OK, so this is a bit of an intuitive argument so I'm not 100% sure about it, and it's totally sideways to the way I normally think about filters, but:
Let's consider the "feedback kernel" ($a_k ... a_n$ - so, not including $a_0$) as an FIR filter. If the magnitude of that kernel in the frequency domain is $< 1$ including inter-bin peaks, the original filter should be stable.
If so: to construct a filter: one could generate a random set of $a_k ... a_n$, find the maximum freq response of that feedback kernel, and scale such that the peak is less than 1, to get a stable feedback kernel.
To reason about this: let's construct an infinite sequence of sequences, $Y^k$:
$$
Y^0_i = a_0x_i\\
Y^j_i = -\sum_{l=k}^na_lY^{j - 1}_{i - l}
$$
That is, each $Y^j$ is the previous $Y^{j - 1}$ convolved with our feedback kernel (the negation doesn't affect the argument). I also think it's true that:
$$y_i = \sum_j Y^j_i$$
Let's say $x_i$ is "time-bounded" (zero outside a finite range) - it follows inductively that every $Y^j$ is also time-bounded, because it's a convolution of the previous.
Let's also say that the feedback kernel's transfer function has magnitude $< G$ for all frequencies. This guarantees that whatever the input, the total energy in each $Y^k$ is strictly less than $G$ times the previous one. This implies a geometrically-decreasing upper bound on the magnitude of the sample values of $Y^j$ as well.
This geometric progression of bounds means that when we sum up $y_i$ it converges if $G < 1$, and to a value bounded by $K/(1 - G)$ for some $K$. The time-boundedness means that it decays over time (hand-waving over the formalism here).
Does this hold up?
a/b
, othersb/a
. Personally, I learneda/b
, both in mathematics and filters, so this may be a case of preference, or previous learning. $\endgroup$