# SLR: Variance of a residual

I am having problems calculating the variance of a residual in an SLR setting,
ie $\text{var}$$(y_i- \hat{y_i}). Here is what I have thus far. If \hat{y_i}= \hat{\beta_0} + \hat{\beta_1}x_i and \ y_i= \beta_0 + \beta_1x_i (where \hat{\beta_0} and \hat{\beta_1} are ordinary least squares estimates of \beta_0 and \beta_1). \text{var}$$(e_i)=\text{var}$$(y_i- \hat{y_i}) \ =\text{var}$$(y_i)+\text{var}$$(\hat{y_i})-2*\text{cov}$$(y_i, \hat{y_i})$
$\ =\sigma^2 + \sigma^2( \frac{1}{\ n} +\frac{(x_i-\bar{x})^2\sigma^2}{\sum(xi-\bar{x})^2})-?$

Now what is screwing me up is the covariance term. When I was studying prediction intervals we assume that $\hat{y_i}$ has no bearing on $\ y_i$. I was told by classmate that this is not the case and that $\text{cov}$$(y_i, \hat{y_i})=\text{var}$$(y_i)$. Which option is correct, if any? And what is the difference between the variance we find for prediction intervals and this one?

When doing prediction intervals, you're doing that calculation for an observation that's not used in the estimation, so (by the regression assumptions themselves) $\hat{y}_i$ has no bearing on $y_i$.
$\text{cov}(y_i, \hat{y}_i) = \text{cov}(\hat{y}_i+y_i-\hat{y}_i, \hat{y}_i) = \text{cov}(\hat{y}_i, \hat{y}_i) +\text{cov}(y_i-\hat{y_i}, \hat{y}_i) = \text{var}(\hat{y}_i)+0$
• Would it be correct to say that instead of the covariance of a specific $y_i$ and $\hat{y}_i$, if I take that of $y$ and $\hat{y}$, then I would get: $\operatorname{Cov}(y, \hat{y})=\sum(y_i - \bar{y})(\hat{y}_i - \bar{\hat{y}}_i)/(n-1)=\sum(y_i - \bar{y})(\hat{\beta}_1 x_i+\hat{\beta}_0 - \hat{\beta}_1\bar{x}-\hat{\beta}_0)/(n-1)=\sum(y_i - \bar{y})(\hat{\beta}_1(x_i-\bar{x}))/(n-1)=\hat{\beta}_1\sum (y_i-\bar{y})(x_i - \bar{x})/(n-1)=\hat{\beta}_1\operatorname{Cov}(x,y)$ – user1879926 Apr 10 at 6:04