# How to approximate prediction interval in linear regression

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 :

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

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 $$x_0^T(X^TX)^{-1}x_0$$ ?

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.
• 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.
• 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. Commented Dec 17, 2018 at 18:55