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DavidJN
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Dimensionality of Monte Carlo

I am unclear on the concept of dimensionality when it comes to using Monte Carlo for option valuation and what I am reading has unfortunately not made things any clearer.

Starting as simply as possible, simulating the terminal price at option maturity of a single underlying in one time step jump from valuation to maturity is presumably an example of a one dimensional problem. Is doing the same thing for multiple underlyings (e.g. a European basket option) a multi-dimensional problem?

What is the dimensionality if I value a single underlying vanilla option by evolving the underlying price over a path using multiple time steps DT = T/N, where T is the option maturity in years and N is the number of time steps. Is that a multi-dimension problem? If so, why is this so?

bearish
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Joined: February 3rd, 2011, 2:19 pm

Re: Dimensionality of Monte Carlo

I am unclear on the concept of dimensionality when it comes to using Monte Carlo for option valuation and what I am reading has unfortunately not made things any clearer.

Starting as simply as possible, simulating the terminal price at option maturity of a single underlying in one time step jump from valuation to maturity is presumably an example of a one dimensional problem. Is doing the same thing for multiple underlyings (e.g. a European basket option) a multi-dimensional problem?

What is the dimensionality if I value a single underlying vanilla option by evolving the underlying price over a path using multiple time steps DT = T/N, where T is the option maturity in years and N is the number of time steps. Is that a multi-dimension problem? If so, why is this so?

I would think of it as the number of dimensions that you are integrating over. So, barring corner cases, the one-step European basket option on N assets would be N-dimensional and likewise the evaluation of an N time step single asset problem. In each case you are integrating a payoff function times the joint density function of N random variables.

Alan
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Re: Dimensionality of Monte Carlo

Personally, I would count the dimensionality as the number of distinct spatial factors (so ignoring time) in the transition density, So a 1D problem with a spatial factor S would have a transition density $p(t', S' | t, S)$. If your Monte Carlo gets from t to t' in one time step or numerous time steps wouldn't matter by this nomenclature: both are 1D, and the time-step count is irrelevant.

A stochastic volatility problem, if the generator was associated to a 2D diffusion, would have transition density $p(t', S',V' | t, S,V)$ and be called 2-dimensinal.  An n-dimensional diffusion, transformed into a Monte Carlo or not, would always be called n-dimensional.

But I can see ambiguities. How many dimensions in a GARCH model? How about a Markov chain?

Cuchulainn
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Re: Dimensionality of Monte Carlo

The issues can be resolved by unambiguously defining terms, as in mathematics.
My take is to separate the problem (n-factor, n-dim, SDE) from its solution (paths, Euler, fdm).  In the latter case I don't see where the word 'dimensionality' would be used.

What is the dimensionality if I value a single underlying vanilla option by evolving the underlying price over a path using multiple time steps DT = T/N, where T is the option maturity in years and N is the number of time steps. Is that a multi-dimension problem? If so, why is this so?

Dimensionality == 1.
In this case the generated path is just a 1-dimensional array of values?
One thing I don't understand about your 'multiple time steps'.... normally dt remains constant for each generated path and there are 10^6 paths' draws?

I agree with Alan; it has to do with the number of state variables.
"Compatibility means deliberately repeating other people's mistakes."
David Wheeler

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DavidJN
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Re: Dimensionality of Monte Carlo

Thanks for looking at this. My thinking too was to equate dimensionality with the number of state variables.

Having said this, might path dependency complicate things? Prepayments, knock-ins/outs and the like?

I was thinking of dt as constant, sorry if not clear.

Cuchulainn
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Re: Dimensionality of Monte Carlo

Thanks for looking at this. My thinking too was to equate dimensionality with the number of state variables.

Having said this, might path dependency complicate things? Prepayments, knock-ins/outs and the like?

I was thinking of dt as constant, sorry if not clear.
no problem.
I suppose we can qualify things even more by saying that state variables are stochastic by definition and hence it excludes deterministic variables that arise in SDEs. For example, an Asian option would be 1 factor because the average term A has no diffusion. As a PDE(S,A) it looks like a 2d problem but not quite. Actually, solving Asian options numerically is more difficult than a fill 2-factor stochastic and where it is easy to burn one's fingers. Some challenges when taking 'standard' FDM are discussed here

https://forum.wilmott.com/viewtopic.php?f=34&t=100474

Similarly, Cheyette would also be 1 factor according to this rule.

