[$] \circ [$] is the standard composition operator.

The reviewer was asleep for the Hessian. Lower [$]#[$] follows Vellani notations.

- FaridMoussaoui
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[$] \circ [$] is the standard composition operator.

The reviewer was asleep for the Hessian. Lower [$]#[$] follows Vellani notations.

The reviewer was asleep for the Hessian. Lower [$]#[$] follows Vellani notations.

Last edited by FaridMoussaoui on February 13th, 2019, 9:07 am, edited 1 time in total.

[$]\nabla^2h = \nabla S[$], that is a Jacobian, seen as the Hessian of a convex function, that is important when one want to define S as a one-to-one map.I think there's a typo on page 3; you say [$]\nabla^2[$] is the Jacobian but it is the Laplacian operator (or Hessian which become singular/negative definite).

BTW is reference [6] available yet?

I don't get what [$] A \circ S[$] does/is: it seem to be used a number of times without a sharp definition.

// Haven't got my head yet around article [1] and how to pull-back a measure..It uses upper case # while the article uses lower case #. Are they the same #?

[6] is written, we should have released it since years, we are all busy :/ I'll try to give a talk about it ASAP.

[$] A \circ S[$] composition

article [1] is magical. It tells you that any map $S$ can be decomposed as [$]S = (\nabla h) \circ T[$], h convex, T Lebesgue measure preserving. It is the analogous for functions of the polar decomposition for matrix [$]M = A U[$].

Not really, but quite close : if [$] S_\# \mu = \nu[$], then [$] S^\# \nu = \mu[$], that is nothing but a variable change [$] \int \varphi \circ S d\mu = \int \varphi d\nu [$](AFAIR)

- Cuchulainn
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I'm kind of allergic to magic It's very measure-theoretic which is computationally not obvious?

BTW if you write [$](\nabla S)[$] then it is the gradient,(vector) not Jacobi (matrix)??

Regarding mapping from n-d space to the unit cube [$](0,1)^n[$] I'm assuming we are talking about the independent variable?

BTW if you write [$](\nabla S)[$] then it is the gradient,(vector) not Jacobi (matrix)??

Regarding mapping from n-d space to the unit cube [$](0,1)^n[$] I'm assuming we are talking about the independent variable?

Last edited by Cuchulainn on February 13th, 2019, 10:17 am, edited 8 times in total.

I just wanted to express my deep admiration for this work of Brenier.I'm kind of allergic to magic

BTW if you write [$](\nabla S)[$] then it is the gradient,(vector) not Jacobi (matrix)??

[$](\nabla S)[$] is a field of vectors if S is a scalar function. It is a Jacobian (i.e. field of matrices) if [$]S : R^D \mapsto \

R^D[$] is a map ?

- Cuchulainn
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I don't agree with the notation! I do agree with the conclusion.

Gradient is

https://nl.wikipedia.org/wiki/Gradi%C3%ABnt_(wiskunde)

Jacobian

https://nl.wikipedia.org/wiki/Jacobi-matrix

Or am I missing something?

Cal it [$]J(S)[$] otherwise everyone thinks it's [$]grad[$]. It confuses me no end,

Gradient is

https://nl.wikipedia.org/wiki/Gradi%C3%ABnt_(wiskunde)

Jacobian

https://nl.wikipedia.org/wiki/Jacobi-matrix

Or am I missing something?

Cal it [$]J(S)[$] otherwise everyone thinks it's [$]grad[$]. It confuses me no end,

Last edited by Cuchulainn on February 13th, 2019, 10:30 am, edited 1 time in total.

Well, this is the same notation : [$]\nabla f := (\partial_1 f, \cdots, \partial_D f) [$] for a scalar function [$]f : R^D \mapsto R[$]. [$]\nabla S := (\partial_i S_j )_{i,j=1,\cdots,D} [$] for the Jacobian of the map [$] S:=(S_1,\ldots,S_D)[$]. It seems correct: it is the gradient applied to each component. For instance, I can think to the Hessian operator as [$]\nabla^2 = \nabla \nabla^T[$], and the Laplacian one as [$]\Delta = \nabla^T \nabla[$] (with a slight notation abuse)

Last edited by JohnLeM on February 13th, 2019, 10:46 am, edited 1 time in total.

- Cuchulainn
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OK, but the define it in the article! IMO you are still using notation I have never seen in this context.

The gradient is a special case of Jacobian and writing the latter using forrmer's notation is scary IMO.

The gradient is a special case of Jacobian and writing the latter using forrmer's notation is scary IMO.

