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So in the text book, it has the following proof of correlation being between $-1$ and $1$: enter image description here

This proof seems solid, but the only thing I am not clear about is:

Why is the variance of the ratio $X$ and $\sigma$, i.e., $Var(X/\sigma$) equal to $Var(X)/\sigma^2_x$?

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    $\sigma_X^2$ is defined to be $Var(X)$ – 2017-02-02
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    @DougM Yes but why is $Var(X/\sigma$) equal to $Var(X)/\sigma^2_x$? – 2017-02-02
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    Proper notation in this case is $\sigma_X^2$, not $\sigma_x^2.$ You should be careful about which is $X$ and which is $x$. The expression $\Pr(X\le x)$ is otherwise incomprehensible, as are some other things. – 2017-02-04

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The variance is the square of the deviation. So $\text{Var } x=\sigma_x^2$ and then the numerator and denominator are equal.

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    Yes I understand this part, but why $Var(X/\sigma$) equals $Var(X)/\sigma^2_x$ in the proof? – 2017-02-02
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    @Kior Recall that $\text{Var }(aX)=a^2 \text{Var X}$. Here is $\sigma$ is just a fixed value so it behaves just the same under the variance. – 2017-02-02
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Definition:

$$\operatorname{Var}(X) = \frac 1 n \sigma_X^2= \sum (X_i - \bar X)^2$$

$$\operatorname{Var} \left(\frac {X}{\sigma_X}\right) = $$ $$\frac 1 n \sum \left(\frac {X_i}{\sigma_X} - \frac {\bar X}{\sigma_X}\right)^2\\ \frac 1 n \sum \left(\frac {X_i-\bar X}{\sigma_X}\right)^2\\ \frac 1{\sigma_X^2} \cdot \frac 1 n \sum (X_i-\bar X)^2\\ \frac 1{\sigma_X^2} \operatorname{Var}(X)=\frac 1{\sigma_X^2} \sigma_X^2 = 1$$

Definition:

$$\operatorname{Cov}(X,Y) = \frac 1 n \sum (X_i - \bar X)(Y_i - \bar Y)$$

The Cauchy-Schwartz inequality:

$$\left(\sum uv\right)^2 \le \sum u^2\sum v^2$$

Substitute $u = (X_i - \bar X)$ and $v = (Y_i - \bar Y)$

$$\left(\sum (X_i - \bar X)(Y_i - \bar Y)\right)^2 \le \sum (X_i - \bar X)^2\sum (Y_i - \bar Y)^2\\ |\operatorname{Cov}(X,Y)| \le \sigma_X\sigma_Y$$

Definition:

$$\rho_{X,Y} = \frac {\operatorname{Cov}(X,Y)}{\sigma_X\sigma_Y}$$

$$|\operatorname{Cov}(X,Y)|\le\sigma_X\sigma_Y \\ |\rho_{X,Y}| \le 1$$

And it might be worth noting that $\sigma_X, \sigma_Y$ are equivalent to distance measures, and $\rho_{X,Y}$ is equivalent to $\cos \theta$

Law of cosines:

$$c^2 = a^2 + b^2 + 2ab\cos \theta$$

Has the parallel:

$$\sigma_{X+Y}^2 = \sigma_X^2 + \sigma_Y^2 + 2\sigma_X\sigma_Y \rho_{X,Y}$$

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    Often one distinguishes between a population mean $\mu_X = \operatorname{E}(X)$ and a sample mean $\bar X = (X_1+\cdots+X_n)/n,$ wheren $X_1,\ldots,X_n$ is a sample rather than the whole population. This is crucial when you try to explain why one sometimes divides by $n-1$ rather than by $n$ in finding the sample variance and when certain random variables have a t-distribution, etc. – 2017-02-04
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    I took the liberty of correcting the omission of the crucial factor $\dfrac 1 n. \qquad$ – 2017-02-04