# Showing $Cov(\epsilon, b_0) = 0$ or $Cov(\epsilon, b_1) = 0$

I am not sure where to start with this. But I know for $$Y_i = \beta_0+\beta_1X_i+\epsilon$$, where $$\beta_0,\beta_1,X$$ are assumed to be constants and $$\hat{Y_i} = b_0+b_1X_i$$ is the simple linear regression model, where $$b_0, b_1$$ are random variables (or distributions under circumstance of repeated sampling).

Could anyone help show this?

• $b_0$ and $b_1$ are the least-squares estimates, right? And $\epsilon$ is assumed independent of $X$? Mar 11, 2020 at 5:46

What is stated in the question is incorrect. Let us generalize, and use the matrix formulation of the linear model $$Y=X\beta + \epsilon$$ with $$\DeclareMathOperator{\V}{\mathbb{V}} \V \epsilon=\sigma^2 I_n$$, $$X$$ an $$n\times p$$ matrix of regressors, of ful column rank. Then as is known the least squares estimator is $$\hat{\beta}= (X^TX)^{-1} X^T Y$$:
\DeclareMathOperator{\C}{\mathbb{C}} \begin{align} \C(\epsilon, \hat{\beta}) &=\C(\epsilon, (X^TX)^{-1} X^T Y) \\ &= \C(\epsilon,Y)X (X^TX)^{-1} \\ &= \C(\epsilon,X\beta+\epsilon) X (X^TX)^{-1} \\ &= \C(\epsilon,\epsilon) X (X^TX)^{-1} \\ &= \sigma^2 I_n X (X^TX)^{-1} = \sigma^2 X (X^TX)^{-1} \end{align}
The premise in the title of the question isn't correct. The estimated $$b_0$$ and $$b_1$$ are by definition functions of the observed data, which are functions of the $$\epsilon_i$$. Unless the question is misstated...