Why is a term that comes out of a variance bracket is squared?

Question

I am in a course on data analysis. The following statement is made in the notes made available to us by our professor:

$$ \text{Var}[a] = \text{Var}[\bar{y} -b\bar{x}] = \text{Var}[\bar{y}] + \text{Var}[b\bar{x}] = \dfrac{\sigma^2}{n} + {\color{red}{(\bar{x}^2)}}\text{Var}[b]$$

I have marked the place where I have doubt in red. This is the expression to determine the variance of point estimates in a simple linear regression model.

Thank you.

Answer 1

\begin{align}\operatorname{Var}(kX)&=E[(kX)^2]-(E(kX))^2 \\ &=E[k^2X^2]-(kE(X))^2\\ &=k^2E[X^2]-k^2E(X)^2\\ &= k^2(E[X^2]-E[X]^2)\\ &=k^2\operatorname{Var}(X)\end{align}

Here, $k=\bar{x}$ is a constant.

Answer 2

In a simple linear regression model, the values of x's are known and hence they are treated as constants. With the help of the x's we find the values of y, a , b , which are random in nature , so they they can be treated as Random variables. This is why you can apply the formula mentioned by Siong Thye Goh.

If I may point out , you have missed one term in your question $\text{Var}[a] = \text{Var}[\bar{y} -b\bar{x}] = \text{Var}[\bar{y}] + \text{Var}[b\bar{x}] - {\color{red}{{2\bar{x}}\text{Cov}[{\bar{y}},b]}} = \dfrac{\sigma^2}{n} + {\color{red}{(\bar{x}^2)}}\text{Var}[b]$

After a lot of tedious calculations, you will find , this term $\quad{{2\bar{x}}\text{Cov}[{\bar{y}},b]}\quad$to be $0$ . It is then that you get the variance of the interception.