I understand that we can use option prices to imply volatilities and ultimately to imply a risk neutral density. I also understand that this implied density is not the same as the "real world density". However, the risk neautral density will often show certain shapes/regularities/irregularities that correspond strongly to "real world" concerns. For example, around earnings or events where there is a likelihood of jumping to 1 of two states/prices, the implied risk neutral density will be bimodal. So there is a connection. But does it make sense to look at the implied density and make a statement like "the option prices imply a 23% chance that the underlying will move below 100" or "the option prices imply a 33% chance that there will be a jump higher after earnings"? What is the mapping between implied densities/probabilities and statements/assessments of this sort?


1 Answer 1


I'll outline how you can estimate the (implied) real-world density function from (observed) option prices. Having found this real-world density, you can then compute all sorts of probabilities and quantify the market's expectation of future prices.

Recall firstly that (European-style) options are priced as risk-neutral expectation of the discounted payoff. Thus,

\begin{align*} C(S_0,K,T) &= e^{-rT} \mathbb{E}^\mathbb{Q}\left[ (S_T-K)^+\right] \\ &= e^{-rT} \int_\mathbb{R} (x-K)^+ f_{S_T}^\mathbb{Q}(x)\ \mathrm{d}x \\ &= e^{-rT} \int_\mathbb{R} (x-K)^+ \frac{f_{S_T}^\mathbb{Q}(x)}{f_{S_T}^\mathbb{P}(x)}\ f_{S_T}^\mathbb{P}(x)\mathrm{d}x \\ &= \mathbb{E}^\mathbb{P}\left[ M_T (S_T-K)^+\right], \end{align*} where the random variable $M_T(x)=e^{-rT}\frac{f_{S_T}^\mathbb{Q}(x)}{f_{S_T}^\mathbb{P}(x)}$ is the stochastic discount factor (SDF) aka pricing kernel.

In a more formal way, you can say $C(S_0,K,T) = \mathbb{E}^\mathbb{Q}\left[ \frac{1}{B_T}(S_T-K)^+\right]= \mathbb{E}^\mathbb{P}\left[ \frac{1}{B_T}\frac{\mathrm{d}\mathbb{Q}}{\mathrm{d}\mathbb{P}}(S_T-K)^+\right]$ and can identify the SDF as Radon Nikodym derivate, $M=\frac{1}{B_T}\frac{\mathrm{d}\mathbb{Q}}{\mathrm{d}\mathbb{P}}$, which also explains the name ‘’stochastic discount factor''.

As you said, the risk-neutral density $f_{S_T}^\mathbb{Q}$ can be approximated by option prices, the easiest approach is the Breeden and Litzenberger (1978) result \begin{align*} f_{S_T}^\mathbb{Q}(x) &= e^{rT}\frac{\partial^2 C(S_0,K,T)}{\partial K^2}\bigg|_{K=x}. \end{align*} Of course, there are problems about obtaining a set of arbitrage-free option prices and option prices with very low/high strikes which are required to estimate the tails of $f_{S_T}^\mathbb{Q}$ accurately.

Assume now you know the distribution of the terminal stock price under the risk-neutral measure $\mathbb{Q}$. If you also knew the SDF, you'd be done and the real-world density is given by $f_{S_T}^\mathbb{P}(x)=e^{-rT}\frac{f_{S_T}^\mathbb{Q}(x)}{M_T(x)}$.

Unfortunately, we don't know the real SDF. There are many asset pricing model proposing and deriving all sorts of SDFs. The simplest case employs a power utility function $u(x) = \frac{x^{1-\gamma}}{1-\gamma}$ for $\gamma\neq1$ (The case $\gamma=1$ yields log-utility). A key property is that under such a utility function, the relative risk aversion coefficient $-x\frac{u''(x)}{u'(x)}=\gamma$ is constant. The SDF is proportional to the marginal utility $u'(x)=x^{-\gamma}$.

Thus, \begin{align*} f_{S_T}^\mathbb{P}(x) &= C\cdot x^{\gamma}f_{S_T}^\mathbb{Q}(x). \end{align*} The constant $C>0$ (which captures the discount factor $e^{-rT}$ and a proportionality constant) needs to make sure that $f_{S_T}^\mathbb{P}$ integrates to one. Thus, we finally arrive at \begin{align*} f_{S_T}^\mathbb{P}(x) &= \frac{x^{\gamma}f_{S_T}^\mathbb{Q}(x)}{\int_\mathbb{R} x^{\gamma}f_{S_T}^\mathbb{Q}(x) \mathrm{d}x}. \end{align*}

A few notes

  • Suppose you know $f_{S_T}^\mathbb{Q}$ in closed-form from some model. You would probably still need to numerically solve the integral in the denominator... Unless you assume $f_{S_T}^\mathbb{Q}$ is log-normal or the mixture of log-normals or something easy. If you however take the Black Scholes model, you can compute a closed-form real world density. If you estimate $f_{S_T}^\mathbb{Q}$ from observed option prices, you, of course, have no other choice than computing the integral numerically.
  • Minor assumptions were the absence of arbitrage and the existence of a constant risk-free rate. More problematic is the CRRA assumption. The more complicated (realistic) your SDF is, the more complicated your real-world density becomes.
  • Bakshi, Kapadia and Madan (2003) give an example how the power utility function is used to estimate real-world densities. Taylor's book ``Asset Price Dynamics, Volatility and Prediction'' includes a chapter on estimating the risk-neutral density and transferring it to a real-world density.
  • 4
    $\begingroup$ perfect answer. thank you. $\endgroup$
    – roz
    Commented Jan 14, 2020 at 18:38

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