Gamma and Convexity of Black-Scholes Prices

Gamma Measures How Delta Changes

Delta is the local change in an option’s price when the underlying spot price changes. But delta itself changes as spot moves. Gamma measures that second response: it is the derivative of delta with respect to spot. This makes gamma relevant both to the shape of an option price as a function of spot and to how frequently a delta hedge must be rebalanced.

Use the Black–Scholes market with constant interest rate $r$, constant volatility $\sigma>0$, no dividends, and a European option with strike $K>0$ and expiry $T$. On a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$, let $B_t$ be Brownian motion. Under the risk-neutral measure $\mathbb{Q}$, the stock has drift $r$ and diffusion coefficient $\sigma S_t$, so its dynamics are $dS_t=rS_t\,dt+\sigma S_t\,dB_t^{\mathbb{Q}}$, where $B_t^{\mathbb{Q}}$ is Brownian motion under $\mathbb{Q}$. The pricing formulas below use $\mathbb{Q}$, not the physical measure $\mathbb{P}$. Write $\tau=T-t>0$ for time to expiry and define

d_1=\frac{\ln(S/K)+(r+\tfrac12\sigma^2)\tau}{\sigma\sqrt{\tau}}, \qquad d_2=d_1-\sigma\sqrt{\tau}, \qquad \phi(x)=\frac{1}{\sqrt{2\pi}}e^{-x^2/2}.

Here $S>0$ is the current spot, $\phi$ is the standard normal density, and $N(x)=\int_{-\infty}^x\phi(u)\,du$ is the standard normal distribution function. The Black–Scholes prices and their deltas are

C=S N(d_1)-Ke^{-r\tau}N(d_2), \qquad P=Ke^{-r\tau}N(-d_2)-S N(-d_1), \qquad \Delta_C=N(d_1),\quad \Delta_P=N(d_1)-1.

Deriving the Call and Put Gamma

Theorem: For $S>0$, $K>0$, $\sigma>0$, and $\tau>0$, the Black–Scholes call and put have the same gamma: $$ \Gamma_C=\frac{\partial\Delta_C}{\partial S} =\frac{\partial^2 C}{\partial S^2} =\frac{\phi(d_1)}{S\sigma\sqrt{\tau}} =\frac{\partial\Delta_P}{\partial S} =\frac{\partial^2 P}{\partial S^2} =\Gamma_P>0. $$

Proof. Holding $K,r,\sigma,\tau$ fixed, differentiation gives $\frac{\partial d_1}{\partial S}=\frac{\partial d_2}{\partial S}=\frac{1}{S\sigma\sqrt{\tau}}$. Since $N'(x)=\phi(x)$, differentiating the deltas yields $\frac{\partial\Delta_C}{\partial S}=\phi(d_1)/(S\sigma\sqrt{\tau})$ and $\frac{\partial\Delta_P}{\partial S}=\phi(d_1)/(S\sigma\sqrt{\tau})$. This proves the claimed gamma formula and its strict positivity because $\phi(x)>0$ for every real $x$.

To verify that these are also the second derivatives of the prices, differentiate the price formulas. The chain rule gives $\frac{\partial C}{\partial S}=N(d_1)+\phi(d_1)S_S^{\!*}-Ke^{-r\tau}\phi(d_2)d_2{}_S$, where $S_S^{\!*}$ here means $S\,\partial d_1/\partial S$; equivalently, the two additional terms are $S\phi(d_1)\frac{\partial d_1}{\partial S}-Ke^{-r\tau}\phi(d_2)\frac{\partial d_2}{\partial S}$. The identity $S\phi(d_1)=Ke^{-r\tau}\phi(d_2)$ follows by taking logarithms of their ratio and using $d_1-d_2=\sigma\sqrt{\tau}$. Since the two derivatives of $d_i$ are equal, the additional terms cancel, leaving $\partial C/\partial S=N(d_1)$. Similarly, differentiating $P$ gives $\partial P/\partial S=N(d_1)-1$; the density terms cancel by the same identity. Differentiating these deltas once more gives the stated second derivatives. $\square$

Worked Example: Computing Call and Put Gamma

Take $S=100$, $K=100$, $r=0.02$, $\sigma=0.2$, and $\tau=1$. Find both deltas and gammas.

