Why Option Gamma Is Not Guaranteed Positive in Every Model
Summary
The document asks whether vanilla call and put options must have positive gamma without assuming a particular pricing model. Its proposed call-price derivation rewrites the payoff using expectations under risk-neutral and stock-numeraire measures, then expresses the price through probabilities above the strike. It uses this setup to argue that gamma’s sign depends on how the relevant probability distribution changes with the underlying price, so positivity does not follow from an arbitrary density alone.
The discussion is conceptual rather than a complete proof. The displayed gamma differentiation is incorrect or incomplete: differentiating a probability tail with respect to spot requires the spot dependence of the distribution, and the formula shown does not establish a general sign. Thus the useful takeaway is that model assumptions and regularity conditions matter; the document itself does not settle exactly which conditions guarantee positive gamma. Its put generalization is asserted but not worked through.
Key ideas
- The question concerns whether vanilla option gamma is positive independently of a pricing model.
- The proposed call valuation represents its price using probabilities under two measures.
- The sign of gamma depends on how the distribution of terminal prices varies with current spot.
- The displayed differentiation is incomplete and does not prove a general result.
- A rigorous conclusion requires explicit assumptions about the pricing model and option-price regularity.
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Full text
# Gamma is always positive on both put and call
# Gamma is always positive on both put and call
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 by roym00 (score 3, accepted)
https://quant.stackexchange.com/a/16095
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).Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.