# How to prove that $Cov(\hat{\beta},\bar{Y}) = 0$ using given covarience properties

To quote:

It is well known that, if $W_1, ..., W_n, Z_1, ..., Z_m$ are random variables and $a_1, ..., a_n, b_1, ..., b_m$ are constants, then

$Cov ( \sum_{i=1}^n a_iW_i, \sum_{j=1}^m b_jZ_j) = \sum_{i=1}^n \sum_{j=1}^m a_i b_j Cov(W_i,Z_j)$.

Use this property (and other properties of covariance) to prove that for the model $Y_i = \alpha + \beta x + \epsilon$, with $\epsilon \sim N(0,\sigma^2)$, we have $Cov(\hat{\beta},\bar{Y}) = 0$,

where $\hat{\beta} = S_{XY}/S_{XX}$.

Currently my most successful attempt was to try to put $S_{XY} \over S_{XX}$ as $\sum_i (x_i - \bar{x})(Y_i - \bar{Y}) \over \sum_i (x_i - \bar{x})(x_i - \bar{x})$ then substituting $Y = \alpha + \beta x + \epsilon$, but even in this best attempt I just end up with a mess and a stray epsilon I can't deal with (one reduces to 0 though the summing to get $\bar{Y}$, but the other ends up as a nasty coefficient).

I'm really not sure what to do.

Edit: Added a run down on some previous issues and the self-study tag.

• If this is for homework or an assignment, could you please add the [self-study] tag and read its wiki? – Andy May 25 '16 at 10:36
• Sorry about that - I'd assumed that the rules were the same as math.stackexchange for them. Added, also editing in some information about what I've already tried. – liqiudilk May 25 '16 at 10:42
• No worries. Just wanted to highlight this given that you are new here. In this sense, welcome to CV :-) – Andy May 25 '16 at 10:43
• Is the regressor $x$ treated as deterministic or as a random variable? – Alecos Papadopoulos May 25 '16 at 13:01

Here are some tips. First, just like variances, covariance can ignore additive constants. You can use this to show: $$cov (\hat {\beta},\overline {Y})=cov (\hat {\beta},\overline {\epsilon})$$
$$\hat{\beta}=\sum_iw_iY_i$$
Where $w_i=\frac {x_i-\overline {x}}{S_{XX}}$. You can ignore constants as well to get
$$cov (\hat {\beta},\overline {Y})=cov (\sum_iw_i\epsilon_i,\overline {\epsilon})$$