For simplicity, let's restrict multiple linear regression case to 2 predictors, $x_1, x_2$. You regress $y$ on each individually and get $\hat{\beta}_1, \hat{\beta}_2$. Now you regress $y$ on both and get $\hat{\gamma}_1, \hat{\gamma}_2$.

So I know if $(x_1 - \bar{x}_1 \perp \bar{x}_2$, then $\hat{\beta}_i = \hat{\gamma}_i$, but if they're not orthogonal, what can be said about the relationship between them?

If in each of the simple linear regression cases, the slope was positive, i.e., $\hat{\beta}_1, \hat{\beta_2} > 0$, can we expect $\hat{\gamma}_1, \hat{\gamma}_2 > 0$?

I just asked this question on the math SE (https://math.stackexchange.com/questions/3791992/relationship-between-projection-of-y-onto-x-1-x-2-individually-vs-projecti), but I'm looking for more of a linear algebra intuition in that question. Here, I'm opening for any kind of intuition, statistical or not.

  • $\begingroup$ See Frisch–Waugh–Lovell theorem and Simpsons paradox. $\endgroup$ – Jesper for President Aug 15 '20 at 17:40

Here is a simple example that provides insight.

y = c(5.8,5.2,4.7,8.7,8.1,7.7,10.2,9.6,9.0)
x1 = c(1,1.5,2,1.8,2.7,3.5,3,4,4.5)
x2 = c(1,1,1,2,2,2,3,3,3)


legend("topleft", c("x2=1", "x2=2", "x2=3"), pch=1, col=1:3)

The simple regressions have significant positive relationships, but the multiple regression shows that the effect of x1 is significant and negative. The graph gives the intuition clearly:

Multiple Regression Graph

Ignoring x1, there are generally higher values of y for larger x2. Similarly, ignoring x2, there are generally larger values of y for larger x1. These observations explain the simple regression results.

In the multiple regression model, the slope coefficients are estimates of the effect of one x while the other is held fixed. And you can easily see in the graph that the values of y are smaller as x1 increases within any of the three groups where x2 is held fixed (at either 1,2, or 3).

  • $\begingroup$ "In the multiple regression model, the slope coefficients are estimates of the effect of one x while the other is held fixed" In the simple linear regression model, isn't the unmodeled $x$ technically fixed too? I don't think I understand what "fixed" entails in this context. $\endgroup$ – roulette01 Aug 15 '20 at 18:42
  • $\begingroup$ @dd2205 “fixed” meaning what? I take “fixed” to mean in the sense that we hold other variables fixed when we take partial derivatives in multivariable calculus. $\endgroup$ – Dave Aug 15 '20 at 19:02
  • 1
    $\begingroup$ The unmodeled x is definitely not fixed in the simple models. That's why the slope coefficients are both positive in my example. The reason for the "fixed" concept lies in the model itself: $E(y|x_1,x_2) = \beta_0 + \beta_1 x_1 + \beta_2x_2$ implies that, eg, $E(y|x_1,x_2=1) = \beta_0 + \beta_1 x_1 + \beta_2 (1) = (\beta_0 + \beta_2) + \beta_1 x_1$. So $\beta_1$ in the multiple regression model is the effect of $x_1$ when $x_2$ is fixed (at any value, not just 1.) $\endgroup$ – BigBendRegion Aug 15 '20 at 19:05
  • $\begingroup$ Yes, you can think of the "held fixed" concept in terms of partial derivatives as well. $\endgroup$ – BigBendRegion Aug 15 '20 at 19:07
  • $\begingroup$ This generalizes to any number of predictors right? For example, if we had $p$ predictors, we can interpret $\beta_1$ as the effect of $x_1$ when $x_2, \ldots, x_p$ are fixed? $\endgroup$ – roulette01 Aug 15 '20 at 19:15

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