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

Question

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.

Answer 1

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.

Answer 2

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*}