The formula for the multiple coefficient of correlation of two independent variables ($x_1$ and $x_2$) and an dependent variables ($y$) is this:


What is the formula for three ($x_1$, $x_2$, $x_3$) or four ($x_1$, $x_2$, $x_3$, $x_4$) independent variables? I would like to know for my regression analysis.


3 Answers 3


Suppose we have a linear regression model $$y=y_{12\ldots p-1}+\varepsilon_{12\ldots p-1}\,,$$

where $y_{12\ldots p-1}=\beta_0+\beta_1 x_1+\beta_2x_2+\cdots+\beta_{p-1}x_{p-1}$ is the part of $y$ explained by $(x_1,x_2,\ldots,x_{p-1})$ and $\varepsilon_{12\ldots p-1}$ is the unexplained part. Parameters $(\beta_0,\beta_1,\ldots,\beta_{p-1})$ are estimated by the method of least squares to obtain the fitted model $\hat y=\hat y_{12\ldots p-1}$.

By definition, the (sample) multiple correlation coefficient of $y$ on $x_1,x_2,\ldots,x_{p-1}$ is $$r=r_{0\cdot 12\ldots p-1}=\operatorname{corr}(y,\hat y)$$

A related quantity is the coefficient of determination, given by

$$r^2=\frac{\operatorname{var}(\hat y)}{\operatorname{var}(y)}=1-\frac{\operatorname{var}\left(\varepsilon_{12\ldots p-1}\right)}{\operatorname{var}(y)}$$

Towards getting a computational formula of $r$, consider the correlation matrix $R=(r_{ij})_{0\le i,j\le p-1}$ of $(y,x_1,\ldots,x_{p-1})$ where $r_{ij}=\begin{cases}\operatorname{corr}(y,x_j)&,\text{ if }i=0 \\\operatorname{corr}(x_i,x_j)&,\text{ else }\end{cases}$ for every $j$.

So the matrix looks like

$$R=\begin{bmatrix}1& r_{01}& r_{02}& \cdots& r_{0\overline{p-1}} \\ r_{01}& 1& r_{12}& \cdots & r_{1\overline{p-1}} \\ r_{02}& r_{12}& 1& \cdots& r_{2\overline{p-1}} \\ \vdots& \vdots & \vdots& \ddots& \vdots \\ r_{0\overline{p-1}}& r_{1\overline{p-1}}& r_{2\overline{p-1}}& \cdots& 1 \end{bmatrix}$$

Let $R_{ij}$ be the cofactor of the $(i,j)$th element of $R$.

Then it can be shown that

$$\color\green{\boxed{r=\sqrt{1-\frac{\det R}{R_{11}}}}}$$

(Nothing changes if there is no intercept in the model.)

The above gives an expression in terms of the simple correlation coefficients $r_{ij}$. The formula in the original post can be derived as a particular case when $p=3$:


If $(r_{ij})^{-1}=(r^{ij})$, then yet another formula is $$\boxed{r=\sqrt{1-\frac1{r^{00}}}}$$

In terms of the dispersion matrix $(s_{ij})_{0\le i,j\le p-1}$ of $(y,x_1,\ldots,x_{p-1})$ and $(s_{ij})^{-1}=(s^{ij})$, we have


For details and other formulae, following references are helpful:


One option is to just take the square root of the $R^2$ obtained when you do linear regression.

You can also do it this way: $R_{y \cdot \textbf{x}} = \sqrt{R_{y\textbf{x}}R_{\textbf{xx}}^{-1}R_{\textbf{x}y}}$, where the matrices are partitions of your sample correlation matrix: $R = \begin{pmatrix} 1 & R_{y\textbf{x}} \\R_{\textbf{x}y} & R_{\textbf{xx}} \end{pmatrix}$.

The idea behind this is to find the linear combination of your independent variables that maximises the correlation.


If there is intercept, for multiple regression, #regressors=2, $R^2 = [correl(y,x1) corr(y, x2)]\begin{pmatrix} correl(x1,x2) & 1 \\1 & correl(x1,x2) \end{pmatrix}^{-1}[correl(y,x1) corr(y, x2)]'$

You can verify above


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.