I have asked in another question how $\text{Var}{(\hat{y}_h)} = \sigma^2 \left(\frac{1}{n} + \frac{(x_h-\bar{x})^2}{S_{xx}}\right)$. Note that $\hat{y}_h$ = $b_0 + b_1X_h$ which is a regression line estimate at some given $X_h$.

This question concerns why the term $Cov(b_0,b_1)$ alone yields the RHS. Substituting $b_0 = Y - b_1X$ we get that $Cov(Y,b_1) - XCov(b_1,b_1)$ = $Cov(\frac{\sum{Y_i}}{n},\sum k_iY_i) - XVar{(b_1)}$.

We can then rearrange to obtain $\sum \frac{k_i Var(Y_i)}{n} - \frac{X\sigma^2}{S_{xx}}$ which quickly yields the desired result. My question is, why does this work? This single term does not seem like it should alone yield the RHS. Have I made an error in algebra?

I have asked in another question how $\text{Var}{(\hat{y}_h)} = \sigma^2 \left(\frac{1}{n} + \frac{(x_h-\bar{x})^2}{S_{xx}}\right)$. Note that $\hat{y}_h$ = $b_0 + b_1X_h$ which is a regression line estimate at some given $X_h$.

This question concerns why the term $Cov(b_0,b_1)$ alone yields the RHS. Substituting $b_0 = Y - b_1X$ we get that $Cov(Y,b_1) - XCov(b_1,b_1)$ = $Cov(\frac{\sum{Y_i}}{n},\sum k_iY_i) - XVar{(b_1)}$. Here X and Y without subscript are arithmetic means.

We can then rearrange to obtain $\sum \frac{k_i Var(Y_i)}{n} - \frac{X\sigma^2}{S_{xx}}$ which quickly yields the desired result. My question is, why does this work? This single term does not seem like it should alone yield the RHS. Have I made an error in algebra?

I have asked in another question how $\text{Var}{(\hat{y}_h)} = \sigma^2 \left(\frac{1}{n} + \frac{(x_h-\bar{x})^2}{S_{xx}}\right)$. Note that $\hat{y}_h$ = $b_0 + b_1X_h$ which is a regression line estimate at some given $X_h$.

This question concerns why the term $Cov(b_0,b_1)$ alone yields the RHS. Substituting $b_0 = Y - b_1X$ we get that $Cov(Y,b_1) - XCov(b_1,b_1)$ = $Cov(\frac{\sum{Y_i}}{n},\sum k_iY_i) - XVar{(b_1))}$.

We can then rearrange to obtain $\sum \frac{k_i Var(Y_i)}{n} - \frac{X\sigma^2}{S_{xx}}$ which quickly yields the desired result. My question is, why does this work? This single term does not seem like it should alone yield the RHS. Have I made an error in algebra?

I have asked in another question how $\text{Var}{(\hat{y}_h)} = \sigma^2 \left(\frac{1}{n} + \frac{(x_h-\bar{x})^2}{S_{xx}}\right)$. Note that $\hat{y}_h$ = $b_0 + b_1X_h$ which is a regression line estimate at some given $X_h$.

This question concerns why the term $Cov(b_0,b_1)$ alone yields the RHS. Substituting $b_0 = Y - b_1X$ we get that $Cov(Y,b_1) - XCov(b_1,b_1)$ = $Cov(\frac{\sum{Y_i}}{n},\sum k_iY_i) - XVar{(b_1)}$.

We can then rearrange to obtain $\sum \frac{k_i Var(Y_i)}{n} - \frac{X\sigma^2}{S_{xx}}$ which quickly yields the desired result. My question is, why does this work? This single term does not seem like it should alone yield the RHS. Have I made an error in algebra?