Can somebody prove that:

$$E[S_t^2 \times \Gamma(t,S_t)] = S_0^2 \times \Gamma(0,S_0)$$

where $S_t$ follows a lognormal process as in the Black-Scholes model, and Gamma is the second derivative $\partial^2 C/\partial S^2$ of the option price with respect to S.

I can see it is true using simulation, but I can't prove it. It seems to be true for the Vega as well.


What you have to do is to show that the dollar gamma satisfies the Black-Scholes PDE. Using Feynman-Kac it then follows that the dollar gamma is an expectation of a "payoff", just like the Black-Scholes claim price is an expectation of a payoff. And if something is the expectation of a payoff then it's a martingale.

Actually, you don't need the Black-Scholes assumption. This will work for any model (LV, SV, LSV...).

I'll leave the above for you to carry out. What I'd like to show is a nice little trick using the homogeneity property of the Black-Scholes price formula: Denoting partial derivatives by subscripts, the homogeneity of the BS call price function means that $$ C = SC_S + KC_K $$ Take again the derivative to $S$ of the above equation, and also take the derivative to $K$ of the above equation. That will give you two equations, and after some cancelling will lead you the the following equality: $$ S^2C_{SS} = K^2C_{KK} $$ The left hand side is the dollar gamma. The right hand side is $K^2$ times the discounted probability density. But the discounted probability density is just $$ C_{KK} = e^{-r(T-t)} E_t [ \delta(S_T-K)] $$ where $\delta$ is the Dirac delta-function. Hence the dollar gamma is a martingale.

Note that the homogeneity trick also immediately shows that the dollar delta is a martingale as well since $C_K = - e^{-r(T-t)} E_t [\theta (S_T - K)] $, where $\theta$ is the Heaviside function.

  • $\begingroup$ How did you conclude the dollar gamma is a martingale just from whatever you stated above? I've only seen the proof with the full PDE. This way seems much simpler $\endgroup$ – Slade May 6 '19 at 2:32
  • $\begingroup$ Great answer @ilovevolatility. One quick question : do interest rates have to be zero to make it strictly true ? $\endgroup$ – dm63 May 6 '19 at 2:38
  • $\begingroup$ @dm63 thanks, I have a bad habit of either setting r=q=0 or working under the forward measure. The answer holds for deterministic rates and dividend yields as well, as they are already contained in the equation $C = SC_S + KC_K$ which is the starting point, and I've modified the expectations above in the answer to include discounting. $\endgroup$ – Frido Rolloos May 6 '19 at 5:31
  • 2
    $\begingroup$ @Slade The dollar gamma is a martingale because it's the discounted expectation of a delta function payoff. The nice thing about the above alternative proof, me thinks, is that it shows that not only under Black-Scholes is the dollar gamma a martingale, but under any model that is homogeneous of degree 1 in $S$ and $K$, this includes Heston and lognormal SABR. $\endgroup$ – Frido Rolloos May 6 '19 at 5:33
  • 2
    $\begingroup$ Very nice answer. $\endgroup$ – Quantuple May 6 '19 at 7:33

The conjecture is true when the interest rate is zero. Note that, from this question, under the Black-Scholes model, \begin{align*} \Gamma(t,S_t) &= \frac{N'(d_1(t))}{S_t \sigma \sqrt{T-t}}\\ Vega(t,S_t) &= S_tN'(d_1(t)) \sqrt{T-t}, \end{align*} where \begin{align*} d_1(t) = \frac{\ln \frac{S_t}{K} + \big(r+\frac{1}{2}\sigma^2\big)(T-t)}{\sigma \sqrt{T-t}}. \end{align*} Then, it is easy to see that \begin{align*} Vega(t,S_t) = \sigma\, (T-t)\, S_t^2\, \Gamma(t,S_t). \end{align*} Consequently, \begin{align*} E\big( \sigma (T-t)\,S_t^2\, \Gamma(t,S_t)\big) &= E\big(Vega(t,S_t)\big) \tag{1}\\ &= E\left(\frac{\partial}{\partial \sigma}E\left(e^{-r(T-t)} (S_T-K)^+\big|\mathscr{F}_t\right) \right). \end{align*} However, we are not able to take the partial differential out as this differential only involves the volatility from $t$ to $T$, and, if we take it out, then the volatility from $0$ to $T$ is involved.

We denote by $\sigma_1=\sigma$ the volatility from $0$ to $t$, and $\sigma_2=\sigma$ the volatility from $t$ to $T$. Moreover, let \begin{align*} \hat{\sigma} = \sqrt{\frac{1}{T}\left(\sigma_1^2 t + \sigma_2^2 (T-t)\right)} = \sigma. \end{align*} Then \begin{align*} E\big(Vega(t,S_t)\big) &= E\left(\frac{\partial}{\partial \sigma_2}E\left(e^{-r(T-t)} (S_T-K)^+\big|\mathscr{F}_t\right) \right)\\ &=\frac{\partial}{\partial \sigma_2}E\left(e^{-r(T-t)} (S_T-K)^+\right)\\ &= e^{rt} \frac{\partial}{\partial \sigma_2}E\left(e^{-rT} (S_T-K)^+\right)\\ &= e^{rt} \frac{\partial}{\partial \hat{\sigma}}E\left(e^{-rT} (S_T-K)^+\right) \frac{\partial \hat{\sigma}}{\partial \sigma_2}\\ &=e^{rt} Vega(0,S_0) \frac{T-t}{T}\\ &= e^{rt} \sigma\, T\,S_0^2\, \Gamma(0,S_0) \frac{T-t}{T}\\ &= e^{rt} \sigma\, (T-t)\,S_0^2\, \Gamma(0,S_0). \end{align*} Therefore, from $(1)$, \begin{align*} E\big(S_t^2\, \Gamma(t,S_t)\big) = e^{rt} S_0^2\,\Gamma(0,S_0). \end{align*}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.