De regresión múltiple o de correlación parcial?

Question

Ni siquiera sé si esta pregunta tiene sentido, pero ¿cuál es la diferencia entre la regresión múltiple y correlación parcial (aparte de las obvias diferencias entre la correlación y la regresión, que no es lo que estoy apuntando a)?

Quiero averiguar lo siguiente:
Tengo dos variables independientes ($x_1$, $x_2$) y una variable dependiente ($y$). Ahora individualmente las variables independientes no están correlacionadas con la variable dependiente. Pero para un determinado $x_1$ $y$ disminuye al $x_2$ disminuye. Así que puedo analizar que por medio de la regresión múltiple o de correlación parcial?

editar esperamos mejorar mi pregunta: Estoy tratando de entender la diferencia entre la regresión múltiple y correlación parcial. Así, al $y$ disminuye en un determinado $x_1$ al $x_2$ disminuye, es que debido al efecto combinado de $x_1$ $x_2$ $y$ (regresión múltiple) o es debido a la eliminación del efecto de la $x_1$ (correlación parcial)?

Answer 1

Lineal múltiple coeficiente de regresión y correlación parcial están directamente vinculados y tienen la misma significación (p-valor). Parcial r es simplemente otra manera de estandarizar el coeficiente, junto con beta coeficiente. Por lo tanto, si la variable dependiente es $y$ y los independientes se $x_1$ $x_2$

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

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

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 $y$.