Why Option Pricing Models Often Assume Finite Variance

# Why do people always seek finite-variance models for option pricing


# Why do people always seek finite-variance models for option pricing












For the purpose of getting fatter tails than the Guassian, I have seen people for example use $\alpha$-stable processes to model the stock. But in that case they end up using 'tempered' versions of the processes, where the tails decay exponentially so as to make the second moment finite. So the standard seems to be that the second moment must be finite. But why is this so? Is it just for tractability of the model or do they believe that finite second moment is an empirical fact?

Additional discussion:

In this paper Taleb explores the possibility of constructing a risk-neutral measure in an infinite-variance setting. As he mentions, this destroys the dynamic-hedging theory, but in practice it does not make a difference.

## Answer by vonjd (score 3, accepted)

https://quant.stackexchange.com/a/18642

I would argue that there is some path-dependency involved. The BS model is considered the big breakthrough and it presented the world with some kind of tractable toy model. After that people saw that you had to adjust the model to account for all kinds of stylized facts (e.g. non-constant volatility for different strikes, over time and so on). Yet finite variance somehow survived this tweaking of the models, may it be for mathematical convenience, may it be that people just didn't think into this direction because there are other, more important issues (like numerical stability and so on).

Another thing is that empirically variance is always finite and even theoretically stock prices can only fall to zero but not indefinitely so finite variance (even though as high as needed) seems like a not too bad idea.

Just another trivial reason is that most of statistics deals with probability distributions with finite variance and people tend to use what they find and what they are used to.

As an aside there is a new paper by Taleb where he develops an option pricing model which can cope with infinite variance:

Risk Neutral Option Pricing With Neither Dynamic Hedging nor Complete Markets

Abstract

> Proof that under simple assumptions, such as constraints of Put-Call Parity, the probability measure for the valuation of a European option has the mean derived from the forward price which can, but does not have to be the risk-neutral one, under any general probability distribution, bypassing the Black-Scholes-Merton dynamic hedging argument, and without the requirement of complete markets and other strong assumptions. We confirm that the heuristics used by traders for centuries are both more robust, more consistent, and more rigorous than held in the economics literature. We also show that options can be priced using infinite variance (finite mean) distributions.

## Answer by Nico (score 1)

https://quant.stackexchange.com/a/18616

On a pure technical aspect, a model does not need to have a finite variance. In the context of option pricing, what you need it a way to replicate the behaviour of the stock price. Once you have it you need to find a corresponding risk-neutral measure. There you will have the first difficulty, with infinite variance, the corresponding hedging strategy is not obious. However, as in the case of jumpe processes, you might still find a risk-neutral measure by means of min entropy for instance (see Fujiwara, Miyahara (2003)). So in theory, you can apply compute option prices using monte carlo.

On the other side, on a practical point of view, infinite variance distribution will cause many difficulties during the calibration of your model (simply because infinite variance are directly observable). Furthermore the standard stochastic calculs like Ito's-Lemma does not apply in this context. Infinte variance distribution model, that bring higher cost in term of complexity, should allow for better performences. However, the class of jump-diffusion processes (see Crepey (2013)) or even infinite levy process (see Tankov (2007)) are in pricipe rich enough to enable a efficient pricing (both in term of price replication as computation speed).

So, in my opinion, the reason why there are not used in practice is because there are not tractable enough the outperform more classical model.

## Answer by NaN (score 1)

https://quant.stackexchange.com/a/18912

I would also say that the pricing of some exotic products require to compute expectations of functions of the random variable at consideration, and these functions may grow more than linearly : you need finite moments in order for the prices of these exotic derivatives to be bounded.