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.

  • 3
    $\begingroup$ I think that gamma is a model dependent concept. I once went to a talk by Musiela in which he discussed sufficient condition to make sure that it was positive. So I think the answer to your question is no. It is true for BS however. $\endgroup$
    – Mark Joshi
    Commented Jan 5, 2015 at 20:25
  • $\begingroup$ @MarkJoshi Darn, thanks though :) I tried googling, but could not find his results. If you stumble on them I would love to know. $\endgroup$
    – htd
    Commented Jan 6, 2015 at 7:38
  • 1
    $\begingroup$ You can look at ElKaroui & al "Robustness of the Black Scholes formula". They give show that calls have positive Gamma in the local volatility model but that it can fail to be true in stoch vol. $\endgroup$
    – AFK
    Commented Jan 8, 2015 at 0:54

1 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).

  • $\begingroup$ How to assert the existence of measure $P$? $\endgroup$
    – Idonknow
    Commented Sep 21, 2019 at 13:08

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