Theorem 4 The discretized signature kernel over $k$, $$ \mathrm{k}^{+}: X^{+} \times X^{+} \rightarrow \mathbb{R}, \quad \mathrm{k}^{+}(x, y)=\left\langle\mathrm{S}^{+}\left(\mathrm{k}_x\right), \mathrm{S}^{+}\left(\mathrm{k}_x\right)\right\rangle $$
- is a positive definite kernel,
- $$\mathrm{k}^{+}(x, y)=\sum_{m \geq 0} \sum_{\substack{1 \leq i_1<\cdots<i_m<|x| \\ 1 \leq j_1<\cdots<j_m<|y|}} \prod_{r=1}^m \nabla_{i_r, j_r} \mathrm{k}(x, y)$$
- $$\mathrm{k}^{+}(x, y)=1+\sum_{\substack{i_1 \geq 1 \\ j_1 \geq 1}} \nabla_{i_1, j_1} \mathrm{k}(x, y) \cdot\left(1+\sum_{\substack{i_2>i_1 \\ j_2>j_1}} \nabla_{i_2, j_2} \mathrm{k}(x, y) \cdot\left(1+\sum_{\substack{i_3>j_2 \\ j_3>j_2}} \ldots\right)\right)$$ where we use the notation $x=\left(x_i\right)_{i=1}^{|x|}, y=\left(y_i\right)_{i=1}^{|y|} \in X^{+}$and $$ \nabla_{i, j} \mathrm{k}(x, y):=\mathrm{k}\left(x_{i+1}, y_{j+1}\right)+\mathrm{k}\left(x_i, y_j\right)-\mathrm{k}\left(x_i, y_{j+1}\right)-\mathrm{k}\left(x_{i+1}, y_j\right) $$ Source (page 14): https://jmlr.org/papers/volume20/16-314/16-314.pdf
I'll explain what I am finding a bit hard using the first two $m$s.
For $m=0$, (1) shows that both the $i_k$ and $j_k$ indices start from $k=1$ and end at $k=m$, but $m$ starts from zero, and, in the second equation, it seems like they are considering the case $i_0$ $j_0$ as $=1$. How does this work exactly? Generally, I don't see where the $1$s are coming from in (2).
Let $m=1$, we get $$\sum_{1\leq i_1< |x|}\sum_{1\leq j_1<|y|}\prod_{r=1}^1f(i_r,j_r)$$
If either $|x|=1$ or $|y|=1$, what do I do? Would I be correct in saying that this notation assumes sequences that are longer than $1$? Finally, I wrote this as two sums, but I could be wrong here. What do you think?
Or more generally, because of the inequalities in the summation, i.e $i_1$ starting from $1$, $i_2$ from $i_1+1$, etc, all up to $|x \text{ or } y|-1$, then at least $\min(|x|, |y|)\geq m+2$ if I want to compute the sums at level $m$?