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I stumbled upon the following formula for the coefficient of determination:

$$1-R_{y(x_1,x_2...x_n)}^2=\left(1-\rho_{y,x_1}^2\right)\left(1-\rho_{y,x_2(x_1)}^2\right)\left(1-\rho_{y,x_3(x_1,x_2)}^2\right)\,\cdots\,\left(1-\rho_{y,x_n(x_1,x_2...x_{n-1})}^2\right).$$

where $R_{y(x_1,x_2,...,x_n)}$ is the coefficient of determination associated with the multiple linear regression between $y$ and ${x_1,x_2,...,x_n}$ and $\rho_{y,x_p(x_1,x_2,...,x_{p-1})}$ is the partial correlation between $y$ and $x_p$ controlling for $x_1,x_2,...,x_{p-1}$. Although this intuitively makes sense, would anyone have a proof of this formula?

I had a go by starting with the regression model: $$y=\mathbf{\beta}^T\mathbf{x}+\epsilon$$ where $\mathbf{\beta}$ is the vector of regression coefficients and $\epsilon$ is the error term. Then $$1-R_{y(x_1,x_2,...,x_n)}^2=E[\epsilon^2]/\sigma_y^2$$ where $\sigma_y^2$ is the variance of y. One can then try look at: $$1-R_{y(x_1,x_2,...,x_n)}=\frac{1}{\sigma_y^2}E[(y-\mathbf{\beta}\mathbf{x})^T(y-\mathbf{\beta}\mathbf{x})]$$ by rewriting $\mathbf{\beta}$ in terms of the correlations and standard deviations between the explanatory variables $\mathbf{x}$ and dependent variable $y$. However this seems very long winded especially as then the result would have to be refactored in terms of the partial correlations which the final formula has. So I was wondering if anyone knows a better, perhaps recursive/inductive approach, starting with 1 variable regression and adding more.

Thanks for the help!

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1 Answer 1

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I think I managed a proof by induction, let me know what you think.

Statement to prove for all positive integers $n$: $$1-R_{y(x_1,x_2...x_n)}^2=\left(1-\rho_{y,x_1}^2\right)\left(1-\rho_{y,x_2(x_1)}^2\right)\left(1-\rho_{y,x_3(x_1,x_2)}^2\right)\,\cdots\,\left(1-\rho_{y,x_n(x_1,x_2...x_{n-1})}^2\right).$$

  1. Prove for $n=1$. This is relatively straightforward and has been done here before: Correlation Coefficient and Determination Coefficient
  2. Assume true for $n=k$: $$1-R_{y(x_1,x_2...x_k)}^2=\left(1-\rho_{y,x_1}^2\right)\left(1-\rho_{y,x_2(x_1)}^2\right)\left(1-\rho_{y,x_3(x_1,x_2)}^2\right)\,\cdots\,\left(1-\rho_{y,x_n(x_1,x_2...x_{k-1})}^2\right).$$
  3. Then attempt to prove for $n=k+1$. Start with the regression model: $$y=\sum_i^n\beta_{y,x_i(\{x\}_{k+1}\setminus x_i)}x_i+\epsilon_{k+1}$$ where $\{x\}_{a}$ is the set of $a$ independent variables. Therefore: $$y=\sum_i^k\left(\beta_{y,x_i(\{x\}_{k+1}\setminus \{x_i\})}x_i\right)+\beta_{y,x_{k+1}(\{x\}_k)}x_{k+1}+\epsilon_{k+1}$$

$$y=\sum_i^k\left(\beta_{y,x_i(\{x\}_{k+1}\setminus \{x_i\})}x_i\right)+\beta_{y,x_{k+1}(\{x\}_k)}\left(x_{k+1}-\sum_i^k\beta_{x_{k+1}x_i(\{x\}_k\setminus x_i)}x_i\right)+\beta_{y,x_{k+1}(\{x\}_k)}\sum_i^k\beta_{x_{k+1}x_i(\{x\}_k\setminus x_i)}x_i+\epsilon_{k+1}$$ $$y-\sum_i^k\left(\beta_{y,x_i(\{x\}_{k+1}\setminus \{x_i\})}x_i\right)-\beta_{y,x_{k+1}(\{x\}_k)}\sum_i^k\beta_{x_{k+1}x_i(\{x\}_k\setminus x_i)}x_i=\beta_{y,x_{k+1}(\{x\}_k)}\left(x_{k+1}-\sum_i^k\beta_{x_{k+1}x_i(\{x\}_k\setminus x_i)}x_i\right)+\epsilon_{k+1} $$ $$\epsilon_k=\beta_{y,x_{k+1}(\{x\}_k)}\left(x_{k+1}-\sum_i^k\beta_{x_{k+1}x_i(\{x\}_k\setminus x_i)}x_i\right)+\epsilon_{k+1}$$

where we have used

$$\sum_i^k\left(\beta_{y,x_i(\{x\}_{k+1}\setminus \{x_i\})}x_i\right)+\beta_{y,x_{k+1}(\{x\}_k)}\sum_i^k\beta_{x_{k+1}x_i(\{x\}_k\setminus x_i)}x_i=\sum_i^k\left(\beta_{y,x_i(\{x\}_{k}\setminus \{x_i\})}x_i\right)$$

This is an intuitive result but probably can be proved by induction as well. We have essentially collapsed the $n=k+1$ regression model into the $n=1$ regression model using the error terms. Therefore using a similar proof to 1. we can see that

$$E[\epsilon_{k+1}^2]=E[\epsilon_k^2](1-\rho_{y,x_k+1(x_1,x_2,...,x_k)})$$

Using the relation between the determination coefficient and the error terms we get:

$$1-R_{y(x_1,x_2...x_{k+1})}^2=(1-R_{y(x_1,x_2...x_{k})}^2)(1-\rho_{y,x_k+1(x_1,x_2,...,x_k)})$$

Then, using 2. on the RHS, we get the desired result:

$$1-R_{y(x_1,x_2...x_{k+1})}^2=\left(1-\rho_{y,x_1}^2\right)\left(1-\rho_{y,x_2(x_1)}^2\right)\left(1-\rho_{y,x_3(x_1,x_2)}^2\right)\,\cdots\,\left(1-\rho_{y,x_k+1(x_1,x_2,...,x_k)}\right)$$

Hence we have proven that if true for $n=k$, then the statement is true for $n=k+1$ and since it is true for $n=1$, it is true for all positive integers $n$.

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