Why does regression capture differences in volatility?

Question

I read the following statement in the book python for data analysis, chapter 11, and I was wondering if someone could give me intuition about why regression has this effect? The purpose of the exercise was to compare a basic correlation between microsoft and apple versus a dynamic regression.

One issue with correlation between two assets is that it does not capture differences in
volatility. Least-squares regression provides another means for modeling the dynamic
relationship between a variable and one or more other predictor variables.

Answer 1

I guess what they are trying to say here is that, assume you have two time series $X$ and $Y$ which are exactly the same i.e. $X=Y$, the correlation is :

$$\rho_{X,Y}= \frac{Cov(X,Y)}{\sigma_X \sigma_Y}\overset{X=Y}{=}\frac{Cov(X,X)}{\sigma_X \sigma_X}=\frac{\sigma_X^2}{\sigma_X^2}=1$$

Now assume a time series $Z=2 \cdot X$, you have:

$$\sigma_Z=2 \sigma_X$$

and

$$Cov(X,Z)=Cov(X,2X)=2 Cov(X,X) = 2 \sigma_X^2$$

So,

$$\rho_{X,Z}= \frac{Cov(X,Z)}{\sigma_X \sigma_Z}=\frac{2 \sigma_X^2}{2\sigma_X^2}=1$$

The fact that $Z$ is twice as volatile as $Y$ does not appear in the correlation measure.

In regression you would fit:

$$Y_t=\alpha_Y + \beta_Y X_t + \epsilon_t ~ \text{and} ~ Z_t=\alpha_Z + \beta_Z X_t + \epsilon_t$$

Your regression would give you $\alpha_Y=\alpha_Z=0$, $\beta_Y=1$ and $\beta_Z=2$, which shows that $Z$ and $Y$ have a different relationship with $X$.