2
$\begingroup$

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.

$\endgroup$
  • 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 Jan 5 '15 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. $\endgroup$ – Henrik Jan 6 '15 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 Jan 8 '15 at 0:54
2
$\begingroup$

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).

| improve this answer | |
$\endgroup$
  • $\begingroup$ How to assert the existence of measure $P$? $\endgroup$ – Idonknow Sep 21 '19 at 13:08

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.