Linear Regression: Deriving the variance and probability distribution of $B_0$ the intercept estimator

Question

So I am trying to derive the variance $V(B_0)$ and the probability distribution of the estimator intercept where $B_0 = \bar{Y} - B_1 \bar{x}$ in simple linear regression. I expressed $B_0$ as a linear combination of $Y_i$ i.e. $\sum_{i=1}^n a_i \cdot Y_i$:

\begin{align*} B_0&=\bar{Y}-B_1\bar{x}\\ &=\frac1n\sum_{i=1}^nY_i-\bar{x}\sum_{i=1}^na_i \cdot Y_i \end{align*}

and now I am stuck trying to derive the variance $V(B_0)$ using the linear combination. How do I go about it?

Also, how do I find the probability distribution using the variance of $B_0$?

Thanks

Answer 1

I think you meant that you expressed $B_1$ as a linear combination $\sum_{i=1}^n a_i Y_i$.

What you have is correct: you have $B_0 = \sum_{i=1}^n \left(\frac{1}{n} - a_i \bar{x}\right) Y_i$. Now recall that the variance of a sum of independent random variables is the sum of the variances to obtain $$\text{Var}(B_0) = \sum_{i=1}^n\left(\frac{1}{n} - a_i \bar{x}\right)^2 \text{Var}(Y_i)$$