Home > Combinatorics and Probability, Functional Analysis, Linear Algebra, Markov Chains, Mathematics > On Patch(ix)es as Kernels of Integral Transforms (RWLA,MCT,GT,AM Part VII)

## On Patch(ix)es as Kernels of Integral Transforms (RWLA,MCT,GT,AM Part VII)

[This post is ongoing, as I think of a few things I will write them down too]

So just a couple of days ago I was asked by a student to give a class on DEs using Laplace transforms, and it was in my research that I realized that what I've been describing by converting a probability distribution on [0,1] to another is in effect a transform (minus the transform pair, which was unclear to me how to obtain, corresponding perhaps to inverting the patch(ix)).  The general form of integral transforms is, according to my book Advanced Engineering, 2nd ed., by Michael Greenberg p. 247:

$F(s) = \int_a^b f(t) K(t,s) dt$, where $K(t,s)$ is called the kernel of the transform, and looks an awful lot like a function by patch(ix) "multiplication," which I described as:

$b(x) = \int_0^1 a(1-y) p(x,y) dy$ you may recall.  In the former context $p(x,y)$ looks like a kernel, but here $a(1-y)$ is a function of $y$ than of $x$, and I sum across $y$.  To rewrite patch(ix)-multiplication as an integral transform, it would seem we need to rethink the patch position on the xy plane, but it seems easy to do (and we do in number 1 below!).

In this post I want to (eventually be able to):

1. Formally rewrite my function-by-patch(ix) multiplication as a "Pasquali" integral transform.

If we are to modify patch multiplication to match the integral transform guideline, simply think of $p(t,s)$ as oriented a bit differently, yielding the fact that $\int_0^1 p(t,s) ds = 1$ for any choice of $t$.  Then, for a probability distribution $b(t)$ in [0,1], the integral transform is $B(s) = \int_0^1 b(t) p(t,s) dt$.  Now $p(t,s)$ is indeed then a kernel.

2. Extend a function-by-patch multiplication to probability distributions and patches on all $\mathbb{R}$ and $\mathbb{R}^2$, respectively.

When I began thinking about probability distributions, I restricted them to the interval [0,1] and a patch on $[0,1] \times [0,1]$, to try to obtain a strict analogy of (continuous) matrices with discrete matrices.  I had been thinking for a while that this need not be the case, but when I glanced at the discussion of integral transforms on my Greenberg book, and particularly the one on the Laplace transform, I realized I could have done it right away.  Thus, we can redefine patch multiplication as

$B(s) = \int_{-\infty}^{\infty} b(t) p(t,s) dt$

with

$\int_{-\infty}^{\infty} p(t,s) ds = 1$

3. Explore the possibility of an inverse-patch via studying inverse-transforms.

3a. Write the patch-inverse-patch relation as a transform pair.

4. Take a hint from the Laplace and Fourier transforms to see what new insights can be obtained on patch(ix)es (or vice-versa).

Vice-versa: Well one of the things we realize first and foremost, is that integral transforms are really an extension of the concept of matrix multiplication: if we create a matrix "surface" and multiply it by a "function" vector we obtain another "function," and the kernel (truly a continuous matrix) is exactly our path connecting the two.  Can we not think now of discrete matrices (finite, infinite) as "samplings" of such surfaces?  I think so.  We can also combine kernels with kernels (as I have done in previous posts) much as we can combine matrices with matrices.  I haven't really seen a discussion exploring this in books, which is perhaps a bit surprising.  At any rate, recasting this "combination" shouldn't be much of a problem, and the theorems I proved in previous posts should still hold, because the new notation represents rigid motions of the kernel, yielding new kernel spaces that are isomorphic to the original.