Proving $\text{Var}{(\hat{y}_h)} = \sigma^2 \left(\frac{1}{n} + \frac{(x_h-\bar{x})^2}{S_{xx}}\right)$ 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?
 A: I think the intuition is that the error is built into the estimates. In other words, we have $$ y = \beta_0 + \beta_1 x + \epsilon $$ where $\epsilon\sim P_\epsilon$ such that $\mathbb{E}[\epsilon]=0$ and $\mathbb{V}[\epsilon]=\sigma^2$. Note that $\beta_i$ are the true parameters of the data, not the estimated ones $b_0$ and $b_1$. But we never see $\epsilon$, we only see $(x,y)$. Thus, the noise gets built into our estimates:
$$ y = (\beta_0 + \epsilon_0) + (\beta_1 + \epsilon_1)x = b_0 + b_1x $$
Thus, we can view all of the variance of prediction (i.e., the variance in the predictive distribution) as coming from variance due to error in parameter estimation. Hence it is no surprise that one can compute $\mathbb{V}[\hat{y}_h]$ from $C=\text{Cov}(b_0,b_1)$. Since $C$ contains all the variance in the parameter estimation due to noise, it contains all the variance (by our assumption on the data, since we e.g. assume no noise in $x$ itself). In this sense, the variance in estimated parameters captures all the predictive variance.
Mathematically, note that
$$ \mathbb{V}[\hat{y}_h]
= \mathbb{V}[ b_0 + b_1x_h ] $$
so intuitively all the predictive variance is contained in $b_0,b_1$, since $x_h$ is not random.

I think it's useful to differentiate between $b_i$ and $\beta_i$. Since the latter are not random, variances involving only them drop out.
We can derive the answer from first principles fairly fast with this in mind.
We will see that the $1/n$ term derives from $b_0$, while the $(x_h-\bar{x})^2/S_{xx}$ comes from $b_1$.
\begin{align}
\mathbb{V}[\hat{y}_h]
&= \mathbb{V}[ b_0 + b_1x_h ] \\
&= \mathbb{V}[ \bar{y}-b_1\bar{x} + b_1x_h ] \\
&= \mathbb{V}[\bar{y} + (x_h - \bar{x})b_1] \\
&= \mathbb{V}\left[
      \frac{1}{n}\sum_i \beta_0 + \beta_1 x_i + \epsilon_i
      + c_h \sum_i \delta_i (q_i + \epsilon_i) 
\right] \\
&= \mathbb{V}\bigg[
      \underbrace{\frac{1}{n}\sum_i q_i}_{0,\text{ non-random}} + \frac{1}{n}\sum_i \epsilon_i
      + \underbrace{c_h \sum_i \delta_i q_i}_{0,\text{ non-random}} + c_h \sum_i \delta_i \epsilon_i 
\bigg] \\
&=\mathbb{V}\bigg[ \sum_i
\frac{1}{n} \epsilon_i + c_h\delta_i \epsilon_i
\bigg]\\
&= \sum_i \mathbb{V}\bigg[ 
\left(\frac{1}{n}  + c_h\delta_i\right) \epsilon_i
\bigg] \;\;\;\;\;\text{(Independence)} \\
&= \sum_i \left(\frac{1}{n}  + c_h\delta_i\right)^2 \underbrace{\mathbb{V}\big[ 
 \epsilon_i
\big]}_{\sigma^2} \\
&= \sigma^2\bigg(n\frac{1}{n^2} + \frac{2}{n}c_h\underbrace{\sum_i\delta_i}_0 + \frac{(x_h-\bar{x})^2}{S_{xx}^2} \underbrace{\sum_i\delta_i^2}_{S_{xx}} \bigg) \\
&= \sigma^2 \left( \frac{1}{n} + \frac{(x_h-\bar{x})^2}{S_{xx}} \right)
\end{align}
where we used independence of $\epsilon_i$, $ b_0 = \bar{y} - b_1 \bar{x}$, 
$q_i = \beta_0 + \beta_1 x_i$ (the true "noiseless" $y_i$),
$\delta_i=x_i-\bar{x}$,
and $b_1 = \sum_i \delta_i y_i / S_{xx}$, with $S_{xx}=\sum_i(x_i - \bar{x})^2$ and $$ c_h = \frac{x_h-\bar{x}}{S_{xx}} $$
Ultimately all the noise was contained in $\epsilon$, which was being implicitly held by $b_0$ and $b_1$.
