Autocall pricing: what does "Lipschitz continuous parameterization" mean?

I've been reading through this research paper (A Monte Carlo Pricing Algorithm For Autocallables That Allows for Stable Differentiation by T. Alm, B. Harrach, D. Harrach, M. Keller) about a method for valuing Autocalls. I've understood everything up to page 12 where they introduced a rotation matrix in order to "obtain a (Lipschitz) continuous parameterization of the bounds". I've searched for dozens of sources explaining this concept but to no avail. I have two questions:

• Can someone please explain what does it mean to have a Lipschitz-continuous parameterization of said bounds (or at least point to some literature that explains it) ?

• What problem does this rotation solve, what's the intuition behind it?

Thanks.

• Do you know what lipschitz continuity is? That would be a ace to start. Oct 5 '20 at 1:41
• I do understand it but not to a very advanced level. I understand the Lipschitz Constant, how it relates to the Mean Value Theorem, and how the graph of a Lipschitz-Continuous function lays outside of a double cone along with every secant line of the function.The intuition I got is that the function doesn't vary too much. Oct 5 '20 at 11:01
• Try wikipedia as a start, very nicely explained there. Lipschitz continuity is the second "strongest" type of continuity for functions after differentiable-continuity: en.wikipedia.org/wiki/Lipschitz_continuity Oct 5 '20 at 11:06
• I've read the article but I didn't find it to answer those two questions Oct 5 '20 at 11:22

I am aware that the question is one year old, but the connection to the rotation has not been answered so far.

Shorter answer: The shape "L" is not the graph of a function, but after rotation you get "V" and this is the graph of a Lipschitz function.

Longer answer: Please look at the survival zone in figure 2.2 in the paper. The borderline of the survival zone (the border between the gray and the white area) is not the graph of a function in the left and center image in fig 2.2. After rotation (right image of fig 2.2), it is the graph of a Lipschitz function.

• It's funny how I just thought about this question again a couple days ago, so your answer is much appreciated even if I asked this a year ago. This cleared up my past misunderstanding, thanks ! Oct 23 at 23:43

It sounds to me that they just mean that each bound can be seen as a function of the parameter(s) in the parametrization and this function is Lipschitz continuous.
An example: Consider the XY-plane. Let $$Y(x)$$ be a function of $$x$$. This function can be seen as describing the upper bound of the area below the graph. This function can then have the Lipschitz property, which means that for all $$x_1$$ and $$x_2$$, we have $$|Y(x_1)-Y(x_2)| < K|x_1 - x_2|$$ for some constant $$K>0$$ independent of the $$x_1$$ and $$x_2$$ chosen.

If you are not so familiar with Lipschitz continuity, you can interpret this that the function is quite nice. It is continuous and even more. You might for example note that all difference quotients are bounded. Take any $$x$$ and some small $$h>0$$, and choose $$x_1 = x+h$$ and $$x_2=x$$.
$$|Y(x+h)-Y(x)| < Kh \Leftrightarrow \frac{|Y(x+h)-Y(x)|}{h} < K$$ So if the function has a derivative in some point its absolute value is bounded by $$K$$. The absolute value of slope between two points on the graph is always bounded by $$K$$

A Lipschitz continuous function does not have to be differentiable though, but it can in some way be seen as being between continuous functions and differentiable functions on the "niceness" scale. Any continously differentiable function is, at least locally (like in a bounded interval), Lipschitz. This is easy to prove from the definition.

You have the corresponding definition in several dimensions if you have functions of several variables and/or vector valued functions: If $$Y$$ is a function from the parameters where the output is vector of bounds, the Lipschitz condition is more or less the same but with vector norms used: $$|| Y(x_1) - Y(x_2) || < K ||x_1 - x_2||$$ for all $$x_1, x_2 \in \mathbb{R}^n$$ and $$Y(x_1), Y(x_2) \in \mathbb{R}^m$$ for some dimensions $$n$$ and $$m$$.

• I understood this but how does this relate to the rotation matrix used? Oct 5 '20 at 20:00