# Phorgy Phynance

## Modeling Currencies

I hope to begin some research into currencies. Before I come out with any result though, I thought I’d ask an open question and hope someone comes by with a response.

First of all, as a former scientist, thinking about currencies is very fun. See, for example, my previous article

This morning, I happened across a recent article that appeared on the arxiv:

The second paragraph really stood out:

One of the problems in foreign exchange research is that currencies are priced against each other so no independent numeraire exists. Any currency chosen as a numeraire will be excluded from the results, yet its intrinsic patterns can indirectly affect overall patterns. There is no standard solution to this issue or a standard numeraire candidate. Gold was considered, but rejected due to its high volatility. This is an important problem as different numeraires will give different results if strong multidimensional cross-correlations are present. Different bases can also generate different tree structures. The inclusion or exclusion of currencies from the sample can also give different results.

This is interesting because financial modeling is often about prices of securities or changes in prices. Currencies are about the relationship between prices. In graph theoretic (or category theoretic) terms, it is tempting to say that currency models should be about directed edges (or morphisms).

Is the best way to model currencies to choose some numeraire as is done in this paper? Or is there a way to study the relationships (morphisms) directly?

Written by Eric

October 28, 2010 at 6:01 pm

## Discrete stochastic calculus and commutators

This post is in response to a question from Tim van Beek over at the Azimuth Project blog hosted by my friend John Baez regarding my paper

The basic fact needed to address the question is that we have a set 0-dimensional objects $\mathbf{e}^\kappa$ and 1-dimensional objects $\mathbf{e}^{\kappa\lambda}$ that obey the following geometrically-motivated multiplication rules:

1. $\mathbf{e}^\kappa \mathbf{e}^\lambda = \delta_{\kappa,\lambda} \mathbf{e}^\kappa$
2. $\mathbf{e}^{\kappa\lambda} \mathbf{e}^\mu = \delta_{\lambda,\mu} \mathbf{e}^{\kappa\lambda}$
3. $\mathbf{e}^\mu \mathbf{e}^{\kappa\lambda} = \delta_{\mu,\kappa} \mathbf{e}^{\kappa\lambda}$

To see the geometrical meaning of these multiplications, it might help to consider discrete 0-forms

$f = \sum_\kappa f(\kappa) \mathbf{e}^\kappa.$

and

$g = \sum_\lambda g(\lambda) \mathbf{e}^\lambda$

Multiplication 1.) above implies

$f g = \sum_\kappa f(\kappa) g(\kappa) \mathbf{e}^\kappa,$

which is completely reminiscent of multiplication of functions where we think of $f(\kappa)$ as the value of the function at the “node” $\mathbf{e}^\kappa.$

Multiplications 2.) and 3.) introduce new concepts, but whose geometrical interpretation is quite intuitive. To see this consider

$f \mathbf{e}^{\kappa\lambda} = f(\kappa) \mathbf{e}^{\kappa\lambda}$

and

$\mathbf{e}^{\kappa\lambda} f = f(\lambda) \mathbf{e}^{\kappa\lambda}.$

Multiplying the function $f$ on the left of the “directed edge” $\mathbf{e}^{\kappa\lambda}$ projects out the value of the function at the beginning of the edge and multiplying on the right projects out the value of the function at the end. Hence, the product of functions and edges do not commute.

In the paper, it was shown that the exterior derivative of a discrete 0-form is given by

$df = \sum_{\kappa,\lambda} \left[f(\lambda) - f(\kappa)\right] \mathbf{e}^{\kappa\lambda}.$

Here, we show that this may be expressed as the commutator with the “graph operator”

$\mathbf{G} = \sum_{\kappa,\lambda} \mathbf{e}^{\kappa\lambda}.$

The result is quite simple and follows directly from $\mathbf{G}$ and the multiplication rules, i.e.

$f\mathbf{G} = \sum_{\kappa,\lambda} f(\kappa) \mathbf{e}^{\kappa\lambda}$

and

$\mathbf{G} f = \sum_{\kappa,\lambda} f(\lambda) \mathbf{e}^{\kappa\lambda}$

so that

$[\mathbf{G},f] = \sum_{\kappa,\lambda} \left[f(\lambda) - f(\kappa)\right] \mathbf{e}^{\kappa\lambda}$

or

$df = [\mathbf{G},f].$

Written by Eric

October 27, 2010 at 3:47 pm