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From volatility surfaces we have a implied distribution of $S_T$. This distribution is the real world distribution or this is a risk neutral distribution?

Joanna
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In this related question How to derive the implied probability distribution from B-S volatilities?, it is shown how to infer the implied probability density of the future prices of a risky asset from a continuum of call prices written on that asset (Breeden-Litzenberger identity).

The developments, which I invite you to read, basically rely on the fact that the call price writes $$ C=e^{-rT} \int_0^\infty (S-K)^+ p(S) dS $$ or equivalently $$ C = e^{-rT} \Bbb{E} \left[ (S_T-K)^+ \right] \tag{1} $$ under some measure where $S_T$ is a random variable with probability density function: $$ d\Bbb{P}(S_T \leq S)/dS = p(S) $$

Now, the fundamental theorem of asset pricing tells us that equation $(1)$ holds under the so-called risk-neutral measure (a measure equivalent to the real-world measure but under which the $t$-value of any self-financing strategy is a martingale when expressed with respect to the risk-free money market account numéraire).

Consequently, the implied density $p(S)$ you compute by evaluating $$ p(S) = e^{rT} \frac{\partial^2 C}{\partial K^2}(K=S) $$

is indeed a risk-neutral pdf (because it relies on the risk-neutral expression of the call price $(1)$.)

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Quantuple
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  • So the expected value of the impled distribution of $S_T$ is always $S_0 e^{rT}$. Correct? – Joanna Dec 14 '16 at 11:37
  • Omitting any other carry effects (e.g. dividends), yes. More accurately the expectation is equal to the forward price. – Quantuple Dec 14 '16 at 12:28
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I'm not a probabilist but I tend to think real-world distribution in financial world doesn't exist (or at least is not a proper term). None of the financial events are really repeatable or IID.

And to your question, it should be risk-neutral density function.

Will Gu
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    That's not true. Of course that the real world measure exists, this has nothing to do with any statistical property such as IIDness. The time series that you work with every day are precisely realisations of stochastic processes under the real world measure. In addition, everything which can be expressed as an expectation under $\Bbb{Q}$ can also be expressed as an expectation under $\Bbb{P}$ since $ \Bbb{E}^\Bbb{Q}t [ X_T ] = \Bbb{E}^P_t \left[ X_T \left. \frac{d\Bbb{Q}}{d\Bbb{P}} \right\vert{\mathcal{F}_T} \right]$. – Quantuple Dec 14 '16 at 09:19
  • So, yes, the real world measure exists. However, it is not relevant for option pricing purpose. To convince yourself go back to the binomial pricing framework (or BS continuous time framework) and see how real world probabilities disappear from the pricing equation due an absence of arbitrage opportunities + static (resp. dynamic) replication argument. – Quantuple Dec 14 '16 at 09:21
  • @Will Gu Without the the real world measure, we haven't anything. –  Dec 14 '16 at 10:55
  • @Quantuple Is there a way o getting the real distribution of $S_T$? – Joanna Dec 14 '16 at 12:10
  • This would require some working modelling assumptions, since after all you are looking at modelling some future state of the economy based on your current knowledge. See the many existing and related SE questions. – Quantuple Dec 14 '16 at 12:30
  • @Quantuple what is SE? Could you please give the link of a such related question? – Joanna Dec 14 '16 at 12:50
  • @John please avoid cross-posting the same question at different places. You have just created a question dedicated to that matter, I have answered there. – Quantuple Dec 14 '16 at 12:57
  • @Quantuple Thanks for your comments. I understand that the real-world measure does not matter in option pricing. I guess I was thinking more like the real-world distribution is subjective. After all, you can literally choose any radon nikodym derivative in change of measure and get a distribution, but there's virtually no way to know if this is real, right? On the other hand, the distribution of number of heads in N coin tosses is known in theory and can be tested by experiment. – Will Gu Dec 14 '16 at 18:24
  • @Will Gu The manner in which one can model the real world measure is subjective in the sense that it remains a mathematical abstraction of a real life phenomenon, hence it depends on the effects you would like to account for. But it's same for any non trivial random outcome you would try to forecast... Even a dice (e.g. Take wear due to fatigue or thermal expansion into account). It all depends on the glasses you want to wear at the end of the day – Quantuple Dec 14 '16 at 19:03
  • I would say instead that "the real-world distribution exists but is unknown" while "the risk-neutral distribution is known but does not exist" – Kiwiakos Dec 14 '16 at 19:23
  • Clever way to put it indeed. – Quantuple Dec 14 '16 at 22:09