and $\sigma_y = E\left[|Y - E(Y)|\right]$. This is not true in general since $$E[|X - E[X]|]^2 \ne E[|X - E[X]|^2].$$
I think in your examples it works out because you have 2 values each variable can take. Try expanding to $X = \{1,3, 4\}$ and you will not get the same result.