# Expectation of Gamma times S$^2$ in Black-Scholes model

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.

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.

• 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 – Slade May 6 at 2:32
• Great answer @ilovevolatility. One quick question : do interest rates have to be zero to make it strictly true ? – dm63 May 6 at 2:38
• @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. – ilovevolatility May 6 at 5:31
• @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. – ilovevolatility May 6 at 5:33
• Very nice answer. – Quantuple May 6 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*}