Suppose we have a linear regression model of the following format : $$ y(x) = \beta_0 + \beta_1 x_1+ \beta_2x_2+\beta_3x_3+\epsilon$$

We know that the prediction interval associated with a level $\alpha$, for a new observation $x_0$, is :

$$ IC_{1-\alpha}=\left[\hat{y}_i \pm t_{\alpha/2, n-p-1}\times \sqrt{\hat{\sigma}^2(1+x_0^T(X^TX)^{-1}x_0)}\right]$$

The asked question is :
Without calculation and using approximations, give a prediction interval at $95\%$ for a given $(x_1,x_2,x_3)$.
Available information :

  • Output of the summary of this regression model given by R (so we have the $\beta s$, $\hat{\sigma}^2$ and $std(\beta)$)
  • The means and std errors of our variables.

We can approximate $t_{\alpha/2, n-p-1}$ by 2, how can we approximate $x_0^T(X^TX)^{-1}x_0$ ?


1 Answer 1


The covariance matrix of $\hat{\beta}$ you can obtain in R by vcov(model). If you are unwilling to do that then can crudely approximate $\hat{\sigma}^2(X'X)^{-1}$ by a diagonal matrix where its diagonal consists out of the variances of each $\hat{\beta}_j$, respectively. These values (standard deviation) you can find in the summary output.

  • $\begingroup$ As we don't have access to R for the question, the second answer is what I was looking for. Could you please elaborate on why the diagonal matrix of the variances of $\hat{\beta}$ is a good/decent approximate of $\hat{sigma}^2(X'X)^{-1}$ ? Thanks. $\endgroup$
    – mlx
    Commented Dec 17, 2018 at 18:48
  • $\begingroup$ It depends whether your model is stable. If you have complete multicoliniearity, then $(X'X)$ is singular, on the other extreme if the $x$s are orthogonal $\sigma^2(X'X)$ is diagonal as the non-diagonal entries are the covariances of the $\hat{\beta}_s$. Hence, given that your model is pretty stable (low standard deviations of $\hat{\beta}$s and significant model) then you can consider neglecting this covariance terms. $\endgroup$
    – V. Vancak
    Commented Dec 17, 2018 at 18:55

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