All these 'extra' constraints do not increase the dimensionality of the problem but the downstream numerics look complexer.
Anyways, that's my 2 cents.
"Compatibility means deliberately repeating other people's mistakes."
David Wheeler

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Alan
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Re: Dimensionality of Monte Carlo

Yeah, both Asian options under GBM and simple GARCH models pose similar issues. From the point of view of sources of randomness, they are 1D. From the point of view of the simplest Markov process representations, they are 2D. Let's call them 1.5D

Cuchulainn
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Re: Dimensionality of Monte Carlo

2d pdes are easier to solve than 1.4999 pde!
"Compatibility means deliberately repeating other people's mistakes."
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bearish
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Re: Dimensionality of Monte Carlo

Perhaps we can conclude from this discussion that the concept of dimensionality is not particularly useful in this context?

Alan
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Re: Dimensionality of Monte Carlo

2d pdes are easier to solve than 1.4999 pde!
Very true. Sometimes you can treat 1.4999 as 2 - $\epsilon$, where $\epsilon$ is a perturbation

Cuchulainn
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Re: Dimensionality of Monte Carlo

2d pdes are easier to solve than 1.4999 pde!
Very true. Sometimes you can treat 1.4999 as 2 - $\epsilon$, where $\epsilon$ is a perturbation
Epsilon is also yugely overloaded, see e.g. $\varepsilon$
"Compatibility means deliberately repeating other people's mistakes."
David Wheeler

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http://www.datasim.nl

Cuchulainn
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Re: Dimensionality of Monte Carlo

Perhaps we can conclude from this discussion that the concept of dimensionality is not particularly useful in this context?
Depends of course the dimension is integral or not!
On a follow-on remark, can we not just ban all nouns completely and replace them by verbs as proposed by Swift's Laputan professors?
"Compatibility means deliberately repeating other people's mistakes."
David Wheeler

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Cuchulainn
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Re: Dimensionality of Monte Carlo

2d pdes are easier to solve than 1.4999 pde!
Very true. Sometimes you can treat 1.4999 as 2 - $\epsilon$, where $\epsilon$ is a perturbation
Same as Hausdorff dimension?
https://en.wikipedia.org/wiki/Hausdorff_dimension
"Compatibility means deliberately repeating other people's mistakes."
David Wheeler

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http://www.datasim.nl

Alan
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Re: Dimensionality of Monte Carlo

2d pdes are easier to solve than 1.4999 pde!
Very true. Sometimes you can treat 1.4999 as 2 - $\epsilon$, where $\epsilon$ is a perturbation
Same as Hausdorff dimension?
https://en.wikipedia.org/wiki/Hausdorff_dimension

More akin to having various d-dimensional integrals, for example $\int e^{-\vec{x}\cdot\vec{x}/2} dx_1 dx_2 \cdots dx_d = (2 \pi)^{d/2}$.
The r.h.s. expression extends the integral to arbitrary d dimensions -- d not necessarily an integer.

Cuchulainn
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Re: Dimensionality of Monte Carlo

Very true. Sometimes you can treat 1.4999 as 2 - $\epsilon$, where $\epsilon$ is a perturbation
Same as Hausdorff dimension?
https://en.wikipedia.org/wiki/Hausdorff_dimension

More akin to having various d-dimensional integrals, for example $\int e^{-\vec{x}\cdot\vec{x}/2} dx_1 dx_2 \cdots dx_d = (2 \pi)^{d/2}$.
The r.h.s. expression extends the integral to arbitrary d dimensions -- d not necessarily an integer.
Integral calculus looks like a good place to start. Stilinger has set out the axioms

https://pdfs.semanticscholar.org/6223/2 ... 1579961958

So, instead of discrete dimension D = (1,2,3, }, D is now a real or complex continuous  variables. I think I can get my head around that concept!

So, indeed a Gaussian integral or radial Laplace operator.

Q. How would you explain the concept to a metal drummer (waves??)
"Compatibility means deliberately repeating other people's mistakes."
David Wheeler

http://www.datasimfinancial.com
http://www.datasim.nl