Assuming I have the following linear regression set-up:

$y_i = \alpha + x_i * \beta + \epsilon_i$

for $i = 1,2,..., n$.

When I run the regession, I get a $\beta$ and $\alpha$ estimates, along with their standard errors. Let $\sigma_{\alpha}$ and $\sigma_{\beta}$ be the standard error of $\alpha$ and $\beta$ respectively.

If I want to compute the standard error of the expression $\hat{\alpha} + x_i * {\hat\beta}$ for each value of $i$, would that be:

$\sqrt{\sigma^2_{\alpha} + x^2_i * \sigma^2_{\beta}}$ ???

Any help would be appreciated! Thanks!

  • $\begingroup$ Well $x_i$ as you said can be treated as a constant. I mean overall standard error. We have the standard error of the two terms, are they additive? Is there a correlation term? $\endgroup$
    – Mayou
    Commented Nov 10, 2014 at 21:20
  • $\begingroup$ What would be your suggestion in computing the standard deviation of that quantity? $\endgroup$
    – Mayou
    Commented Nov 10, 2014 at 21:40
  • $\begingroup$ Actually, this is not really what I am after. I am looking for the standard error of that specific quantity. $\endgroup$
    – Mayou
    Commented Nov 10, 2014 at 21:52
  • $\begingroup$ well you could calculate $var(\hat{\alpha} + x_i \hat{\beta})$ if you knew $E(\hat{\alpha}\hat{\beta})$. You can find formulas for $\hat{\alpha}$ and $\hat{\beta}$, so perhaps you can calculate this expectation analytically. $\endgroup$ Commented Nov 11, 2014 at 0:17

1 Answer 1


Perhaps calculating this variance will help

\begin{align} var(\hat{\alpha} + x_i \hat{\beta})&= var(\hat{\alpha})+x_i^2 var(\hat{\beta}) + 2cov(\hat{\alpha},\hat{\beta}x_i)\\ &= var(\hat{\alpha})+x_i^2 var(\hat{\beta}) + 2x_icov(\hat{\alpha},\hat{\beta})\\ &= var(\hat{\alpha})+x_i^2 var(\hat{\beta}) - 2x_i \frac{\sigma^2\bar{x}}{S_{xx}} \end{align}

where $\sigma^2=var(\epsilon_i)$. The last line comes from here. If $\sigma^2$ is unknown, an unbiased estimator for it is

$$\hat{\sigma}^2 = (S_{yy} − \beta S_{xy})/(n − 2)$$

Sanity check: Note that the covariance term is negative. This makes sense. In simple linear regression the point $(\bar{x},\bar{y})$ is always on the regression line. Now imagine increasing the slope, but fixing $(\bar{x},\bar{y})$, in this case the y intercept would decrease. Therefore, it makes sense that the estimate of the slope and the estimate of the intercept would be negatively correlated.


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