Inferring Risk-Neutral Jump Probabilities from Option Prices
Summary
The document examines a stock model with continuous Black–Scholes movement and a jump at a specified time, restricted initially to two possible proportional jump sizes. It asks whether option prices can reveal the probabilities of those jump outcomes and whether those probabilities represent investors’ real-world beliefs. The responses point out that the proposed return expression omits the usual drift adjustment associated with log returns and that option prices alone do not identify physical probabilities.
A suggested calibration approach treats jump sizes and volatility as unknown, fits the model to option prices, and applies the martingale condition to obtain risk-neutral jump probabilities. These probabilities are pricing weights, not forecasts of actual event likelihoods; they can reflect the market price of downside insurance and may exceed physical downside probabilities. The two-outcome jump model is explicitly a simplifying assumption, and the document gives no empirical calibration or estimate of the gap between risk-neutral and real-world probabilities.
Key ideas
- Option prices can be used to calibrate a specified jump model, including its jump sizes and risk-neutral probabilities.
- Risk-neutral jump probabilities satisfy the martingale pricing condition and are not physical event probabilities.
- The proposed two-outcome jump structure is a simplifying modeling assumption.
- The return expression in the question omits a variance adjustment needed for log returns under geometric Brownian motion.
- Downside insurance pricing can make risk-neutral downside probabilities larger than real-world probabilities.
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# Which distribution do I get?
# Which distribution do I get?
Let's assume the stock moves according to a classic Black-Scholes model, and makes a proportional jump with an unknown proportion. Say, it is either +1% or -3% of the stock value, and we know for sure that no other outcomes are possible. After that we assume the stock to move with the same volatility. Interest rates are zero. Hence, the total return on the stock $\xi$ until expiry $T$ becomes $$ \xi = \eta_1\sigma\sqrt{t_j} + j + \eta_2\sigma\sqrt{T-t_j} $$ where $t_j$ is the time of the jump, $\eta_i$ are iid standard normal random variables, and $j \in \{0.01, -0.03\}$. If I have market prices for options for the expiry $T$, I can use them to find distribution of $j$. My question is what is the meaning of this distribution, that is - what can I use it for? In particular, is that really what market things are the real probabilities for the jump? My guess is no, as we'll get the risk-neutral probability of the jump, hence its mean would be zero, even though people may be sure it is 99% possible.
## Answer by AFK (score 1)
https://quant.stackexchange.com/a/18867
First your equation for returns is false. Forgetting about the jump, it does not reduce to Gbm returns. The variance term from Ito's formula is missing. In the case of a jump, a similar term should appear.
Secondly, the distribution is obviously not "what market things are the real probabilities": you chose to impose specific sizes for the jump, the market did not.
Finally you cannot expect to imply real world probabilities from option prices since the latter do not depend on the drift.
## Answer by q.t.f. (score 1)
https://quant.stackexchange.com/a/30895
If believe you intend that $t_j$ is a fixed time known in advance, for example the day of a contentious election. And you postulate that the two possible jump sizes are known in advance and agreed uniformly by the market. This is an artificial assumption -- real markets would price in a continuous range of possible jump sizes -- but is convenient mathematically.
Then you are right, the martingale condition requires that the risk-neutral probability of up-jump and down-jump are such that the expected value is zero.
To try to get information from the options market, you should take the jump sizes as initially unknown and calibrate those from the options prices. You could still try the simplistic assumption of a two point distribution -- a fixed up jump size and a fixed down jump size. Optimise a fit to options prices over different choices for volatility and for these jump sizes. Again there is a martingale condition that translates these jump sizes to probability of up-jump and down-jump. But now you are getting those from the market, not ad hoc assumption.
Again the probabilities are risk-neutral not real world. Risk neutral prices tend to charge more for insuring against bad outcomes, so expect the risk-neutral price for the down jump to be an overestimate of the real world probability of that event. How much of an overestimate? Hard to say...Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.