I'm trying to find the solution to $cov(y_i, \frac{\sum{y_i}}{n})$, where $y_i = \beta_0 + \beta_1x_i + \epsilon_i$ (a regression model).

Here's what I have so far...

$cov(y_i, \frac{\sum{y_i}}{n}) = \frac{1}{n}cov(y_i, \sum{y_i}) = \sum\frac{1}{n}var(y_i) = \frac{1}{n}\sum\sigma^2$, since $var(y_i) = \sigma^2$. But I don't think this is the right solution. Can someone please help me? Thanks!

  • $\begingroup$ Your notation is sloppy and confuses the problem. First, you can't "solve" an expression. Do you mean "simplify"? Second, you are using a free term "$y_i$" in the same expression with a sum that has a dummy index $i$ and the sum contains a term $y_i$. Please make your notation more rigorous--it may be part of what is causing some of your difficulties... $\endgroup$
    – MPW
    Commented Apr 5, 2014 at 23:56

1 Answer 1


First, I would expand the y's in terms of the underlying variables:

$\sum\limits_{i=1}^n y_i = \sum\limits_{i=1}^n [\beta_0 + \beta_1 x_i + \epsilon_i]=n\beta_0 + \beta_1\sum\limits_{i=1}^n x_i + \sum\limits_{i=1}^n \epsilon_i$

Therefore, $\frac{\sum\limits_{i=1}^n y_i }{n} = \bar y= \beta_0 + \beta_1\bar x + \frac{1}{n}\sum\limits_{i=1}^n \epsilon_i$. Now, a particular $y_t = \beta_0 + \beta_1 x_t + \epsilon_t$.

We now have enough for calculating the covariance:

$cov(y_i,\bar y)=E[y_i \bar y]-E[y_i]E[\bar y]=E[\beta_0^2+\beta_0 \beta_1\bar x+\frac{\beta_0}{n}\sum\limits_{i=1}^n \epsilon_i + \beta_0\beta_1x_i+\beta_1^2\bar x x_i+\frac{\beta_1x_i}{n}\sum\limits_{i=1}^n \epsilon_i +\beta_0\epsilon_i+\beta_1^\bar x \epsilon_i+\frac{\epsilon_i}{n}\sum\limits_{i=1}^n \epsilon_i]-E[\beta_0+\beta_1 x_i+\epsilon_i]E[\beta_0 + \beta_1\bar x + \frac{1}{n}\sum\limits_{i=1}^n \epsilon_i]$

Simplifying the above, we get: $\beta_0^2+\beta_0 \beta_1\bar x + \beta_0\beta_1x_i+\beta_1^2\bar x x_i+E[\frac{\epsilon_i^2}{n}]-[\beta_0+\beta_1 x_i][\beta_0 + \beta_1\bar x]= E[\frac{\epsilon_i^2}{n}] = \frac{\sigma^2}{n}$


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