I got stuck in a homework question:
In Linear regression model with assumption $\varepsilon_{i} \sim \cal{N}(0, \sigma^{2})$, iid.
$$Y_{i} = X_{i}^{\intercal}\theta^{*} + \varepsilon_{i}, ~ i = 1, \cdots, n, ~ \mathbb{E}\varepsilon_{i} \sim \cal{N}(0,\sigma^{2})$$
I am asked to find an elliptical confidence set for the mean response $\mathbb{E}(Y)$ in this model, with $\sigma^{2} > 0$ unknown.
Below is what I tried. I got stuck since my answer contains $\sigma^{2}$ and not sure if that is the right path.
Typically the mean response confidence interval is like $\hat{y}_{h} \pm t_{\frac{\alpha}{2}, n-p} se(\hat{y}_{h})$. Compare to that, I claim $\hat{\theta}$ is the estimated parameter, and $\hat{Y} = X^{\intercal}\hat{\theta}$ is the estimated value. Remaining is to find variance.
$$Var(\hat{Y}) = Var(X^{\intercal}\hat{\theta}) = Var(X^{\intercal}(XX^{\intercal})^{-1}XY)\\ =X^{\intercal}(XX^{\intercal})^{-1}X Var(Y) X^{\intercal} (XX^{\intercal})^{-1}X = \sigma^{2} X^{\intercal}(XX^{\intercal})^{-1}X$$ since $Var(Y) = Var(\varepsilon) = \sigma^{2} I_{n}$
Now that $(\hat{y}/se(\hat{y})^{2} \leqslant F_{\alpha/2, 1, n-p}$, I used the idea that square of a t-distributed r.v. follows a 1,n-p F-distribution. In fact the right hand side with misuse of notation I mean the F-value with 1,n-p df and $\alpha/2$ cutoff.
My major question is about left hand side, if I similarly plug in the expression, this is what I got eventually:
$$\hat{\theta}^{\intercal}XX^{\intercal}\hat{\theta} \frac{1}{\sigma^{2}} (X^{\intercal}(XX^{\intercal})^{-1}X)^{-1} \\ = \frac{1}{\sigma^{2}} Y^{\intercal}(X^{\intercal}(XX^{\intercal})^{-1}X) Y (X^{\intercal}(XX^{\intercal})^{-1}X)^{-1}$$
Of course the confidence set is the set where this expression is bounded by that F-value.
Am I totally wrong from the beginning? Or there is something I can do to fix my calculation? It just doesn't look right at this moment.
(Kind of have idea now. Please allow me some time to update it when I finish my homework).
I think it should be:
$$\mathbb{E}(Y) = X^{\intercal}\theta^{*} \in \{ X^{\intercal} \theta \in \mathbb{R} ^ {n}: || (XX^{\intercal}) ^{ \frac{1}{2}} (\theta - \hat{\theta}||^{2} \leqslant \frac{p}{n-p} S(\hat{\theta}) Q(1-\alpha, F(p,n-p) ) \}$$ where $S(\hat{\theta}) = ||Y - X^{\intercal}\hat{\theta}||^{2}$ and $\hat{\theta} = (XX^{\intercal})^{-1}XY$. $Q$ is the quantile inverse function.