Justifying the distribution for the maximum likelihood estimator in a linear regression example

Question

Data $(x_1, y_1), \dots, (x_n, y_n)$ is modelled with $x_i$ being non random and $y_i$ being observed values of $$Y_i = \alpha + \beta (x_i - \bar x) + \sigma \epsilon_i$$ with $\epsilon_i \sim N(0,1)$.

I have calculated the MLE's of the parameters $\alpha, \beta$ and $\sigma^2$ with $\hat \alpha = \bar y$ and $$\hat \beta = \frac{\sum y_i(x_i - \bar x)}{\sum (x_i - \bar x)^2}$$ we now have a corresponding estimator for $\beta$ given by $$B = \frac{\sum Y_i(x_i - \bar x)}{\sum (x_i - \bar x)^2}$$ which apparently (according to my notes) is normally distributed with a mean of $\beta$ and a variance of $$\sigma^2 \over \sum(x_i - \bar x)^2$$ and I can't justify to myself why this is the case! Could someone please help explain this?

Answer 1

Step 1: General: Recognize that $(x_i - \bar x)$, and its square, and $\frac{1}{\sum (x_i - \bar x)^2}$ are constants, not random variables. Further note that $\alpha$ and $\beta$ and $\sigma$ are also constants. First write $B$ as a constant times an expression containing a random variable. Focus carefully on the part that's not just a constant and then deal with that constant after you have that part worked out.

Step 2: Expectation: Remember that the expectation of a sum is the sum of expectations. Notice that you can take the expectation inside the summation and then move $(x_i - \bar x)$ outside that expectation. Notice you have another expression for $Y_i$ that you can use. Look again at step 1 and split up the expectation and pull out constants in the appropriate way. What's left is trivial to find the expectation of.

Step 3: Variance: Since the $Y$s are independent, the variance of a sum is the sum of the variances. You should also know a fact about the variance of a constant times a random variable to use here (more than once).

It's really nothing more than the basic properties of expectation and variance and using the facts already there in your question.