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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 $\sigma$?

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    $\begingroup$ What is the definition of "volatility" if you consider a general Levy process? If you define it as the implied volatility of the (respective) put option: then yes. It is trivial. The price is a monotone function of the implied Black volatlity. $\endgroup$
    – Christian Fries
    Jun 10, 2013 at 9:29
  • $\begingroup$ @ChristianFries Point taken. I meant $\sigma$. I will modify the question. $\endgroup$
    – Lost1
    Jun 10, 2013 at 9:52

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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 $\sigma$.

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  • $\begingroup$ The paper is behind a pay-wall. I assume he treats the jump-diffusion case. And yes - it should be monotonous in $\sigma$ as it is just Brownian motion between jumps. This should carry over to more complex Levy models. $\endgroup$
    – Richi Wa
    Feb 6, 2014 at 15:05
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    $\begingroup$ Paper for those whose institutions don't have a submission. $\endgroup$
    – Yu-Ho Haeppoelae
    Feb 6, 2014 at 16:41
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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.

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    $\begingroup$ The price process is correctly compensated so that $e^{-rt}S_t$ is a martingale. This argument you present above is not correct. In fact if you give me a compound poisson + Brownian motion (using Levy-Ito decomposition) I can prove for you while the higher volatility BM gives higher option price by coupling the process such that the compound Poisson process coincide. This means we essentially see the compound Poisson process as a constant. Then we apply Jensen's inequality. $\endgroup$
    – Lost1
    Jun 10, 2013 at 21:42
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    $\begingroup$ also what said seems to suggest $dX_t = dY_t + \sigma dB_t - \dfrac{1}{2} \sigma^2 dt$ where the $X_t$ is the log price, $Y_t$ is a jump process. I am saying with certainty that this is monotone in $\sigma$. Extra volatility of the Brownian actually means we have compensate it more with a negative jump. (although this intuition is actually wrong, it is due to convexity of $(K-S_t)^+$). I am more interested in the case where $dX_t = \sigma dY_t - \phi(\sigma) dt$, where $\phi$ is the log moment generating function of $Y$. $Y$ itself is a Levy, so $\sigma$ appears before the Y. $\endgroup$
    – Lost1
    Jun 10, 2013 at 21:50
  • $\begingroup$ You're right -- the compensation kills that argument. $\endgroup$
    – Brian B
    Jun 11, 2013 at 12:49
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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 $Y.$

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  • $\begingroup$ I guessed this much, but it is unclear to me how this can be proven in a straightforward manner. $\endgroup$
    – Lost1
    Nov 7, 2015 at 18:07

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