$\beta= r* \frac{s_y}{s_x} $ Any more intuition behind this?

Question

In my textbook, it says, we can see with some algebraic manipulation that: $\beta= r* \frac{s_y}{s_x} $

However, the text does not actually show this manipulation. Can anyone show it or the intuition behind this formula?

Answer 1

Let $c$ be the covariance between $y$ and $x$. The corresponding correlation is defined as $r = c/(s_xs_y)$. So then $$ \beta = \frac{c}{s_xs_x} = \frac{c}{s_xs_x}\Big(\frac{s_y}{s_y}\Big) = \frac{c}{s_xs_y}\Big(\frac{s_y}{s_x}\Big) = r\Big(\frac{s_y}{s_x}\Big). $$

As for intuition, one way to think about why this makes sense is to think about the units involved. The correlation coefficient is dimensionless. The slope is in $y$-units per $x$-units -- for example, if we regress income in dollars on age in years, the slope of that regression has units of dollars-per-year. Now note that the ratio $s_y/s_x$ is also in $y$-units per $x$-units, since $s_y$ is in units of $y$ and $s_x$ is in units of $x$ So multiplying the dimensionless correlation coefficient by $s_y/s_x$ results in something that is in the units that we know the slope has. Obviously this is not a full intuition for why the right-hand expression equals the slope, but it's at least an intuition for why it might be similar.

Answer 2

This is the slope of a simple OLS regression with intercept. The full derivation without linear algebra can be found here.

These are the two parameter estimates in the case of a linear regression model of the form: $\large Y\,=\,\beta_o \,+\,\beta_1\,X:$

$$\ \hat\beta_1\,=\, \underbrace{\text{cor}(Y, X)}_{r}\,\frac{\text{SD}(Y)}{\text{SD}(X)}\,=\,\frac{\text{cov} (Y,X)}{\text{var}(X)}\small $$

remembering that $ \text{cor} (X,Y) = \frac{\text{cov}(X,Y)}{\text{SD}(X) \text{SD}(Y)}.$

is the slope, while the intercept can be calculated knowing that the regression line passes through the means:

The intercept

$$\large \hat\beta_o\,=\,\bar Y\,-\,\hat\beta_1\,\bar X $$

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To be clear, I believe that your formula uses $r$ as the Pearson correlation coefficient, and $S_y= \text{SD}(Y)$ (and $S_x= \text{SD}(X)$.