# Calculate WEIGHTED confidence interval for a linear fit

I am trying to calculate the confidence band about a regression line using the top answer here: Understanding shape and calculation of confidence bands in linear regression

I don't entirely understand the answer, but I am calculating the confidence interval as:

$$\sqrt \frac{\Sigma_i^n (Y_i - \hat{Y})^2}{n-2} \sqrt{\frac{1}{n} + \frac{(X-\bar{X})^2}{\Sigma_i^n (X_i-\bar{X})^2}}$$

I assume that this will give me a 1-sigma confidence interval. My question is: Is there a way to include the uncertainty on my data points in my confidence interval expression above? (My best fit line is weighted by the uncertainty on my data points.)

• It would be more helpful if you could clarify whether the uncertainty in your data pertains to $X_i$ or $Y_i$. Additionally, could you explain how you estimated the effect through weighting? Commented Nov 7, 2023 at 12:21

$$\hat y_i = \beta_0 + \beta_1 x_i =\vec \beta'\vec x_i$$ with $$\vec x_i = \begin{pmatrix} 1\\ x_i \end{pmatrix}$$ and $$'$$ meaning transpose.
Now the standard deviation for the estimator of $$\hat\sigma(\vec \beta'\vec x) = \sqrt{\vec x'\hat{\Sigma}_\beta \vec x}$$ and Glen_bs answer i linked provides us with $$\hat{\Sigma}_\beta = \hat{\sigma_e^2} X'WX$$, $$X$$ being the design matrix and $$W$$ the diagonal matrix of weights. The design matrix for the simple regression line is $$X = \begin{pmatrix} 1 & x_1 \\ 1 & x_2 \\ \vdots & \vdots \\ 1 & x_n \end{pmatrix}$$ Your answer already used: $$\hat{\sigma_e^2} = \frac{\Sigma(y_i - \hat y_i)^2}{n-rank(X)}$$, where $$rank(X) = 2$$ in your case.
Put it all together your (pointwise) confidence interval at value $$x$$ will be $$[\vec \beta'\vec x - t_{\alpha/2, n - 2}\cdot \hat\sigma(\vec \beta'\vec x), \vec \beta'\vec x + t_{\alpha/2, n - 2}\cdot \hat\sigma(\vec \beta'\vec x)]$$