Stable digital filter with leading zeroes
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 $&lt; 1$ 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 $&lt; G$ 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 $G &lt; 1$, 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?