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In practice, I have seen articles and financial textbooks on calibration of processes directly under the risk neutral world without showing that the measure is equivalent to a physical measure $P$. They seem to make an assumption that in the physical world, the market is arbitrage free, and that there exists an equivalent measure $P$, whatever the process defined on $Q$ looks like in $P$. Is there a reason for this and why people don't bother with even checking that there exists an equivalent measure $P$?

noidea
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    The argument goes the other way around. The only real measure is $\Bbb{P}$: the financial time series we all observe are realisations of stochastic processes under $\Bbb{P}$. Yet, obivously, $\Bbb{P}$ remains unknown in practice. By making the theoretical assumptions of (i) no arbitrage (ii) complete market, we can show that there exists a unique probability measure $\Bbb{Q}$, equivalent to $\Bbb{P}$ (in the mathematical sense), under which the discounted value of self-financing portfolios are martingales, which allows us to price instruments by taking expectations under $\Bbb{Q}$. – Quantuple Sep 11 '17 at 12:23
  • If there is either arbitrage or the market is incomplete, then this result does not hold (either there is no such measure, or it is not unique). You can be sure that a lot of people bothered demonstrating this. – Quantuple Sep 11 '17 at 12:25
  • While I agree with @Quantuple's first comment, the second I believe is misleading...the market may not be complete, but by observing the option prices that do exist one can observe the "Q" measure that the market has chosen. – user9403 Sep 11 '17 at 12:46
  • @Quantuple I'm not sure I follow. $X$ process on $(\Omega, \mathcal F, P) \implies$ no arbitrage given the process $X$ on $(\Omega, \mathcal F, P)$ IFF $\exists Q$~$P$ discounted process $X^$ is a $Q$-martingale. So if one starts with constructing $X^$ such that it is a martingale how can one possibly know that we can not find an arbitrage with the process $X$ in the real world? – noidea Sep 11 '17 at 13:05
  • @noidea, If $\Bbb{Q} \sim \Bbb{P}$ and there is no arbitrage under $\Bbb{Q}$ then there is no arbitrage under $\Bbb{P}$ by definition of the equivalence notion, see AFK's answer here: https://quant.stackexchange.com/questions/12778/arbitragefree-pricing-q-vs-p. – Quantuple Sep 11 '17 at 13:46
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    @user9403, sorry if this is misleading. I agree that by calibrating to option prices, we will agree on the risk-neutral dynamics of the underlying. But, if the market is incomplete, that does not mean we will necessarily agree on how to translate that back to the $\Bbb{P}$-world. Take stochastic volatility à la Heston for instance. True we'll have the same set of 5 parameters under $\Bbb{Q}$, but depending on our view of the market price of volatility risk, we'll potentially have different RN derivatives for the change of measure to express this dynamics back under $\Bbb{P}$ (...) – Quantuple Sep 11 '17 at 13:50
  • (...) mathematically this is because there is an infinite number of RN derivatives that make discounted asset prices martingales, each impacts the drift of the vol process differently. Financially, this is because there is no unique way to hedge against the volatility uncertainty. – Quantuple Sep 11 '17 at 13:52
  • @Quantuple Yes, I agree with that, but then you made the assumption of equivalence of measures,, which turns us back to the original question, circle argument, no? :-/ – noidea Sep 11 '17 at 13:56
  • @noidea, I'm not sure I understand your point. You mean that your original question should be: why can we say that there is no arbitrage and how much can we trust this assumption? (since we agree that no arb under $\Bbb{Q}$ is the same as no arb under $\Bbb{P}$ if the measures are equivalent)? If so (or if not) could you please edit your question to make things more clear? The argument is that $\Bbb{P}$ exists but is unknown. Making some assumptions, we are able derive a measure $\Bbb{Q}$, which doesn't exist but is equivalent to $\Bbb{P}$ and that we can pin down mathematically, hence known. – Quantuple Sep 11 '17 at 13:58
  • @Quantuple First they choose a martingale process $X^$ and claim they are in $Q$. Then they assume that a process $Y$ is chosen in such a way that it explains the market with an arbitrage free property in the physical world. Now they use asset pricing theorem. How can they be sure that at least one of the equivalent measures $Q$ which makes the discounted price process $Y^$ a $Q$-martingale, equal to $X^*$? – noidea Sep 11 '17 at 14:04
  • You can't say that like that indeed, not unless the process $Y$ is the expression of the process $X$ under the measure $\Bbb{P}$. This requires applying the Girsanov theorem, which itself relies on the expression of the Radon-Nikodym derivative of the change of measure between $\Bbb{Q}$ and $\Bbb{P}$. Usually, we define the process under $\Bbb{P}$ (i.e. $dX_t/X_t = \mu dt + \sigma dW_t^\Bbb{P}$). We known that the discounted prrocess should emerge as a martingale under $\Bbb{Q}$ which fixes the RN derivative (if the market is complete) (i.e. $dX_t/X_t = r dt + \sigma dW_t^\Bbb{Q}$) – Quantuple Sep 11 '17 at 14:06
  • @Quantuple Ok, now I feel that I have not misunderstood too much, they seem to make the implicit assumption that $X$ is one of the processes that makes the market arbitrage free in the physical world, which is actually a very harsh assumption I believe? And regarding your edited comment addition. They use stochastic time changed pure levy jump processes, which will not necessary be a levy process after the time change. Hence, it is really difficult to know if it is a "legit" assumption at all. This is from Schoutens Levy processes in finance, pricing financial derivatives 2003. – noidea Sep 11 '17 at 14:12
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    Well it's as harsh as any modelling assumption I suppose: "all models are wrong, some models are useful". The real test at the end of the day is to hedge with your model and see if you manage to defend the margin that you charged at inception. – Quantuple Sep 11 '17 at 14:21
  • @Quantuple That is so true. – noidea Sep 11 '17 at 14:46

1 Answers1

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Fundamentally, option pricing is an extrapolation exercise. Fitting a q-measure model to the observed option prices gives a way of performing the extrapolation.

If q-measure model gives reasonable dynamics to the option prices observed and the underlying then the process of dynamic hedging with the options will work. If it doesn't, there will be systematic biases and the model will be poor.

Practitioners don't worry about the real-world process because it's unknowable and modelling it better rarely helps with the modelling and hedging.

Mark Joshi
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  • Thank you for your answer. Before accepting it I have one small question. Why do practitioners want the process to be a martingale in the $Q$-measure if they do not bother with that it is equivalent to the real world measure? Couldn't they just choose any kind of process and fit it to the option data if they do not bother with the real world at all? If they can not confirm equivalence of measures, the use of fundamental theorem of asset pricing fails anyway, no? – noidea Sep 12 '17 at 09:03
  • well people generally want a model with no internal arbitrages, a martingale measure guarantees that – Mark Joshi Sep 12 '17 at 10:57
  • Thank you. Perhaps Offtopic but: your proof pattern book seems interesting. I will order it in the future! – noidea Sep 12 '17 at 11:19