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In a linear regression model, as Y is the target and X are predictive features,the CI for $E[Y_{new}|X_{new}]$ is as follows:

Assuming $X_{new}$ is given, $X_{new}$ is the new data given for a target label $Y_{new}$, and that $C = (X^t X)^{-1}$

$$\hat{E}[Y_{new}|X_{new}] = \hat{Y}_{new} = \hat{\beta}^t X_{new}$$

It can be show that $E[\hat{Y}_{new}] = \beta_{0}^t X_{new}$

thus $\hat{Y}_{new}$ is an unbiased estimator.

$$Var(\hat{Y}_{new}) = Var(\hat{\beta}^t X_{new})$$

$X_{new}$ is a vector, hence if we want to get it out of the Var(), we neet to put it on the right and on the left as transpose, like this:

$$Var(\hat{Y}_{new}) = X^t_{new}Var(\hat{\beta})X_{new} = \sigma_{\epsilon}^2 X_{new}^t C X_{new}$$

$$\hat{s.e}(\hat{Y}_{new}) = \sqrt{\hat{\sigma}^2_{\epsilon}X^t_{new}CX_{new}}$$

Thus, a confedence interval for the expectation $E[Y_{new}]$ is:

$$\hat{Y}_{new} \pm z_{\alpha / 2} \hat{s.e}(\hat{Y}_{new})$$

This is the CI for the expected value. How do I find the CI for the $Y_{min}$ itself? cause that would also include the noise.

This pictue shows more or less the difference, between CI for the expected value of $Y_{min}$ and the CI for $Y_{min}$ (The difference is basically the noise):

enter image description here

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