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To prove that $-1\leq \operatorname{Corr}(X,Y) \leq 1$ , I am using the identity
$\operatorname{Var}(\frac{X}{\sqrt{\operatorname{Var}(x)}}-\frac{Y}{\sqrt{\operatorname{Var}(Y)}})\geq 0$. Where does this identity come from? How do I prove it?

Bernard
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elbecker
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    Variance is always nonnegative. Are you familiar with how to prove this? – Alex R. Sep 21 '20 at 16:54
  • @AlexR.Oh I see, thank you. That's what I was confused on. Variance is always non negative because it's the expectation of a squared value – elbecker Sep 21 '20 at 18:58

1 Answers1

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Let's consider the rv $Z=Y+aX$ and let's calculate its variance.

$$V(Z)=V(Y)+a^2V(X)+2a cov(X,Y)\geq0$$

As per the fact that variance cannot be negative, the above expression is a 2nd degree inequality in $a$ and as per the fact that the 2nd degree term has positive coefficient the $\Delta$ must be $\leq0$

Thus

$$4cov^2(X,Y)-4V(X)V(Y)\leq0$$

$$cov^2(X,Y)\leq V(X)V(Y)$$

$$|cov(X,Y)|\leq \sigma_X\sigma_Y$$

$$-\sigma_X\sigma_Y \leq cov(X,Y)\leq \sigma_X\sigma_Y$$

As you know, $\rho_{XY}=\frac{cov(X,Y)}{\sigma_X\sigma_Y}$

So your statement is proved

tommik
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