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The sign of the partial correlation coefficient is consistent with the sign of the corresponding estimated parameter. For example, in the estimated regression equation $\widehat{Y}=\hat{b}_{0}+\hat{b}_{1} X_{1}+\hat{b}_{2} X_{2}$, the sign of $r_{YX_{1}\cdot X_{2}} $ is consistent with the sign of $\hat{b}_{1}$, and the sign of $r{_{YX_{2}\cdot X_{1}}} $ is consistent with the sign of $\hat{b}_{2}$.

I don't understand why we have $r_{YX_{1}\cdot X_{2}} $ is consistent with the sign of $\hat{b}_{1}$ and the sign of $r{_{YX_{2}\cdot X_{1}}} $ is consistent with the sign of $\hat{b}_{2}$.

I know that $$r^{2}_{YX_{1}\cdot X_{2}}=\frac{\left(X^{\prime}_{1}N_{(\mathbf{1_{n}},X_{2})}Y\right)^{2}}{\left(X^{\prime}_{1}N_{(\mathbf{1_{n}},X_{2})}X_{1}\right)\cdot\left(Y^{\prime }N_{(\mathbf{1_{n}},X_{2})}Y\right)}，\quad r^{2}_{YX_{2}\cdot X_{1}}=\frac{\left(X^{\prime}_{2}N_{(\mathbf{1_{n}},X_{1})}Y\right)^{2}}{\left(X^{\prime}_{2}N_{(\mathbf{1_{n}},X_{1})}X_{2}\right)\cdot\left(Y^{\prime }N_{(\mathbf{1_{n}},X_{1})}Y\right)}，$$ Where $$N_{(\mathbf{1_{n}},X_{1})}=I_{n}-P_{(\mathbf{1_{n}},X_{1})},$$$$ P_{(\mathbf{1_{n}},X_{1})}=(\mathbf{1_{n}},X_{1})\cdot\left[(\mathbf{1_{n}},X_{1})^{\prime}(\mathbf{1_{n}},X_{1})\right]^{-1}\cdot(\mathbf{1_{n}},X_{1})^{\prime};$$$$\quad N_{(\mathbf{1_{n}},X_{2})},\quad P_{(\mathbf{1_{n}},X_{2})}\quad \text{similarly}.$$

So I think $r_{YX_{1}\cdot X_{2}} $ is consistent with the sign of $X^{\prime}_{1}N_{(\mathbf{1_{n}},X_{2})}Y$ and the sign of $r{_{YX_{2}\cdot X_{1}}} $ is consistent with the sign of $X^{\prime}_{2}N_{(\mathbf{1_{n}},X_{1})}Y.$

Do the following relationship hold? $$\text{sign}(X^{\prime}_{1}N_{(\mathbf{1_{n}},X_{2})}Y)=\text{sign}(\hat{b}_{1})\quad\text{and}\quad\text{sign}(X^{\prime}_{2}N_{(\mathbf{1_{n}},X_{1})}Y)=\text{sign}(\hat{b}_{2}).$$

$$(X^{\prime}_{2}N_{(\mathbf{1_{n}},X_{1})})\cdot\widehat{Y}$$$$=(X^{\prime}_{2}N_{(\mathbf{1_{n}},X_{1})})\cdot(\hat{b}_{0}+\hat{b}_{1} X_{1}+\hat{b}_{2} X_{2})$$$$=(X^{\prime}_{2}N_{(\mathbf{1_{n}},X_{1})}X_{2})\cdot \hat{b}_{2} \Rightarrow \text{sign}(X^{\prime}_{2}N_{(\mathbf{1_{n}},X_{1})}\widehat{Y})=\text{sign}(\hat{b}_{2}).$$ If $$\text{sign}(X^{\prime}_{2}N_{(\mathbf{1_{n}},X_{1})}\widehat{Y})=\text{sign}(X^{\prime}_{2}N_{(\mathbf{1_{n}},X_{1})}{Y}),$$ then the conclusion will be hold.

Elisa
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  • https://stats.stackexchange.com/q/44279/3277; https://stats.stackexchange.com/a/76819/3277 (they share the numerator) – ttnphns May 14 '23 at 13:49
  • It is just the definition. If you have doubt on this, please post the source of your definition of "partial correlation coefficient" first. – Zhanxiong May 14 '23 at 13:59
  • My source (Applied Statistical Linear Models by Kutner et al., page 270): the square root of a coefficient of partial determination is called a coefficient of partial correlation. It is given the same sign as that of the corresponding regression coefficient in the fitted regression function. – Zhanxiong May 14 '23 at 14:02

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