# Simplify variance expression, taking into account covariances:$\operatorname{var}(Y_i-\bar Y-\hat {\beta}_1(x_i-\bar x))$

Find $\operatorname{var}(Y_i-\bar Y-\hat {\beta}_1(x_i-\bar x))$ where $\hat {\beta}_1=S_{xy}/S_{xx}$ is the least square estimator and $Y_i$ a random variable.

I know that I can't simply split the variances because I need to account for covariance. In fact:

$$\operatorname{cov} (Y_i,\bar Y)= \frac 1 n \sigma^2$$

$$\operatorname{cov} (Y_i,\hat {\beta}_1 (x_i-\bar x))= \frac{(x_i-\bar x)^2}{S_{xx}} \sigma^2$$

$$\operatorname{cov} (\hat {\beta}_1(x_i-\bar x),\hat {\beta}_1(x_i-\bar x))= \frac{(x_i-\bar x)^2}{S_{xx}} \sigma^2$$

Edit: I know I can rewrite this as $\operatorname{cov} ((Y_i-\bar Y-\hat {\beta}_1 (x_i-\bar x),(Y_i-\bar Y-\hat {\beta}_1 (x_i-\bar x))$ and work from there, but is this the only way?

• Do you know matrix algebra? The relationship between matrices and linear transformations? What matrix multiplication, inversion, and transposition are? If so, there's a way to do this using those. $\qquad$ – Michael Hardy Sep 20 '16 at 23:02
• Thanks, yes I know basic level algebra, the second chapter of the notes covers multiparameter regression with matrices and covariance matrices but I'm not completely comfortable with them yet. I guess I will need to read up some more – GRS Sep 20 '16 at 23:08
• Maybe I'll be back$\,\ldots$ It's worth knowing that the sample covariance between $(Y_i : i = 1,\ldots, n)$ and $(x_i : i=1,\ldots,n)$ is $0. \qquad$ – Michael Hardy Sep 20 '16 at 23:18