# Price of Call Option with or without jumps

Suppose two assets in the Black Scholes world have the same volatility, but different drifts and that one has downward jumps at random times. How does this affect the option prices?

I would have thought that downward jumps would decrease the value of the call option because you have more chance of being out of the money (ie below the strike). Apparently, the answer is the reverse.

Does anyone have an explanation for this?

• ill check the wording of the question tomorrow Jul 2, 2020 at 21:12

The call for the stock that can jump downward will be more valuable due to put-call parity. Suppose you have two stocks, both with a price of $100 and the same diffusive volatility. Stock A does not jump, whereas stock B can at some random time jump (for example) to zero. Clearly a put on stock B will be worth more, but the call must therefore also be worth more due to parity: $$\text{Call}(S_{0}, K, T) = \text{Put}(S_{0}, K, T) + S_0 - K e^{-rT}$$ The economic explanation for this is that both stocks have the same price. If the stock that can jump downward is worth the same as the stock that cannot jump, it must have more probability mass on the upside. In the Merton jump model, the stock that can jump to zero has a risk-neutral drift, conditional on no jump, of $$r + \lambda$$, where $$\lambda dt$$ is instantaneous probability of the jump to zero. With this drift, the stock's unconditional drift is $$r$$. The call price in this case is obtained by replacing $$r$$ with $$r+\lambda$$, which results in a higher call price. (Merton discusses this case specifically in his 1976 JFE paper.) • yes but it says that they have the same risk neutral drift, not different? Im not sure how CP parity proves this, how do you know that the$S_0 - Ke^{-rT}$wont be enormous? Jul 5, 2020 at 6:18 • @Permian: When you say "it says", what is "it"? Put-call parity is a distribution-free, no-arbitrage property of European option prices, a consequence of comparing payoffs at expiration. It has nothing to do with Black-Scholes or jumps. Fix$S_0$,$K$,$T$, and$r$. Given those parameters, parity says that if one option is more expensive, the other must be also. Jul 5, 2020 at 12:00 • when you say its clear that the put wll be more expensive is clear, i agree. However, the same argument (not using CP parity) would imply that the call would be less valuable, but its not Jul 5, 2020 at 13:28 • The call is more expensive because conditional on no jump, the call will end up deeper in the money --- the risk-neutral drift is greater than r to offset the eventual loss due to the jump. This extra drift to offset the effect of jumps is often called the jump compensator. Jul 5, 2020 at 13:37 • Just to get my intuition right, in order for the stock with down jumps to be worth the same it needs to have an (conditional on no jump) greater risk neutral drift, with the downwards jumps "countering" this effect so that the unconditional drift is the same. So essentially this means that the stock has a greater volatility (well, not diffusion but a propensity to move up and down by larger amounts) which naturally leads to higher prices for both puts and calls. Is that a correct way to look at it? Jul 12, 2020 at 18:45 There's a lot left unspecified in this question, since it is stated without precision, but the effective idea of the answer given here is that those jumps introduce extra variation into the forward distribution of the underlying. And such variation is the bread-and-butter of option value. That said, the ambiguity in the question leaves room for other interpretations. In particular if you as a market-maker sold an at-the-money call for$100, and them immediately after your sale everyone found out that the underlying had a 50-50 probability of jumping down by half tomorrow, you would be very happy, because the underlying would drop in value by 25% or so and the option would go far out of the money.

So, what the person who said the value increases meant was, given two ATM options on separate underlyings with the same continuous volatility, and where the second underlying also had some downward jumps, the latter option will have higher fair value.

Mathematically, this ends up being associated with the second underlying having higher risky drift.

• 1) How do you explain by explanation as being wrong? 2) I dont have understand how you can have higher risk neutral drift, surely the risk neutral drift is the risk free rate? Jul 3, 2020 at 6:09
• @Permian is correct...I should have written "risky" drift. Fixed. Jul 4, 2020 at 13:05
• I dont understand the difference in higher risky drift, how does this connect to the greater chance of jumping down and then out of the money increased probability Jul 4, 2020 at 14:37
• How can the price go up if the chance of being out of the money increases? are we assume some delta hedge? Jul 4, 2020 at 14:40
• Ive slightly reworded my question, could you check your answer still holds Jul 13, 2020 at 6:06

You can check out those discussions in Merton paper when introducing jumps "Option pricing when underlying stock returns are discontinuous". In the very last part he discusses the influence of considering jumps compared to the usual Black Sholes model. From what i remember it'sall about considering your option is ATM or not , that will usually make the BS model call prices higher.

• can you explain why my line of thinking is wrong? Jul 4, 2020 at 14:33

I'm not too sure if I interpret the question correctly, but I am inclined to say that as the call is long gamma, 'jumps' (second order moves) would always result in higher value in the delta hedged portfolio, and therefore should be built into the price of the option. So the call should be more expensive if it has jumps.