First, $\ln(S/K)=\ln(1)=0$, and $(r+\tfrac12\sigma^2)\tau=(0.02+\tfrac12(0.2)^2)1=0.02+0.02=0.04$. Thus $d_1=0.04/(0.2\cdot1)=0.2$, and $d_2=0.2-0.2=0$. Therefore $\Delta_C=N(0.2)$ and $\Delta_P=N(0.2)-1$, approximately $0.5793$ and $-0.4207$, respectively.

The common gamma is $\phi(0.2)/(100\cdot0.2\cdot1)=e^{-0.2^2/2}/(\sqrt{2\pi}\cdot20)=e^{-0.02}/(20\sqrt{2\pi})$, approximately $0.01955$. Thus a small one-unit spot change produces a first-order delta change of about $0.01955$ for either option.

Check: The difference between the two deltas is $N(0.2)-(N(0.2)-1)=1$, independent of spot. Differentiating that constant difference gives zero, consistent with equal call and put gammas.

Positive Gamma and Convexity

A twice differentiable function with nonnegative second derivative is convex. The theorem therefore implies that, at every positive time to expiry, both Black–Scholes prices are convex functions of spot. The relationship is direct: gamma is the curvature of the price-versus-spot graph.

Proposition: For $\tau>0$, both the Black–Scholes call price $C(S)$ and put price $P(S)$ are strictly convex in $S>0$. At expiry, the call payoff $(S-K)^+$ and put payoff $(K-S)^+$ are convex, but their ordinary second derivatives are not finite functions at $S=K$.

Proof. For $\tau>0$, the theorem gives $C''(S)=P''(S)=\phi(d_1)/(S\sigma\sqrt{\tau})>0$ at every $S>0$. A differentiable function with positive second derivative has a strictly increasing first derivative, which is the calculus criterion for strict convexity. At expiry, each payoff is the maximum of two affine functions: $(S-K)^+=\max(S-K,0)$, and $(K-S)^+=\max(K-S,0)$. The maximum of affine functions is convex: for either affine function $L$ and $0\le\lambda\le1$, $L(\lambda x+(1-\lambda)y)=\lambda L(x)+(1-\lambda)L(y)\le\lambda f(x)+(1-\lambda)f(y)$; taking the maximum over the two choices of $L$ proves the convexity inequality. Each payoff has a change in slope at $S=K$, so its ordinary second derivative is not defined there. $\square$

This distinction matters near expiry. The payoff has a kink at the strike, while a positive-time Black–Scholes price is smooth. As expiry approaches, the price’s curvature becomes concentrated around the strike rather than remaining a smooth finite curvature of the payoff at that point. Gamma is therefore especially important for options whose spot is near the strike and whose remaining time is short.

Worked Example: Which Strike Has the Greatest Gamma?

Fix $S=100$, $r=0.02$, $\sigma=0.2$, and $\tau=1$, and compare strikes that give $d_1=1$, $d_1=-1$, or $d_1=0$. Solving the definition of $d_1$ for $K$ gives $K=S\exp((r+\tfrac12\sigma^2)\tau-d_1\sigma\sqrt{\tau})$. The common additive term is $(0.02+0.02)1=0.04$.

For $d_1=1$, $K=100e^{0.04-0.2}=100e^{-0.16}$, approximately $85.21$. For $d_1=-1$, $K=100e^{0.04+0.2}=100e^{0.24}$, approximately $127.1$. For $d_1=0$, $K=100e^{0.04}$, approximately $104.1$.

The denominator $S\sigma\sqrt{\tau}$ is $100\cdot0.2\cdot1=20$ in all three cases. Since $\phi(1)=\phi(-1)=e^{-1/2}/\sqrt{2\pi}$, the first two strikes have the same gamma, $e^{-1/2}/(20\sqrt{2\pi})$, approximately $0.01210$. The $d_1=0$ strike has gamma $1/(20\sqrt{2\pi})$, approximately $0.01995$.

Check: The normal density satisfies $\phi'(x)=-x\phi(x)$, so it increases for $x<0$ and decreases for $x>0$; its maximum is at zero. With the denominator fixed, gamma is maximized at $d_1=0$. This condition identifies the maximum-gamma strike; it need not be the spot-level strike $K=S$.

