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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

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

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    $\begingroup$ Hint: Not the good approach to prove this. Use instead the Cauchy-Schwartz inequality. $\endgroup$ – dv_bn May 3 '16 at 13:13
  • $\begingroup$ I think you are getting 1 because you are considering only two possible values for the uniforms. Try using a more general uniform variable. $\endgroup$ – Greenparker May 3 '16 at 13:46
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The reason that the corr$(X,Y)$ is between -1 and 1 is due to Cauchy-Schwarz In equality, which says $$Var(Y) \geq \dfrac{Cov(X,Y)^2}{Var(X)}. $$

You can find the proof on the wiki page. In addition, your claim is in general untrue, but for your simple example it turns out to be true.

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.

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    $\begingroup$ My examples were working (I think) not because they had 2 values, but because there wasn't at least one value that was more than equidistant away. Intuitively, variance/standard deviation gives more weight to values further away from the mean relatively (whereas the expected values in the numerator don't). $\endgroup$ – George May 3 '16 at 15:46
  • $\begingroup$ @George Ya that would explain it. With 2 values, you would always be "equidistant" from the mean. $\endgroup$ – Greenparker May 3 '16 at 15:47

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