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

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?

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Re: "model says volatility is a constant value, which is apparently not the reality considering volatility smile". Since implied volatility is always estimated with respect to a model, another interpretation is that the Black-Scholes forumula is false or inaccurate because the backed-out implied volatilities are not all constant. – Quant Guy Mar 11 '12 at 21:46

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@QuantGuy Good point, but to be fair to athos, realized volatility, measured as quadratic variation, has also been shown to be inversely related to price (known as the leverage effect). – Tal Fishman Mar 12 '12 at 14:51
Fair point. Touche! – Quant Guy Mar 12 '12 at 15:06

## 3 Answers

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)
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thanks @nicolas. hmmm, it's my first time heard of "Kerviel superiority criterion", i'm wondering what is that? tried to google but to no avail, could you please give a hint, e.g. some links? thanks... – athos Mar 11 '12 at 15:25

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– Bob Jansen Mar 11 '12 at 17:13
@athos yeah the last two ones are a joke. But actually, semi serious. the name is a joke, but some people might appreciate the concept. that's why you have cops like risk department, who is sometimes allowed to make its work, and regulators, who, if they knew what to do and where to look, might be doing theirs too. – nicolas Mar 11 '12 at 18:23
+1 for "Kerviel superiority criterion". (snicker) – Brian B Mar 12 '12 at 20:38

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.

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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.

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+1 for the hedging comment.  Black-Scholes may be convenient as a pricing tool, but if the goal of your model is to come up with accurate deltas, different models may make a huge difference, and these models should be judged based on how well they explain market movements. – Tal Fishman Mar 12 '12 at 15:02
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