"Model Calibration" article in Encyclopedia of Quantitative Finance states that

. . . a common approach for selecting a pricing measure $\mathbb{Q}$ is to choose, given a set of liquidly traded derivatives with (discounted) terminal payoffs $(H^i)_{i \in I}$ and market prices $(C_i)_{i \in I}$, a pricing measure $\mathbb{Q}$ compatible with the observed market prices

where $\mathbb{Q}$ denotes

a probability measure on the set $\Omega$ of possible trajectories $(S_t)_{ t \in [0,T ]}$ of the underlying asset such that the asset price $\frac{S_t}{N_t}$ discounted by the numeraire $N_t$ is a martingale.

But we know that market prices $(C_i)_{i \in I}$ are generated by fallible human beings! Each of them has rather limited knowledge about "possible trajectories $(S_t)_{ t \in [0,T ]}$ of the underlying asset". Otherwise they wouldn't need the model we are trying to calibrate, would they?

So The Calibration Process receives some prices $(C_i)_{i \in I}$, some arbitrarily choosen mathematical model (i.e. Heston) and produces as an output the calibrated model which supposedly able to give us predictions about the future $(S_t)_{ t \in [0,T ]}$

Why do we believe that The Calibration Process is different from GIGO process?

  • 3
    $\begingroup$ It's an interesting question. Have you read the excellent (IMO): How Derivatives and Risk Models Really Work: Sociological Pricing and the Role of Co-Ordination (papers.ssrn.com/sol3/papers.cfm?abstract_id=2365294). The idea is that in a perfect world were complex instruments can be perfectly replicated by elementary building blocks like vanilla options (the $C_i$'s), then you need to have the right prices for the latter. Of course, the world is not perfect and additional features need to be priced in, hence the need for a good model... $\endgroup$
    – Quantuple
    Commented Apr 19, 2017 at 13:26
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    $\begingroup$ This also reminds me of the, not less excellent, Blank Swan by Elie Ayache. Calibrating is merely a way to mark a wrong model to the market. And although this will give you an appropriate "static" view (if you calibrate to vanilla options), this does not mean you'll capture the appropriate "dynamics". $\endgroup$
    – Quantuple
    Commented Apr 19, 2017 at 13:27
  • $\begingroup$ In my opinion, the the answer to your question is simply that it is no different for the exact reasons cited in the comments above. Excellent sources cited by @Quantuple. $\endgroup$
    – amdopt
    Commented Apr 19, 2017 at 13:37
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    $\begingroup$ @amdopt I suspect that it is no different. But having accepted that we can't help but agree that Mathematical Finance is a new branch of Alchemy. It is a frightening conclusion $\endgroup$
    – zer0hedge
    Commented Apr 19, 2017 at 15:57
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    $\begingroup$ @Quantuple "A model is merely your reflection of reality and, like probability, it describes neither you nor the world, but only a relationship between you and that world." D. Lindley "The philosophy of statistics"(2000) $\endgroup$
    – zer0hedge
    Commented May 25, 2017 at 11:15

1 Answer 1


I am intrigued by this question because it gets at the heart of so many grey areas of the financial system in which it becomes almost impossible to know how many assets derive their values from some unseen or ill-prescribed, but presumed extant, underlying process.

Calibration can be interpreted as means of deriving an expectation which is the probabilistic point estimate, subject to certain parameters, $p_1,\,p_2,\,...p_n$, i.e.,:

$\mathbb{E}[X_T] = f(X_t,\,p_1,\,p_2,\,...p_n)$

This is, in essence, The Strong Law of asset pricing. However, the law of arbitrage supersedes the strong law when it is possible to show that:

$f(\mathbb{E}[(H_T^i)_{i \in I}]) \ne (C_{i,T})_{i \in I}$

only if/when it is possible to partake in both $H^i$ and $C_i$, such as in the assumption regarding complete markets.

However, in the absence of a complete market, or when faced with complicated pay-off scenarios, any expectation of a $\mathbb{Q}$ martingale may indeed be parametric at best (i.e., the expectation must taken through calibration).

My skepticism is perhaps best demonstrated by the following passage out of Baxter's and Rennie's Financial Calculus (kudos to @DaneelOlivaw for making me aware of this):

Almost everything appeared safe to price via expectation and the strong law, and only forwards and close relations seemed to have an arbitrage price. Since 1973, however, and the infamous Black-Scholes paper, just how wrong this is has slowly come out. Nowhere in this book will we use the strong law again. […] All derivatives can be built from the underlying −− arbitrage lurks everywhere.

Perhaps... but my personal, fallible experience tells me otherwise. While the no-arbitrage range of possible values for an equity option may be known presuming that the price of the equity is known, what is the fair value of an equity? I.e., how can we construct a replicating payoff for this equity in a way that is not a tautology (i.e., a thing which defines itself but nothing more)? To my knowledge, there exist no market for accounting values of assets and liabilities. More explicitly, how can we show the value of a thing, $C_t$, as follows:

$C_{i,t} = \int_t^T f(\mathbb{E}[H_{i,t}]P_t) \, dt$; $P_t := e^{-rt}$

when $C_{i,t}$ is a function of human perception regarding the unknown future values of $T$ and $H_{i,t}$, even if we take the risk-neutral expectation and short-rate as gospel?

Given that no perfect model for human behavior exists (otherwise that model would equal reality and its creator, a god), an imperfect (practical) answer to mitigating GIGO is to derive an expectation which make use of the fewest possible parameters. Fewer parameters means less calibration, which means decreased odds of over-fitting.

A thing which is descriptive of the past, present, and/or future, and which is also not highly calibrated has a better likelihood of being prescriptive than a thing which is more highly descriptive but also more highly calibrated. Is there a model for that? And don't say degrees of freedom...

Does this imply that highly-specified models (e.g., Heston) which calibrate expectations to observations are less robust? Not necessarily if the fit is not garbage (i.e., not spurious; i.e., it states something which is true regarding the nature of uncertainty), but in aggregate, I believe so.

I take the broad corpus of economic literature's failure to predict anything but the past as evidence.

  • $\begingroup$ "I take the broad corpus of economic literature's failure to predict anything but the past as evidence." I like this statement. $\endgroup$
    – will
    Commented Apr 20, 2017 at 22:15
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    $\begingroup$ I particularly agree with the tautology part of your answer. And this is why there will always be a "smile problem" as Ayache puts it. $\endgroup$
    – Quantuple
    Commented Apr 21, 2017 at 9:19
  • $\begingroup$ @Quantuple. Thank you. I actually haven't read Blank Swan, but I have just added it to my (growing) read list. $\endgroup$ Commented Apr 21, 2017 at 21:57

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