Let me give you some background on my issue.
Let's say I have a simple one covariate linear regressor as follows:
$$\log\alpha =\alpha_0+\alpha_1x_{i1}.$$
Obviously, to find $\alpha$, we use simple algebra to exponentiate both sides and get this:
$$\alpha = \exp(\alpha_0+\alpha_1x_{i1})$$
I have a working program on R that is able to easily estimate $\alpha_0$ and $\alpha_1$ based on the $x$ values given and the likelihood function. Using R, I can also obtain the standard errors of the estimates $\hat{\alpha}_0$ and $\hat{\alpha}_1$.
My issue is the following: If I want to obtain let's say a 95% C.I for $\hat{\alpha}$, I need the standard error of $\hat{\alpha}$ itself. Part of me wants to say that I can get the bounds by doing this:
Take the lower bound of a 95% C.I for $\hat{\alpha}_0$ and the lower bound of a 95% C.I For $\hat{\alpha}_1$, substitute them in here: $\hat{\alpha}_{(L)}=\exp(\hat{\alpha}_{0(L)}+\hat{\alpha}_{1(L)} x_{i1})$, where $L$ means lower bound, and then do the same things for an upper bound.
The other part of me (the part that believes if it's too good to be true it probably is), thinks that there's something else I have to do. If anyone could shed any light on this, that'd be awesome.
Thanks!
Almost right, just don't forget the covariance between the $\hat{\alpha}$, i.e., let $\hat{\alpha}=(\hat{\alpha}_0, \hat{\alpha}_1)$ and $ x= (1,x_i)$ ,hence \begin{align} & \operatorname{Var}(\hat{y}_i)=\operatorname{Var}(x^T\hat{\alpha}) =x^T \operatorname{Var}( \hat{\alpha})x \\[10pt] = {} & x^T \Sigma_{\alpha}x=\|\Sigma^{1/2}x\|_2^2=\operatorname{Var}(\hat{\alpha}_0)+x_i ^ 2\operatorname{Var}(\hat{\alpha}_1)+2x_i\operatorname{Cov}(\hat{\alpha}_0,\hat{\alpha}_1). \end{align} Now, as $\exp$ is just a one-to-one transformation of the CI of $y_i$, then the estimated $1-\alpha$ level CI of $y_i$ is $$ \exp\left\{ \hat{y}_i \pm Z_{ 1- \alpha/2} \left\| \hat{\Sigma}^{1/2}x \right\| \right\} $$