Suppose you have the following model, where all variables are continuous:

$$y = \alpha + \beta_0x_0 + \beta_1x_1 + \beta_2 x_0 x_1 + \epsilon$$

The standard error for the effect of $x_1$ is

$$\text{se}(\hat\beta_1 + \hat\beta_2x_0)= \sqrt{\text{var}(\beta_1) + 2\text{cov}(\beta_1, \beta_2) + \text{var}(\beta_2)x_0^2} $$

The last term in the sum says that as the values of the modulator $x_0$ get large in magnitude the precision of the estimates decreases. This is especially odd if your data is mostly at large magnitudes of $x_0$.

Why is this the case? Is it related to some other properties of linear models or is it a one-off oddity?

  • $\begingroup$ You formula has several typos in it. The correct expression should be the root of the variance of $\hat\beta_1 + x_0\hat\beta_2,$ which immediately shows why the SE varies with $x_0.$ It's unclear why this implies interactions are harder to estimate at high values of $x_0$ -- the SE just as well could decrease as $x_0$ increases through its natural range. $\endgroup$
    – whuber
    Sep 15, 2020 at 16:37
  • $\begingroup$ How can the SE decrease? Higher values of $x_0$ increase the last term monotonically as they get larger in magnitude, and that's their only impact on SE. $\endgroup$
    – badmax
    Sep 15, 2020 at 16:55
  • 1
    $\begingroup$ It's often the case that the covariance is negative. It can even be far greater than the variance of $\beta_2$ in magnitude. It's also unclear what you mean by "high" values: for instance, is $-1$ higher or lower than $-2$? This issue will be more apparent to you once you fix up the typos in the right hand side of your equation: the covariance needs to be multiplied by $x_0.$ $\endgroup$
    – whuber
    Sep 15, 2020 at 18:39
  • $\begingroup$ Thanks, I will add this as an answer. $\endgroup$
    – badmax
    Sep 16, 2020 at 0:06

1 Answer 1


Thanks to whuber who provided the answer in the comments.

The problem was the expression for the standard error was not correct. It should be

$$\text{se}(\hat\beta_1 + \hat\beta_2x_0)= \sqrt{\text{var}(\beta_1) + 2x_0\text{cov}(\beta_1, \beta_2) + \text{var}(\beta_2)x_0^2} $$


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