# multiplying regression beta (covariance) coefficients

I consider 3 variables $$x$$, $$y$$, $$z$$ and study the relationship of the beta coefficient from OLS regression of two variables connected via a third.

Whilst it is not true that $$\beta_{(x,y)}\times\beta_{(y,z)}\equiv\beta_{(x,z)}$$ (no proof shown - but would be appreciated), does anyone know if there is a simple relationship between $$\beta_{(x,y)}$$, $$\beta_{(y,z)}$$ and $$\beta_{(x,z)}$$?

Another way to look at it is to use the relationship between beta's and correlation coefficient:

1- $$\beta_{(x,y)}$$ = $$\rho_{x,y}$$($$\sigma_{x}$$/ $$\sigma_{y}$$)

2- $$\beta_{(y,z)}$$ = $$\rho_{y,z}$$($$\sigma_{y}$$/ $$\sigma_{z}$$)

3- $$\beta_{(x,z)}$$ = $$\rho_{x,z}$$($$\sigma_{x}$$/ $$\sigma_{z}$$)

1 mutliplied by 2 - we get: $$\beta_{(x,y)}\times\beta_{(y,z)}$$ = $$\rho_{x,y}$$($$\sigma_{x}$$/ $$\sigma_{y}$$) $$\times$$ $$\rho_{y,z}$$($$\sigma_{y}$$/ $$\sigma_{z}$$) = $$\rho_{x,y}$$ .$$\rho_{y,z}$$ ($$\sigma_{x}$$/ $$\sigma_{z}$$)

therefore this can be proved if we can prove that correlation coeffiencients are compoundable - meaning $$\rho_{x,z}$$ $$\equiv$$ $$\rho_{x,y}$$ $$\rho_{y,z}$$ ?

• Please precisely define the meaning of the coefficients. Commented May 27, 2020 at 13:22
• I wanted to know if the relationship of the beta coefficient from OLS regression of two variables connected via a third. Beta(x,y) = covariance(x,y)/variance(x) Commented May 27, 2020 at 13:29
• - for anyone else the answer comes from the study of path analysis - faculty.cas.usf.edu/mbrannick/regression/Pathan.html Commented May 27, 2020 at 16:28
• For a simple, convincing counterexample, let $y$ be a any random variable independent of $x$ and let $z$ be a linear function of $x,$ so that the two betas on the left side of the equation are zero but the beta on the right side is $\pm1.$
– whuber
Commented Nov 1, 2023 at 22:13

## 1 Answer

As you indicate in the comment, for the statement $$\beta_{(x,y)}\times\beta_{(y,z)}\equiv\beta_{(x,z)}$$ to be true we would need $$\frac{Cov(X,Y)}{Var(X)}\frac{Cov(Y,Z)}{Var(Y)}\equiv\frac{Cov(X,Z)}{Var(X)}$$ or $$Cov(X,Y)\frac{Cov(Y,Z)}{Var(Y)}\equiv Cov(X,Z)$$ As a counterexample suffices to show that a statement not true, you might just run

x <- rnorm(100)
y <- rnorm(100)
z <- rnorm(100)

cov(x,y)*cov(y,z)/var(y)
cov(x,z)

As such I am also not aware of a simple relationship.