Assume we have inferred risk neutral density of stock price at time T from option prices. Assume we have obtained a parameterized density p(S). How can we infer real world measure? I know about Girsanov's Theorem but I am not sure whether I can use it for this.


1 Answer 1


Let $\Bbb{P}$ denote the physical measure and $\Bbb{Q}$ the risk-neutral one.

First of all, it's important to realise that while $\Bbb{P}$ exists but is not tractable (it is the measure under which we observe realisations of the various market quantities), $\Bbb{Q}$ is a pure mathematical construct that cannot be observed in the real world but can be tied to $\Bbb{P}$ under some assumptions.

These assumptions are: picking a valuation model and postulating the absence of arbitrage opportunities (= fair valuation). Without this, there is no way to relate the 2 measures.

This caveat can be further understood by noting that (using a mathematical concept known as change of measure) \begin{align} \Bbb{P}[S_T \leq K] &= \Bbb{E}^\Bbb{P} \left[ \Bbb{1}\left\{ S_T \leq K \right\} \right] \\ &= \Bbb{E}^\Bbb{Q} \left[ \Bbb{1}\left\{ S_T \leq K \right\} \left. \frac{d\Bbb{P}}{d\Bbb{Q}} \right\vert_{\mathcal{F}_T} \right] \end{align} where the expectation on the RHS is the one you should compute under the risk-neutral measure $\Bbb{Q}$ to end up on the CDF under the real-world measure $\Bbb{P}$.

Now although the probability density function of $S_T$ under $\Bbb{Q}$ $$ q(T, S) = \frac{d\Bbb{Q}[S_T < S]}{dS} $$ can be computed in a model-free way from vanilla option prices (Breeden-Litzenberger identity), the Radon-Nikodym derivative $$ \left. \frac{d\Bbb{P}}{d\Bbb{Q}} \right\vert_{\mathcal{F}_T} $$ remains model-specific. Hence without a model, you are stuck.

Actually, even with a model, estimating the parameters appearing in the Radon-Nikodym derivatives (market prices of risk) can prove quite tricky if not impossible, see discussion here.

Some people worked on a model-free methods for moving from $\Bbb{P}$ to $\Bbb{Q}$ by e.g. minimising the Kullback-Leibler divergence (or relative entropy) between the two densities subject to some relevant constraints (see work of Derman and Zou) . I haven't found that to work very well in practice though.

  • $\begingroup$ not sure I fully understand your answer. Let's recap. what I have is an estimated (from option prices) risk neutral density of stock price at T in parametric form q(S_T, params), where params are parameters of the density (not the underlying stochastic process). So q is a continuous function over the reals. RD derivative would also have to be continuous function with same domain, so I can do p=q*RD, right? Now you are saying that RD would be model specific? What would RD look like if written out? $\endgroup$ Jan 4, 2018 at 17:12
  • $\begingroup$ Everything you say is fine. The RN derivative is model-specific and involves parameters known as market prices of risk (each model having its own view on what is the 'market' hence these market prices of risks involve both market quantities and model parameters). See the links in the answer I gave where RN is made explicit for BS: $$ \left. \frac{d\mathbb{Q}}{d\mathbb{P}} \right\vert_{\mathcal{F}_t} = \mathcal{E}(-\lambda W_t^{\mathbb{P}}) $$ where $\lambda$ is the Sharpe ratio of the stock in that case (thereby relating quantities under P and quantities under Q). $\endgroup$
    – Quantuple
    Jan 4, 2018 at 17:20
  • $\begingroup$ I read this actually, but it's still not clear. What you provide above seems to introduce a problem. The density q is a deterministic function of S_T. The expression for the RN-derivative is stochastic. Multiplying q with it would lead to a function p that is also stochastic. However, p is simply a density, just like q. $\endgroup$ Jan 4, 2018 at 19:28
  • $\begingroup$ The RN derivative is a random variable indeed. Like the terminal stock price $S_T$. Both are functions of $W_T^\Bbb{Q}$. Maybe it would help you to write the expectation in integral form in terms of the random variable $W_T^\Bbb{Q}$. This will show that the knowledge of $q(T,S_T)$ is not enough. $\endgroup$
    – Quantuple
    Jan 4, 2018 at 19:37
  • $\begingroup$ I don't believe that addresses my last question. You have earlier confirmed that p=q*RNderivative. Given the expression you have provided for RNderivative, I have raised the issue that this would imply deterministic function = deterministic function * random function, which is impossible. What gives? $\endgroup$ Jan 5, 2018 at 0:07

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