Risk-Neutral Option Densities and Their Limits for VaR

# on "recovering probability distributions from option prices" - how to subtract influence of stochastic volatility?


# on "recovering probability distributions from option prices" - how to subtract influence of stochastic volatility?












This is based on a 1995 paper by Rubinstein/Jackwerth by the above title where the authors produces a distribution of stock prices inferred from option prices. But their approach only produces a joint distribution of stock prices and whatever other contributing factors, most prominent being volatility.

My question is: is there another way to make out this distribution which also incorporates the expected changes in volatility based on say vix futures? A reference to paper or code will be helpful. Is the difference not sufficient to agonize over? I am interested in computing VaR, that too at time horizons other than the option expiry.

Thanks

## Answer by Brian B (score 4)

https://quant.stackexchange.com/a/503

You cannot derive the probability distribution you require, because for VaR you need a real-world probability distribution. From the options prices, it is only possible to obtain a risk-neutral distribution.

Now, if you are willing to assume some kind of parametric relationship between the risk-neutral and real-world distributions, then you might find the options prices useful. The resulting mathematics for a stochastic volatility model is somewhat tricky, however. You can find most of it in Jim Gatheral's books. A sloppy treatment would just take the risk-neutral distribution and shift its mean.

Obtaining the approximate risk-neutral distribution is fairly simple. Let p(S) be the time-T risk-neutral probability density. Then we see that (TeX notation alert) \begin{equation} C := Call(T) = B(0,T) \int_0^\infty Max(0,S-K) p(S) dS \end{equation} \begin{equation} \frac{dC}{dK} = B(0,T) \int_0^\infty 1[S>=K] (-1) p(S) dS \qquad\text{[differentiate under integral] } \end{equation} \begin{equation} \frac{dC}{dK} = B(0,T) \int_K^\infty (-1) p(S) dS \end{equation} \begin{equation} \frac{d^2C}{dK^2} = B(0,T) p(K) \qquad \text{ [Fundamental thm of calculus]} \end{equation} Alternatively, you could say that p(S) is the density, and is the derivative of the cumulative distribution function P(S), and write

\begin{equation} C := Call(T) = B(0,T) \int_0^\infty Max(0,S-K) p(S) dS \end{equation} \begin{equation} \frac{dC}{dK} = B(0,T) \int_0^\infty 1[S>=K] (-1) p(S) dS \qquad\text{[differentiate under integral] } \end{equation} \begin{equation} \frac{dC}{dK} = B(0,T) \int_K^\infty (-1) p(S) dS \end{equation} \begin{equation} \frac{dC}{dK} = B(0,T) (-1) ( P(\infty) - P(K)) \end{equation} \begin{equation} \frac{d^2C}{dK^2} = B(0,T) p(K) \end{equation}

Either way you end up finding the density

\begin{equation} p(x) = \frac{1}{B(0,T)} \frac{d^2C(x)}{dx^2} \end{equation} where $x$ is the strike. So an approximate density comes from using the actual option prices available to you. You can spline interpolate, or if you have a regular grid of strikes spaced by dK you can make a histogram of values \begin{equation} \frac{ C(K+dK) - 2C(K) +C(K-dK) }{ dK^2} \end{equation} and divide by the discount factor to find your risk-neutral distribution.

## Answer by quant_dev (score 2)

https://quant.stackexchange.com/a/416

It would help if you made your question more clear (or included a link to a copy of the paper). If you know call (or put) option prices for a single stock across all strikes $K$ for common expiry $T$, you can then derive the distribution of the stock price at time $T$ in the $T$-forward measure (in which the numeraire is the zero-coupon bond with maturity $T$, and the price of the put option is $V_K(t) = D(T) E[(K-S(T))^+]$, $D(T)$ being the discount factor to expiry $T$). You can do this by differentiating $V_K(T)$ twice over $K$ and dividing by $D(T)$. This will give you simply the distribution of the stock price in the $T$-forward measure, nothing more and nothing else. In other measures (in particular in the real-world measure) this distribution will be different.