# Is the term 'standardised regression coefficient' ambiguous?

Partial r is just another way of standardizing the coefficient, along with beta coefficient (standardized regression coefficient)$$^1$$. So, if the dependent variable is $$y$$ and the independents are $$x_1$$ and $$x_2$$ then

$$\text{Beta:} \quad \beta_{x_1} = \frac{r_{yx_1} - r_{yx_2}r_{x_1x_2} }{1-r_{x_1x_2}^2}$$

$$\text{Partial r:} \quad r_{yx_1.x_2} = \frac{r_{yx_1} - r_{yx_2}r_{x_1x_2} }{\sqrt{ (1-r_{yx_2}^2)(1-r_{x_1x_2}^2) }}$$

Those are not compatible definitions. Does anyone have a source for the formula in the CrossValidated answer?

• These are defining different things. The two quotations just happen to use the word "standardizing" in different senses, but both are perfectly clear.
– whuber
Commented Apr 22 at 13:47

They are compatible in that you can show that the "standardized coefficient" $$\hat{\beta^*}$$, defined by your first quotation, is indeed identical to the expression $$\hat{\beta}_{x_1}$$ in your cited answer. This has been proven in my answer as an intermediate step for validating the equivalence between two definitions of $$R^2$$. Although your second quotation involved the concept of "partial R", I am assuming it is actually irrelevant because your first quotation solely concerns with the "standardized regression coefficient".
Specifically, using the notations in my answer, I showed that for a general multiple linear regression with $$p$$ independent variables (where $$\hat{\gamma}$$ denotes the OLS estimate for the original regression parameters): \begin{align*} & \hat{\beta^*} = s_Y^{-1}\Lambda\hat{\gamma}, \tag{1}\label{1} \\ & \hat{\beta^*} = r_{XX}^{-1}r_{YX}, \tag{2}\label{2} \end{align*} where $$\Lambda = \operatorname{diag}(s_{X_1}, \ldots, s_{X_p})$$, $$r_{XX}$$ is the order $$p$$ sample correlation matrix of $$(X_1, \ldots, X_p)$$, $$r_{YX}$$ is the $$p$$-long column vector of sample correlations between $$Y$$ and $$(X_1, \ldots, X_p)$$, and $$s_{X_i}, s_Y$$ are sample standard deviations of $$X_i$$ and $$Y$$. In your case, $$p = 2$$, and what you are trying to confirm is just that the first component of $$\eqref{1}$$ and the first component of $$\eqref{2}$$ are the same (which is of course true given that as two vectors, $$\eqref{1}$$ and $$\eqref{2}$$ are proven to be the same).