I am not sure to understand your question. Yes all is mapped into a fixed set in this paper, for instance [$](0,1)^D[$], that is a localization principle. In other words, instead of solving the Fokker planck equation (1), describing a probability density [$]\mu(t,x),x\in R^D[$], we solved equivalently the equation satisfied by [$]S(t,x) : (0,1)^D \mapsto R^D[$], transporting the Lebesgue measure of the unit cube (not. [$]dx[$]) into [$]\mu(t,x)[$], i.e. [$]S(t,\cdot)_\# dx = d\mu(t,\cdot)[$].Regarding mapping from n-d space to the unit cube [$](0,1)^n[$] I'm assuming we are talking about the independent variable?

- Cuchulainn
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I mean e.g. solving the heat equation u_t = u_xx on the real line. Define newI am not sure to understand your question. Yes all is mapped into a fixed set in this paper, for instance [$](0,1)^D[$], that is a localization principle. In other words, instead of solving the Fokker planck equation (1), describing a probability density [$]\mu(t,x),x\in R^D[$], we solved equivalently the equation satisfied by [$]S(t,x) : (0,1)^D \mapsto R^D[$], transporting the Lebesgue measure of the unit cube (not. [$]dx[$]) into [$]\mu(t,x)[$], i.e. [$]S(t,\cdot)_\# dx = d\mu(t,\cdot)[$].Regarding mapping from n-d space to the unit cube [$](0,1)^n[$] I'm assuming we are talking about the independent variable?

I am not familiar with the term "localization principle". Maybe that's my confusion. Is it a term from physics?

Is it possible for motivation to give an example (even 1d/2d) to show how equations (5) and (6) work?

If you consider a fixed-in-time change of variable, you are done. This is precisely an error that the AI community does today. Recall that Fokker-Planck equations are considered with Dirac measures as initial conditions. Try to solve numerically the heat equation in one dimension [$]u_t = u_{xx}, u(0,x) = \delta_{x_0}(x)[$], where [$]x_0[$] can be anywhere, with your variable change [$]y = x/(1+x)[$] to see this.I mean e.g. solving the heat equation u_t = u_xx on the real line. Define newindependent variabley = x/(1+x) to get a pde u_t = a(y)(a(y)u_y)_y on (0,1) where a(y) = 1-y*y.

- FaridMoussaoui
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The change of variable is for the probabilty (transition) density (as he is solving the Fokker Planck equation).

For example for the GMB process [$] dX = X r dt + X \sigma dW [$], we have [$] X \in R_{+} [$].

We look for a map to localise from [$] R_{+} [$] to [$] (0~1)[$].

The map is defined in terms of the cumulative probability function which is in [$] (0~1)[$]

PS: sorry, this is duplicate to Jean-Marc answer. I was typing when he posted his answer.

For example for the GMB process [$] dX = X r dt + X \sigma dW [$], we have [$] X \in R_{+} [$].

We look for a map to localise from [$] R_{+} [$] to [$] (0~1)[$].

The map is defined in terms of the cumulative probability function which is in [$] (0~1)[$]

PS: sorry, this is duplicate to Jean-Marc answer. I was typing when he posted his answer.

I am not familiar too with this term, but it describes quite well what we wanted to do.I am not familiar with the term "localization principle". Maybe that's my confusion. Is it a term from physics?

Is it possible for motivation to give an example (even 1d/2d) to show how equations (5) and (6) work?

For one dimensional, there are plenty of example in my old paper.

For two dimensional, the CRAS paper contains one numerical example for Heston. Here is another example for log normal process.

No problem Farid, it is better to have several point of views. It is almost what you are saying : in one-dimension, the map is a quantile, and its inverse map is a cumulative. In several dimensions, the correct generalization of a quantile is a map [$] S = \nabla h : (0,1)^D \mapsto R^D[$], h convex (Brenier again).The change of variable is for the probabilty (transition) density (as he is solving the Fokker Planck equation).

For example for the GMB process [$] dX = X r dt + X \sigma dW [$], we have [$] X \in R_{+} [$].

We look for a map to localise from [$] R_{+} [$] to [$] (0~1)[$].

The map is defined in terms of the cumulative probability function which is in [$] (0~1)[$]

PS: sorry, this is duplicate to Jean-Marc answer. I was typing when he posted his answer.

Farid, as I wrote, I think this is going to be really difficult to implement alone. A suggestion could be : find a public research group that could help to release a version for academical purposes, so that anybody could toy with it. Any suggestions ?The article alone is not enough to implement the algorithm (in my point of view). As usual, you have to look to the references.

- Cuchulainn
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