Take the 2-minute tour ×
Quantitative Finance Stack Exchange is a question and answer site for finance professionals and academics. It's 100% free, no registration required.

I am more of a probabilist than a financial mathematician. I am currently working on the features of American put options under a particular stochastic volatility model.

Like most stochastic volatility models, it is incomplete. (In fact, it would be nice if someone tell me a complete, stochastic volatility model, is there any?) In my current treatment, I have just treated the model as a maths toy. I have chosen an arbitrary risk neutral measure and try to say something about the value of options. (Of course, the proofs holds in any EMM.)

Though the question I asked here is not extremely closely related to what I am doing, I would still like to know:

How does someone choose an EMM? Do you restrict yourself to a subclass of EMM and give yourself some parameters which you try to fit using given data?

share|improve this question
2  
There won't be a non-degenerate complete stochastic volatility model. An informal way of thinking about it is that the space of randomness is two dimensional (two BM's driving things), and the space of attainable claims is just one-d –  quasi Jan 15 at 23:34
    
@quasi i see, unless you can trade volatility as an asset, this is starting to ring a bell. –  Lost1 Jan 16 at 0:54
add comment

3 Answers

up vote 2 down vote accepted

Hum, that's one of the most important questions in financial engineering, that why no answer is proposed.

If you have available data as option prices, you may calibrate a parametric EMM but nothing can tell that it's the best EMM (cause there is no best EMM).

So make a choice and defend your choice by saying 'it's simple and allows beautiful result' like every body use to do.

share|improve this answer
    
In my work, we just assume it is an EMM and what I do is true for all EMM of that particular model anyway. In particularly, I am looking at BNS model. I think the difficulty is actually picking a 'correct' Levy process to model the volatility, which is hard. Doing this is definitely a non trivial problem in statistical inference, which does not interest me, in particular –  Lost1 Jan 19 at 14:05
add comment

A stochastic volatility model for a single risky asset can't be complete because you have two sources of randomness. But you can easily make it complete by adding a derivative whose value depends on the volatility. For example, if you add a variance swap in the Heston model then it becomes complete. This allows you to calibrate the model.

But your question is even more relevent when considering models with jumps (like discontinuous Levy process) because jumps are a "sum" of Poisson processes so each size of jump adds another source of randomness. You would need to add an infinite number of derivatives to make the market complete which is absurd. From the point of view of probability changes this is because you can find equivalent probabilities that change the drift but also the distribution of the jumps.

The originial approach by Merton is to consider that you don't want the distribution of jumps to change so there is a unique risk neutral probability doing that. The probability change is the same as in the Black-Scholes model.

Other approaches use utility functions but now you are just asking yourself which utility function to use. This seems worse since you step away from the no arbitrage philosophy based on assumptions on markets and go back to modelling people's aversion to risk as if people were rational optimizing machines.

share|improve this answer
add comment

There are several ways to choose a particular EMM. I believe that the most popular approach is to use a "distance" between $\mathbb{P}$ and $\mathbb{Q}$. Most papers use a minimal entropy approach(for example, Fujiwara and Miyahara, Esche and Schweizer, or Hubalek and Sgarra) or a relative q-entropy approach (for example, Jeanblanc, Klöppel, & Miyahara)

In applications, the Escher transform is also used very frequently to establish a relationship between $\mathbb{P}$ and $\mathbb{Q}$. The transform is sort of an exponential tilting of the probability measure.

share|improve this answer
    
What is the reason to pick up $\Bbb Q$ to be closest to $\Bbb P$ w.r.t. some metric? –  Ilya Apr 8 at 15:32
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.