Gamma is always positive on both put and call

Question

I recently met the claim that for standard put and calls the gamma of the options are always positive. Is this a general result?

I am hoping not to assume any model, especially not Black-Scholes.

Answer 1

I'll use a European call option as an example, I think you can easily generalize it for a put option.

Given underlying $S(t) = S_t$, maturity $T$, strike $K$ and risk-free rate $r$, the price of a call option as time $t$ under the rik-neutral measure $Q$ is

$$ \begin{align} C(t, S_t) & = \mathbb{E}^Q \left[ e^{-r(T-t)} \max (S_T - K, 0) \right] \\ & = \mathbb{E}^Q \left[ e^{-r(T-t)} (S_T - K) \cdot \mathbb{1}_{S_T \geq K} \right] \\ & = \mathbb{E}^Q \left[ e^{-r(T-t)} S_T \cdot \mathbb{1}_{S_T \geq K} \right] - \mathbb{E}^Q \left[ K e^{-r(T-t)} \cdot \mathbb{1}_{S_T \geq K} \right] \\ & = e^{-r(T-t)} \mathbb{E}^Q \left[ S_T \cdot \mathbb{1}_{S_T \geq K} \right] - K e^{-r(T-t)} \mathbb{E}^Q \left[ \mathbb{1}_{S_T \geq K} \right] \\ \end{align} $$

where $\mathbb{1}_{S_T \geq K}$ is a function values $1$ when $S_T \geq K$ and $0$ otherwise. For the first expectation, we can change the probability measure to make it more manageable. Call $P$ the new measure; the Radon-Nikodym derivative between the $P$ and $Q$ is

$$dQ = \frac{S_t}{S_T} e^{r(T-t)} dP$$

Therefore you get

$$ \begin{align} C(t, S_t) & = e^{-r(T-t)} \mathbb{E}^Q \left[ S_T \cdot \mathbb{1}_{S_T \geq K} \right] - K e^{-r(T-t)} \mathbb{E}^Q \left[ \mathbb{1}_{S_T \geq K} \right] \\ & = e^{-r(T-t)} \mathbb{E}^P \left[ S_T \cdot \mathbb{1}_{S_T \geq K} \cdot \frac{S_t}{S_T} e^{r(T-t)} \right] - K e^{-r(T-t)} \mathbb{E}^Q \left[ \mathbb{1}_{S_T \geq K} \right] \\ & = e^{-r(T-t)} \mathbb{E}^P \left[ \mathbb{1}_{S_T \geq K} S_t e^{r(T-t)} \right] - K e^{-r(T-t)} \mathbb{E}^Q \left[ \mathbb{1}_{S_T \geq K} \right] \\ & = S_t \mathbb{E}^P \left[ \mathbb{1}_{S_T \geq K} \right] - K e^{-r(T-t)} \mathbb{E}^Q \left[ \mathbb{1}_{S_T \geq K} \right] \\ \end{align} $$

Expanding the expectations as integrals you get:

$$ \begin{align} C(t, S_t) & = S_t \int_K^{\infty} f^P(S_T) dS_T - K e^{-r(T-t)} \int_K^{\infty} f^Q(S_T) dS_T \\ & = S_t P_1 - K e^{-r(T-t)} P_2 \end{align} $$

where $P_1, P_2$ highlight that the integrals are probabilities.

Now the Greeks:

$$ \begin{align} \Delta & = \frac{\partial C}{\partial S_t} = \int_K^{\infty} f^P(S_T) dS_T = P_1 \\ \Gamma & = \frac{\partial^2 C}{\partial S_t} = \frac{\partial \Delta}{\partial S_t} = f^P(S_t) \frac{\partial f^P(S_t)}{\partial S_t} \\ \end{align} $$

The derivative in $\Gamma$ is the key. I don't think you can prove $\Gamma$ to be positive for any probability density (i.e. any model).