Can you break up a regression slope coefficient into a product of slopes? $\beta_{A,C}=\beta_{A,B}\beta_{B,C}$

Question

Any regression slope coefficient $\beta$ is defined as:

$\beta_{X,Y}=Cov(X,Y)/Var(Y)$

It seems intuitive that you can break up a regression slope coefficient like this:

$\beta_{A,C}=\beta_{A,B}\beta_{B,C}$

(i.e. in a similar sort of manouevre to the chain rule in calculus.)

Is this correct?

Thanks!

Answer 1

$Cov(X,Y) = \sigma_X\sigma_Y\rho_{X,Y}.$ Where $\rho$ is the correlation. Usually, you see this written the other way round to define correlation.
$\beta_{X/Y} = \frac{\sigma_X}{\sigma_X}\rho_{X,Y}$

$\beta_{A/B}\beta_{B/C} = \frac{\sigma_A}{\sigma_C}\rho_{A,B}\rho_{B,C}$

$\beta_{A/B}\beta_{B/C} = \beta_{A/C}$ would imply that $\rho_{A,B}\rho_{B,C} = \rho_{A,C}$ which is generally not true.

Suppose that A and C are perfectly correlated, and B is independent, and all three have the same variance.

$\beta_{A/B} = 0, \beta_{B/C} = 0, \beta_{A/C} = 1$

Answer 2

It does not seem to be correct. Let $A,B,C$ be random variables such that $P(A=1)=P(A=0)=P(A=-1)=1/3$, $C=A$, and $B=A^2$. Observe that $E[A]=E[C]=0$ and $E[B]=2/3$. Then, letting $\rho$ denote the covariance, $$ \rho_{A,C} = Var(A)= \frac{1}{3}(1-0)^2+\frac{1}{3}(0-0)^2+\frac{1}{3}(1-0)^2=\frac{2}{3} $$ On the other hand, $$ \rho_{A,B} = \rho_{B,C} = \frac{1}{3}(1)(1-\frac{2}{3})+\frac{1}{3}(0)(0-\frac{2}{3}) + \frac{1}{3}(-1)(1-\frac{2}{3})=0 $$ which makes it impossible that $$ \beta_{A,C}=\beta_{A,B}\beta_{B,C}. $$