When Does Greater Lévy-Driven Uncertainty Raise Put Values?
Summary
The discussion asks whether a European put price remains monotone as the scale parameter of a general Lévy process increases. Under Black–Scholes, the post notes that monotonicity follows from the Gaussian model and can also be understood through conditional Jensen’s inequality. It cautions that the same argument does not automatically extend when Brownian motion is replaced by a general Lévy process.
The responses offer competing intuition rather than a complete proof. One suggests consulting Merton’s jump-diffusion option formula, whose series terms may each be monotone in the scale parameter. Another sketches a counterexample in which a jump component drives the asset toward zero and added Brownian risk can increase the chance of survival, reducing put value. A further response gives a sufficient convex-order idea: if one positive payoff factor can be represented as another times an independent, positive, mean-one factor, convex option payoffs increase in expectation. These comments do not settle monotonicity for every Lévy specification; the effect depends on how the process changes with the parameter.
Key ideas
- Put-price monotonicity in the Black–Scholes model is linked to Gaussian structure and Jensen’s inequality.
- The Gaussian argument does not directly establish the result for a general Lévy process.
- The responses propose both a possible series-based argument and a jump-driven counterexample intuition.
- An independent positive mean-one multiplier can increase the value of a convex payoff through Jensen’s inequality.
- The discussion does not prove a universal monotonicity result across Lévy models.
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# Answer by Yu-Ho Haeppoelae (score 3)
# Is the price of European put option monotone in volatility if we replace BM in Black-Scholes with a general Levy process?
Under the Black-Scholes model, we have the European put option is $\mathbb{E} [e^{-rt}(K-S_t)]$, where we take $\log(S_t)=X_t$ and $dX_t= \sigma dW_t - \dfrac{1}{2}\sigma^2 dt + rdt$. Here the option price is monotone in $\sigma$.
To show this we can appeal to Black-Scholes formula. though there is an easier, which directly appeal to the Gaussianity of $\log (S_t)$, the fact that a Gaussian random variable can be written as a sum of two Gaussian random variables, and uses conditional Jensen inequality. This trick would even work even if we work with stochastic volatility, as long as the volatility is driven by a process independent of the Brownian Motion.
However, this trick fails instantly we replace $W_t$ with another Levy process and replace $\dfrac{1}{2}\sigma^2$ with the log moment generating function of the Levy process.
My question is, suppose we replace the $W_t$ by a general Levy process. would this remain true? does there exists any literature on this subject. The gut-feeling is yes, but I have failed to prove this myself. Does anyone know any literature written on this subject?
EDIT: As Christian pointed out, volatility is not actually an appropriate word to use here. What I really mean is that, is the price monotone in $\sigma$?
## Answer by Yu-Ho Haeppoelae (score 3)
https://quant.stackexchange.com/a/10168
Have you checked the paper by Merton from the seventies? He gives the price of an option as an infinite series (eq. 16), whose every term seems monotonous in $\sigma$.
## Answer by Brian B (score 2)
https://quant.stackexchange.com/a/8199
I'm pretty certain it is not monotone -- consider the following argument:
A Levy process is decomposable into a sum of brownian motions, jump processes and poisson processes. Consider a case where the poisson component is driving price to zero $\left(\log(S) \rightarrow -\infty\right)$ with near certainty, making option value close to intrinsic. Adding in a highly volatile brownian motion will start to put more probability on survival, decreasing the option value.
They key here is that, viewing a Levy process as the sum of two complex processes, we can architect the interaction between them to obtain pathological behavior.
## Answer by Mark Joshi (score 1)
https://quant.stackexchange.com/a/21561
I actually discuss this question at length in chapter 1 of More Mathematical Finance.
The essential point is that if you can write $$ X=YZ $$ with $Y,Z$ independent $E(Z)=1$ and $Z>0$ then $X$ is more uncertain than $Y.$ It then follows from Jensen's inequality that the price of an option on $X$ that has a convex pay-off will be at least as high as the same option on $Y.$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.