How to remove historical values from exponential filter Suppose I have two time series $A:=[0,0,0,4,5,6]$ and $B:=[1,2,3,4,5,6]$. I implement the following filter (with initial value at the first element):

$$ f(t) = f(t-1) + \alpha*(f(t)-f(t-1))$$

with $\alpha \in [0,1]$.

Is there any way I can use filtered values of $B$ to recover the filtered values of $A$? For example suppose I can store all historical filtered values of $B$ for any choice of $\alpha$ in memory, is it possible to reconstruct what the filtered values for $A$ would have been for some choice of $\alpha$?

Intuitively this would involve subtracting the some filtered value(s) of $B$ at time step $3$ from the filtered value(s) of $B$ at time step $6$, to recover an approximation of $A$.
Yes. Your filter $f(t)$ is an LTI system, so you can use that to your advantage. Let:

$$
C(t) = B(t) - A(t)
$$

Then, if $y_A(t)$, $y_B(t)$, and $y_C(t)$ are the system's responses to $A(t)$, $B(t)$, and $C(t)$ as inputs, then you can straightforwardly do what you want:

$$
A(t) = B(t) - C(t) \rightarrow y_A(t) = y_B(t) - y_C(t)
$$

So, pass $B(t)$ through the filter, then pass $C(t) = [1, 2, 3, 0, 0, 0]$ through the filter and find the difference between the two filter outputs. That's exactly what you would get if you passed $A(t)$ through the filter on its own.