# 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.

• what is $t_d$? it is not clear from your description. – Gordon Jul 14 '15 at 13:52
• @Gordon: was a typo – Ilya Jul 14 '15 at 13:53
• You still need to specify the probability of the jumps, for example, 50% for each possible jump. – Gordon Jul 14 '15 at 14:02
• @Gordon: why do I need to do that? My point is that I have some market prices, hence I can imply distribution from the market prices. I don't know what the distribution would be in advance – Ilya Jul 14 '15 at 14:06

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.

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...