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I don't even know if this question makes sense, but what is the difference between multiple regression and partial correlation (apart from the obvious differences between correlation and regression, which is not what I am aiming at)?

I want to figure out the following:
I have two independent variables ($x_1$, $x_2$) and one dependent variable ($y$). Now individually the independent variables are not correlated with the dependent variable. But for a given $x_1$ $y$ decreases when $x_2$ decreases. So do I analyze that by means of multiple regression or partial correlation?

edit to hopefully improve my question: I am trying to understand the difference between multiple regression and partial correlation. So, when $y$ decreases for a given $x_1$ when $x_2$ decreases, is that due to the combined effect of $x_1$ and $x_2$ on $y$ (multiple regression) or is it due to removing the effect of $x_1$ (partial correlation)?

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Multiple linear regression coefficient and partial correlation are directly linked and have the same significance (p-value). 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

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

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

You see that the numerators are the same which tell that both formulas measure the same unique effect of $x_1$. I will try to explain how the two formulas are structurally identical and how they are not.

Suppose that you have z-standardized (mean 0, variance 1) all three variables. The numerator then is equal to the covariance between two kinds of residuals: the (a) residuals left in predicting $y$ by $x_2$ [both variables standard] and the (b) residuals left in predicting $x_1$ by $x_2$ [both variables standard]. Moreover, the variance of the residuals (a) is $1-r_{yx_2}^2$; the variance of the residuals (b) is $1-r_{x_1x_2}^2$.

The formula for the partial correlation then appears clearly the formula of plain Pearson $r$, as computed in this instance between residuals (a) and residuals (b): Pearson $r$, we know, is covariance divided by the denominator that is the geometric mean of two different variances.

Standardized coefficient beta is structurally like Pearson $r$, only that the denominator is the geometric mean of a variance with own self. The variance of residuals (a) was not counted; it was replaced by second counting of the variance of residuals (b). Beta is thus the covariance of the two residuals relative the variance of one of them (specifically, the one pertaining to the predictor of interest, $x_1$). While partial correlation, as already noticed, is that same covariance relative their hybrid variance. Both types of coefficient are ways to standardize the effect of $x_1$ in the milieu of other predictors.

Some numerical consequences of the difference. If R-square of multiple regression of $y$ by $x_1$ and $x_2$ happens to be 1 then both partial correlations of the predictors with the dependent will be also 1 absolute value (but the betas will generally not be 1). Indeed, as said before, $r_{yx_1.x_2}$ is the correlation between the residuals of y <- x2 and the residuals of x1 <- x2. If what is not $x_2$ within $y$ is exactly what is not $x_2$ within $x_1$ then there is nothing within $y$ that is neither $x_1$ nor $x_2$: complete fit. Whatever is the amount of the unexplained (by $x_2$) portion left in $y$ (the $1-r_{yx_2}^2$), if it is captured relatively highly by the independent portion of $x_1$ (by the $1-r_{x_1x_2}^2$), the $r_{yx_1.x_2}$ will be high. $\beta_{x_1}$, on the other hand, will be high only provided that the being captured unexplained portion of $y$ is itself a substantial portion of $y$.


From the above formulas one obtains (and extending from 2-predictor regression to a regression with arbitrary number of predictors $x_1,x_2,x_3,...$) the conversion formula between beta and corresponding partial r:

$$r_{yx_1.X} = \beta_{x_1} \sqrt{ \frac {\text{var} (e_{x_1 \leftarrow X})} {\text{var} (e_{y \leftarrow X})}},$$

where $X$ stands for the collection of all predictors except the current ($x_1$); $e_{y \leftarrow X}$ are the residuals from regressing $y$ by $X$, and $e_{x_1 \leftarrow X}$ are the residuals from regressing $x_1$ by $X$, the variables in both these regressions enter them standardized.

Note: if we need to to compute partial correlations of $y$ with every predictor $x$ we usually won't use this formula requiring to do two additional regressions. Rather, the sweep operations (often used in stepwise and all subsets regression algorithms) will be done or anti-image correlation matrix will be computed.


$^1$ $\beta_{x_1} = b_{x_1} \frac {\sigma_{x_1}}{\sigma_y}$ is the relation between the raw $b$ and the standardized $\beta$ coefficients in regression with intercept.

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  • $\begingroup$ Thank you. But how do I decide which one to go with, e.g. for the purpose described in my question? $\endgroup$ – user34927 Nov 17 '13 at 21:38
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    $\begingroup$ Obviously, you are free to choose: the numerators are the same, so they convey the same information. As for your (not fully clarified) question, it seems to be about topics "can regr. coef. be 0 when r isn't 0"; "can regr. coef. be not 0 when r is 0". There's a lot questions about that on the site. Just for example, you might read stats.stackexchange.com/q/14234/3277; stats.stackexchange.com/q/44279/3277. $\endgroup$ – ttnphns Nov 17 '13 at 22:00
  • $\begingroup$ I tried to clarify my question.. $\endgroup$ – user34927 Nov 17 '13 at 22:44
  • $\begingroup$ Fixing X1 ("x1 given") = removing (controlling) the effect of X1. There is no such thing as "combined effect" in multiple regression (unless you add the interaction X1*X2). Effects in multuple regression are competitive. Linear regression effects are actually partial correlations. $\endgroup$ – ttnphns Nov 17 '13 at 23:00
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    $\begingroup$ Wait a bit, @user34927. to prove that the DV (Y) is significantly correlated with one of two IVs (X1) if the effect of the other IV (X2) is removed The effect removed from where? If you "remove" X2 from both Y and X1 then the corr. between Y and X1 is the partial correlation. If you "remove" X2 from X1 only then the corr. between Y and X1 is called the part (or semi-partial) correlation. Were you really asking about it? $\endgroup$ – ttnphns Nov 19 '13 at 15:02
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Just bumped to this tread by chance. In the original answer, in the formula for $\beta_{x_1}$ the factor $\sqrt{SSY/SSX_1}$ is missing, that is $$ \beta_{x_1} = \frac{r_{yx_1} - r_{y x_2} ~r_{x_1 x_2}} {1-r^2_{x_1 x_2}} \times \sqrt{\frac{SSY}{SSX_1}}, $$ where $SSY=\sum_i (y_i-\bar y)^2$ and $SSX_1 = \sum_i {(x_{1i} - \bar{x}_1)^2}$.

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  • $\begingroup$ You are giving the formula of $b$. My answer was about $\beta$. $\endgroup$ – ttnphns Sep 23 '17 at 6:52

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