How Regression Reveals Relative Volatility Beyond Correlation
Summary
The document contrasts correlation with least-squares regression when comparing two asset series. Correlation is standardized by both series’ volatility, so it describes how they move together without showing the scale of one series relative to the other. If one series is a constant multiple of another, their correlation remains perfect even when their volatility levels differ.
Regression retains that scale relationship in its slope coefficient. The example compares a series with the original and one doubled in magnitude: both have the same correlation with the original, while their fitted slopes differ. This makes regression useful for describing the magnitude of a linear relationship alongside its direction. The explanation is a simple exact-multiple case; it does not address noisy data, changing relationships, or other modeling considerations in dynamic regressions.
Key ideas
- Correlation measures standardized co-movement and is unchanged by multiplying one series by a positive constant.
- A regression slope reflects the scale of the dependent series relative to its predictor.
- Perfect correlation alone does not imply equal volatility.
- The example illustrates a linear scale difference, not the broader behavior of dynamic regression models.
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Full text
# Why does regression capture differences in volatility?
# Why does regression capture differences in volatility?
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 by SRKX (score 6, accepted)
https://quant.stackexchange.com/a/11215
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 $X$.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.