My question is related to this question (and its answer).
This has been pointed in the comment - I think the answer is not what OP asked - the answer first use R to compute linear regression, and then "hand computed" based on the knowledge of the linear regression.
Unlike the linear regression estimator with standard condition, the estimator of logistic regression is not immediately seen as a normal distribution. Using classical MLE estimator theory, we know that asymptotically it is normally distributed.
To that end, suppose we are just interested in the asymptotic result, I can first get the (normal) distribution of $\hat \theta_n$, and then the (normal) distribution of $\mu(x^T \hat \theta_n)$ using Delta method, where $x$ is a fixed vector, and $\mu$ is the inverse of the logit function. I can then utilize the variance of $\mu(x^T \hat \theta_n)$ to get confidence interval, or standard error, etc. Is this procedure correct? Is this standard error same as the se.fit from R's glm result?
Suppose now I'm interested in non-asymptotic result. I am only aware of non-asymptotic upper bound of $\left|x^{\prime}\left(\hat{\theta}_n-\theta^*\right)\right|$ in the existing literature.
How do I get an upper bound for $\left|\mu(x^T\hat{\theta}_n)-\mu(x^T\theta^*)\right|$, which is prepare for getting the standard error of a fitted value of logistic regression?