I understand, that the basic model of simple linear regression assumes homoscedasticity, i.e. the variances $\sigma^2$ around the regression line are equal for all predictor levels/values.

With this assumption in mind, I wonder why confidence intervals for a predicted $y$ value are not equally large for all predictor levels. The formula for such a confidence interval (e.g. 95%) suggests, that the interval gets larger, when the difference between the predictor value of interest $x$ and the mean predictor value $\overline x$ increases (see the enumerator under the root):

$\hat y \pm t_{n-2}(0.975) \cdot s \cdot \sqrt{1 + \frac{1}{n} + \frac{(x- \overline x)^2}{\sum_{i=1}^n \left(x_i - \overline x \right)^2}}$

The increasing size of CIs toward the edges of a distribution can also be seen in standard geom_smooth() plots by the ggplot2package in R.

Simple linear regression plot with CI

Now, how can the variances around the regression line be equal at all predictor levels, but the corresponding CIs are larger at the edges?
Is this a problem of sample and population, insofar as the difference in CI range shrinks, when the sample gets larger (When $n \rightarrow \infty$, you can drop the entire root term in the CI formula), while at the same time there is higher evidence for homoscedasticity when $n$ gets larger?

Or do I get things completely wrong and the two things are totally independent of each other?

  • 1
    $\begingroup$ See @gung's answer here. Nothing to do with heteroscedascity. $\endgroup$ Commented Jul 2, 2013 at 15:29
  • $\begingroup$ Or even heteroscedasticity. $\endgroup$ Commented Jul 2, 2013 at 15:36
  • 1
    $\begingroup$ You assume equal variances around the true regression line, not your fitted one. $\endgroup$ Commented Jul 2, 2013 at 15:40

1 Answer 1


The standard least squares regression line will always go through the point ($\bar{x}$,$\bar{y}$). So in the center of the line the only uncertainty is from the variability of $\bar{y}$, the estimated slope of the line will not affect your prediction at the mean of the x's. As you move away from the center your predictions now depend also on the estimated slope, think of pinning the line to the center and then rotating the line a small amount, near the center the line does not move very far, but further away from the center that rotation translates to a bigger difference in the predicted values. Note that in the formula you quote, when predicting at $\bar{x}$ the rightmost piece goes to 0, but get larger the further from the center you predict, this just mathematically represents the same idea.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.