Comparing Gamma Across Maturities

Strike comparisons at fixed maturity have a simple structure because the denominator of gamma is fixed and only $\phi(d_1)$ changes. Maturity comparisons are more delicate: changing $\tau$ changes both the denominator $S\sigma\sqrt{\tau}$ and $d_1$. Consequently, there is no maturity ranking based on the denominator alone. It is also important to say what is held fixed. A fixed strike comparison and a comparison that keeps $d_1$ fixed follow different paths through the formula.

Worked Example: A Fixed-Strike Maturity Comparison

Keep $S=K=100$, $r=0.02$, and $\sigma=0.2$, and compare $\tau=0.25,1,4$. Here $\ln(S/K)=0$ and $r+\tfrac12\sigma^2=0.04$, so $d_1=0.04\sqrt{\tau}/0.2=0.2\sqrt{\tau}$.

For $\tau=0.25$, $d_1=0.2\sqrt{0.25}=0.1$, and the denominator is $100\cdot0.2\cdot\sqrt{0.25}=10$. Thus gamma is $\phi(0.1)/10=e^{-0.005}/(10\sqrt{2\pi})$, approximately $0.03970$.

For $\tau=1$, $d_1=0.2\sqrt1=0.2$, and the denominator is $100\cdot0.2\cdot1=20$. Gamma is $\phi(0.2)/20=e^{-0.02}/(20\sqrt{2\pi})$, approximately $0.01955$.

For $\tau=4$, $d_1=0.2\sqrt4=0.4$, and the denominator is $100\cdot0.2\cdot2=40$. Gamma is $\phi(0.4)/40=e^{-0.08}/(40\sqrt{2\pi})$, approximately $0.009207$. In this fixed-strike example gamma decreases across the three listed maturities.

Check: If instead $d_1=0$ is held fixed while maturity changes, gamma is $1/(S\sigma\sqrt{2\pi\tau})$. With $S=100$ and $\sigma=0.2$, it is $1/(10\sqrt{2\pi})$, $1/(20\sqrt{2\pi})$, and $1/(40\sqrt{2\pi})$ at maturities $0.25,1,4$, approximately $0.03989$, $0.01995$, and $0.009974$. This comparison uses a different strike for each maturity; it must not be confused with holding the strike fixed.

Gamma and Hedge Rebalancing

A delta hedge uses the option’s delta as the number of shares held per option, with the sign determined by whether the option is held long or short. If spot changes by a small amount $h$, differentiability gives $\Delta(S+h)=\Delta(S)+\Gamma(S)h+o(h)$ as $h\to0$. Thus gamma estimates the change in the required share position per unit spot move. A large positive gamma means a long call or put’s delta rises as spot rises; for a short option position, the corresponding hedge adjustment has the opposite sign.

The local statement is not an exact finite-move rule. For a finite spot change, gamma generally changes along the path from $S$ to $S+h$, so the initial gamma times $h$ is only a first-order approximation. The same curvature also explains the second-order price approximation $V(S+h)\approx V(S)+\Delta(S)h+\tfrac12\Gamma(S)h^2$, where $V$ is the call or put price. These are local approximations, not substitutes for repricing after a substantial move.

Common Pitfall: Comparing Only the Denominator

A tempting but incorrect comparison is to conclude that longer maturity always means lower gamma because the denominator contains $\sqrt{\tau}$. This ignores the numerator $\phi(d_1)$, which also changes with maturity. For instance, as $\tau$ changes, $d_1$ may move toward zero, raising $\phi(d_1)$, or away from zero, lowering it. The corrected comparison computes both $d_1$ and $S\sigma\sqrt{\tau}$ for each maturity, as in the worked example.

A second common error is to assign a finite ordinary gamma to the payoff at expiry. Away from $S=K$, each payoff is linear and has second derivative zero; at $S=K$, the slope jumps, so the ordinary second derivative does not exist. Positive-time Black–Scholes gamma is well-defined under the theorem’s assumptions, but that does not make the expiry payoff twice differentiable at its kink.

Check Your Understanding

Use the gamma formula and its stated assumptions to answer the following.

Why do a call and a put with the same $S,K,r,\sigma,\tau$ have equal gamma even though their deltas differ?
For fixed $S,\sigma,\tau$, which value of $d_1$ maximizes gamma, and why?
What changes in the gamma calculation if $d_1$ is held fixed while maturity varies, rather than holding strike fixed?
Why is positive gamma equivalent here to strict convexity of the positive-time option price as a function of spot?
Why is the first-order hedge adjustment $\Gamma h$ only an approximation for a finite spot move?