my question is the following one: I don't manage to prove that, in Black-Scholes model, single-signed Gamma options have values that are monotonic in the volatility. I am looking for an exhaustive and general proof, since I know how to prove it for vanilla options.

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Consider any option, vanilla or exotic. In between fixing dates it satisfies the Black & Scholes PDE (for simplicity zero interest rate and dividends) $$ \frac{1}{2} \sigma^2 S^2 \frac{\partial^2 U}{\partial S^2}(S,t)+\frac{\partial U}{\partial t}(S,t)=0 $$ Let ${\cal V}(S,t) = \frac{\partial U}{\partial \sigma}(S,t)$ be the option vega. Differentiating the BS PDE wrt $\sigma$ you get $$ \frac{1}{2} \sigma^2 S^2 \frac{\partial^2 {\cal V}}{\partial S^2}(S,t)+\frac{\partial {\cal V}}{\partial t}(S,t)+\sigma S^2 \frac{\partial^2 U}{\partial S^2}(S,t)=0 $$ So ${\cal V}(S,t)$ also satisfies the BS PDE, with continuous payoff $\sigma S^2 \frac{\partial^2 U}{\partial S^2}(S,t)$. Also ${\cal V}(S,t)$ is continuous wrt time on every fixing date (fixings do not depend on $\sigma$), therefore ${\cal V}(S_0,0)$ is the expectation of the continuous payoff $\sigma S^2 \frac{\partial^2 U}{\partial S^2}(S,t)$, that is, $T$ being the final maturity of the option, $$ {\cal V}(S_0,0) = \mathbb{E}\left[\int_0^T \sigma S_t^2 \frac{\partial^2 U}{\partial S^2}(S_t,t) dt \right] $$ Hence, if the gamma is $> 0$ everywhere, then the vega is $> 0$. It is easy to adapt to the case of non zero rates and dividends.

Added 8 Apr 2021: By differentiating twice the BS PDE wrt to $S$ we see that the dollar gamma $\gamma(S,t) = S^2 \frac{\partial^2 U}{\partial S^2}(S,t)$ also satisfies the BS PDE $$ \frac{1}{2} \sigma^2 S^2 \frac{\partial^2 \gamma}{\partial S^2}(S,t)+\frac{\partial \gamma}{\partial t}(S,t)=0 $$ in between fixing dates. If the option is vanilla so that there are no intermediary fixing dates, this proves that $\gamma(S_t, t)$ is a martingale and we recover the well known formula for vanilla options $$ {\cal V}(S_0,0) = T \sigma \gamma(S_0, 0) $$

  • $\begingroup$ Thank you very much, it was really useful and it solved my problem!!! $\endgroup$ – Eduardo Contreras Apr 7 at 18:50

I'll give a heuristic "proof" for general European claims which will cause mathematicians to feel sick, but which physicists / practitioners would probably be quite happy work with:

Write the Black-Scholes PDE as $$ \frac{\partial F}{\partial\tau}(\tau) = \mathcal{A} F(\tau) $$ with $\tau = T- t$, and the operator $\mathcal A$ is defined as $$ \frac{1}{2}\sigma^2 S^2 \frac{\partial^2 }{\partial S^2} + (r-q) S \frac{\partial }{\partial S} - r $$

The formal solution to the PDE is $$ F(\tau) = e^{\tau \mathcal A} F(0) $$ where $F(0)$ is the payoff of the claim.

We can treat $e^{\tau \mathcal A}$ as an operator that depends on the constant parameters ($\sigma$, $r$, $q$). So let's differentiate both sides of the formal solution of the BS PDE wrt the parameter $\sigma$:

\begin{align} \frac{\partial F}{\partial \sigma} (\tau) &= \left(\frac{\partial e^{\tau \mathcal A}}{\partial \sigma} \right) F(0) \\ &= \tau\sigma S^2 \frac{\partial^2 }{\partial S^2}( e^{\tau \mathcal A} F(0)) \\ &= \tau\sigma S^2 \frac{\partial^2 F }{\partial S^2}(\tau) \end{align}

With a bit more work the above can also be done if the parameters are not constant, but deterministic functions of time.

EDIT: I just saw Antoine's good answer below. My answer should be treated as an intuitive shortcut, Antoine's answer is the more rigorous one and hence the once that should be accepted by the OP.

  • $\begingroup$ It was really useful to have an example! Thanks! :) $\endgroup$ – Eduardo Contreras Apr 7 at 18:51

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