Confidence interval for difference of means in regression

Question

Suppose I have a quadratic regression model $$ Y = \beta_0 + \beta_1 X + \beta_2 X^2 + \epsilon $$ with the errors $\epsilon$ satisfying the usual assumptions (independent, normal, independent of the $X$ values). Let $b_0, b_1, b_2$ be the least squares estimates.

I have two new $X$ values $x_1$ and $x_2$, and I'm interested in getting a confidence interval for $v = E(Y|X = x_2) - E(Y|X=x_1) = \beta_1 (x_2 - x_1) + \beta_2 (x_2^2 - x_1^2)$.

The point estimate is $\hat{v} = b_1 (x_2 - x_1) + b_2 (x_2^2 - x_1^2)$, and (correct me if I'm wrong) I can estimate the variance by $$\hat{s}^2 = (x_2 - x_1)^2 \text{Var}(b_1) + (x_2^2 - x_1^2)^2 \text{Var}(b_2) + 2 (x_2 - x_1)(x^2 - x_1^2)\text{Cov}(b_1, b_2)$$ using the variance and covariance estimates of the coefficients provided by the software.

I could use a normal approximation and take $\hat{v} \pm 1.96 \hat{s}$ as a 95% confidence interval for $v$, or I could use a bootstrap confidence interval, but is there a way to work out the exact distribution and use that?

Answer 1

The general result you are looking for (under the stated assumptions) looks like this: For linear regression with $p$ predictor variables (you have two, $X$ and $X^2$) and an intercept, then with $n$ observations, $\mathbf{X}$ the $n \times (p+1)$ design matrix, $\hat{\beta}$ the $p+1$ dimensional estimator and $a \in \mathbb{R}^{p+1}$

$$ \frac{a^T\hat{\beta} - a^T \beta}{\hat{\sigma} \sqrt{a^T(\mathbf{X}^T\mathbf{X})^{-1}a}} \sim t_{n-p-1}.$$

The consequence is that you can construct confidence intervals for any linear combination of the $\beta$ vector using the same $t$-distribution you use to construct a confidence interval for one of the coordinates.

In your case, $p = 2$ and $a^T = (0, x_2 - x_1, x_2^2 - x_1^2)$. The denominator in the formula above is the square root of what you compute as the estimate of the standard error (provided that this is what the software computes ...). Note that the variance estimator, $\hat{\sigma}^2$, is supposed to be the (usual) unbiased estimator, where you divide by the degrees of freedom, $n-p-1$, and not the number of observations $n$.