Let $x[n]$ be a time-series, and two filters:
- $A[n] = a_0 x[n] + a_1 x[n-1] + ... a_q x[n-q]$
- $B[n] = b_0 x[n] + b_1 x[n-1] + ... b_r x[n-r]$
I think the answer is No, but is there a way to have this problem (sign of 2 filters should be both positive):
$$A[n] > 0 \quad AND \quad B[n] > 0$$
expressed in terms of the study of the sign of a single linear filter
$$C[n] = c_0 x[n] + c_1 x[n-1] + ... c_s x[n-s]$$
If not possible, is there a way to have nearly equivalence (that holds for most values of $n$)?
A[n] = x[n] - x[n-1] # derivative (or a smoothed version with moving average) B[n] = x[n] - 2 * x[n-1] + x[n-2] # acceleration (idem)
I'd like to find when both derivative and acceleration of $x[n]$ are positive by studying the sign of only 1 linear filter.
Or said in another way: is it possible to create a linear filter $C[n] = c_0 x[n] + c_1 x[n-1] + ... c_s x[n-s]$ such that $C[n] > 0$ if and only if derivative and acceleration of $x[n]$ are both positive?