Is it possible to demonstrate that one pricing model is better than another?

Question

Take the classic GBM (geometric Brownian motion) model for equities as an example:

ds = mu * S * dt + sigma * S * dW.

It is the basis for the classic Black-Scholes formula.

The model says volatility is constant, which is apparently not true considering the volatility smile. However, many practitioners use the formula, although they apply some interpolation scheme. For example, if the stock price is \$100, to price an option with strike price \$130, people may

1. Ask big banks what Black-Scholes volatility they are using for strike prices of \$100, \$120, and \$140.
2. Interpolate for a stock price of \$130.
3. Plug that vol into Black-Scholes and calculate the option price.

Since everyone is applying the same formula, there's no risk or bad consequences to using an inaccurate formula, as long as it's "smartly" used, as in the example, with some interpolation to handle the volatility smile.

What's more, if there's any mispricing, it seems it's also hard to say what's the cause -- if a new model projected a different option price and the options on the market gradually converged to this value, it can be any reason, maybe the Black-Scholes model is not wrong but the users' interpolation is not accurate, maybe the whole environment changed so convergence is just by chance?

In this case, if there's another model, for example, a modification to the GBM model leading to a formula slightly different from Black-Scholes, how could one argue it's better?

Answer 1

There are many different ways a pricing model can be better :

• It can allow to reproduce the observed market price (Fit criterion)

• It takes into account a specific recognized behaviour of the underlying S, say the forward smile dynamic. If you write a product whose value is mostly derived from said behaviour, you dont want to miss that aspect. (Don't fill me up with 0 unpriced risk criterion )

Then 2 quite similar criteria can be additionally noted

• it generates more PL (Kerviel superiority criterion)
• it gets you more client (Building a great franchise criterion)

Answer 2

In the way that you have posed the question, I would say that we are here discussing a derivative-pricing model rather than a predictive model.

That's an important distinction because a predictive model would be assessed by its ability to generate money.

In contrast, I think of derivative pricing as a fancy way of doing interpolation/extrapolation on prices of vanilla instruments to derive the 'fair' price of a derivative product. It does not attempt to be predictive.

However, the main principle that underlies all derivatives pricing is the ability to use those vanilla instruments as a dynamic hedge.

This implies that a good model is one which generates a hedging strategy that works well and which therefore allows derivatives traders to sell the derivative product at a premium and know that they can effectively capture that premium by hedging with the vanillas.

Answer 3

Might be a bit overlapping with nicolas' answers, but here it goes:

Id say you would have to look at the prediction-power of the model at hand. What if you do a backtest where you set a time t in the future?

Set a price range for the stock at time t, and check with market data how often the price have been within the range. Then, for each model calculate the probability that the price would be within this range.

You should also probably test with different ranges, and put more weight on the ranges that are as wide as you would care. (There is no problem of defining a model saying that a stock has a value between 0 and infinity. It will always be correct but not very precise)

Also, if one is interested in the options, and one have derived hedging strategies for each model, one can backtest how good the hedging strategies are.