What is the relationship between average of a rolling intercept and the intercept from a regression over the entire period? If i calculate rolling (e.g. 3 periods back) intercepts for a time series using OLS, is the average of these rolling intercepts then in some way related to the intercept from an OLS of the entire series?
Below is an example - the average of the rolling intercepts is 1.05, while the intercept from regression on the whole series is 2.21.

 A: There is no direct relationship between these two estimation methods, and their interaction depends on the underlying data in a complex way.  To see this, we can derive the formulae for the two estimators.  Using the whole dataset, the OLS intercept estimator in simple linear regression can be written as:
$$\hat{\beta}_0 = \frac{1}{n} \sum_{i=1}^n \Bigg[ y_i - x_i \cdot \frac{(\sum_{i=1}^n x_i y_i) - \tfrac{1}{n} (\sum_{i=1}^n x_i) (\sum_{i=1}^n y_i)}{(\sum_{i=1}^n x_i^2) - \tfrac{1}{n} (\sum_{i=1}^n x_i)^2} \Bigg].$$
Suppose you have $n$ data points and you compute the average of the rolling-intercepts, each with a window of $m<n$ data points.  For each $k=0,...,n-m$ the individual OLS rolling-intercepts (each taken over values $i = k+1,...,k+m$) can be written as:
$$\begin{equation} \begin{aligned}
\hat{\beta}_0^{(k)} 
&=\frac{1}{m} \sum_{i=k+1}^{k+m} \Bigg[ y_i - x_i \cdot \frac{(\sum_{i=k+1}^{k+m} x_i y_i) - \tfrac{1}{m} (\sum_{i=k+1}^{k+m} x_i) (\sum_{i=k+1}^{k+m} y_i)}{(\sum_{i=k+1}^{k+m} x_i^2) - \tfrac{1}{m} (\sum_{i=k+1}^{k+m} x_i)^2} \Bigg]. \\[6pt]
\end{aligned} \end{equation}$$
Taking the average of these rolling-intercepts gives:
$$\begin{equation} \begin{aligned}
\hat{\beta}_0^* 
&\equiv \frac{1}{n-m} \sum_{k=0}^{n-m} \hat{\beta}_0^* \\[6pt]
&= \frac{1}{m(n-m)} \sum_{k=0}^{n-m} \sum_{i=k+1}^{k+m} 
\Bigg[ y_i - x_i \cdot \frac{(\sum_{i=k+1}^{k+m} x_i y_i) - \tfrac{1}{m} (\sum_{i=k+1}^{k+m} x_i) (\sum_{i=k+1}^{k+m} y_i)}{(\sum_{i=k+1}^{k+m} x_i^2) - \tfrac{1}{m} (\sum_{i=k+1}^{k+m} x_i)^2} \Bigg] \\[6pt]
&= \frac{1}{m(n-m)} \sum_{i=1}^{m} \sum_{k=\max(0,i-m)}^{\min(n-m,i-1)} 
\Bigg[ y_i - x_i \cdot \frac{(\sum_{i=k+1}^{k+m} x_i y_i) - \tfrac{1}{m} (\sum_{i=k+1}^{k+m} x_i) (\sum_{i=k+1}^{k+m} y_i)}{(\sum_{i=k+1}^{k+m} x_i^2) - \tfrac{1}{m} (\sum_{i=k+1}^{k+m} x_i)^2} \Bigg]. \\[6pt]
\end{aligned} \end{equation}$$
As can be seen, this second estimator has a complicated form, and does not reduce to a function of the OLS estimator for the whole dataset.
