Differentiating the Black–Scholes PDE¶
Greeks can be characterized by differentiating the Black–Scholes PDE satisfied by the option price.
Black–Scholes PDE¶
For a European option \(V(t,S)\),
Delta equation¶
Let \(\Delta=V_S\). Differentiate the PDE with respect to \(S\):
Computing each term: - \(\frac{\partial}{\partial S}(S^2 V_{SS}) = 2S V_{SS} + S^2 V_{SSS}\) - \(\frac{\partial}{\partial S}(S V_S) = V_S + S V_{SS}\)
Substituting \(\Delta = V_S\), \(\Gamma = V_{SS}\), and \(\Delta_S = \Gamma\):
Simplifying:
Or in operator form:
where \(\mathcal{L}\) is the Black–Scholes operator. The terminal condition is \(\Delta(T,S) = \Phi'(S)\) (which may be distributional for kinked payoffs).
Gamma equation¶
Let \(\Gamma = V_{SS}\). Differentiate the delta PDE with respect to \(S\):
Simplifying:
The terminal condition is \(\Gamma(T,S) = \Phi''(S)\). For a vanilla call, \(\Phi''(S) = \delta(S-K)\) (Dirac delta), explaining why gamma concentrates near the strike as \(t \to T\).
Practical use of PDE identities¶
These PDEs are useful for: - Boundary behavior analysis: determining far-field limits \(S \to 0\) and \(S \to \infty\) - Maximum principle arguments: establishing bounds on Greeks - Regularity theory: understanding smoothness away from maturity - Numerical schemes: designing stable finite-difference methods for Greeks
What to remember¶
- Differentiating the pricing PDE yields PDEs for Greeks.
- Delta satisfies a modified Black–Scholes PDE with an extra drift term.
- Gamma satisfies its own PDE with terminal condition given by \(\Phi''\).
- For stable computation, combine PDE identities with transformations or expectation formulas.
Exercises¶
Exercise 1. Starting from the Black--Scholes PDE \(\partial_t V + \frac{1}{2}\sigma^2 S^2 V_{SS} + rSV_S - rV = 0\), differentiate with respect to \(S\) and verify that the delta PDE contains the extra terms \(\sigma^2 S \Gamma\) compared to the original PDE.
Solution to Exercise 1
Starting from the Black--Scholes PDE:
Differentiate every term with respect to \(S\). Since partial derivatives commute, \(\frac{\partial}{\partial S}\frac{\partial V}{\partial t} = \frac{\partial}{\partial t}V_S = \frac{\partial \Delta}{\partial t}\).
For the diffusion term:
For the drift term:
For the discount term: \(\frac{\partial}{\partial S}(-rV) = -rV_S = -r\Delta\).
Combining everything:
The \(r\Delta\) terms cancel, leaving:
Comparing with the original PDE \(\frac{\partial V}{\partial t} + \frac{1}{2}\sigma^2 S^2 V_{SS} + rSV_S - rV = 0\), we see that \(\Delta\) satisfies the same operator \(\mathcal{L}\) plus an extra term \(\sigma^2 S\Gamma\). Rewriting:
The extra term \(\sigma^2 S\Gamma\) arises from differentiating the product \(S^2 V_{SS}\), which introduces terms that do not appear in the original PDE applied to \(\Delta\) alone.
Exercise 2. The gamma PDE has terminal condition \(\Gamma(T,S) = \Phi''(S)\). For a vanilla call \(\Phi(S) = (S-K)^+\), explain why \(\Phi''(S) = \delta(S-K)\) is distributional. How does the parabolic PDE "regularize" this Dirac delta into a smooth function for \(t < T\)?
Solution to Exercise 2
For the vanilla call payoff \(\Phi(S) = (S-K)^+\), we compute derivatives in the distributional sense. The first derivative is:
which is the Heaviside step function. This is a bounded measurable function but has a jump discontinuity at \(S = K\). Differentiating again:
This is the Dirac delta, a distribution rather than a classical function, because \(\Phi'\) has a jump of size 1 at \(S = K\). For any test function \(\varphi\):
which confirms the distributional identity.
The parabolic PDE regularizes this as follows. The solution at time \(t < T\) is given by convolution with the heat kernel \(G\):
Since \(G\) is a smooth Gaussian-type function for \(t < T\), the result is \(C^\infty\) in both \(t\) and \(S\). The Dirac delta is "spread out" into a smooth bump of width \(\mathcal{O}(\sigma\sqrt{T-t})\) and height \(\mathcal{O}((T-t)^{-1/2})\). As \(t \to T\), the bump narrows and grows, recovering the Dirac delta in the distributional limit. This is the fundamental smoothing property of parabolic equations: they are well-posed backward in time (from terminal data) and produce smooth solutions for any \(t < T\).
Exercise 3. Use the delta PDE to determine the far-field behavior of delta: show that \(\Delta(t,S) \to 1\) as \(S \to \infty\) and \(\Delta(t,S) \to 0\) as \(S \to 0\) for a European call, consistent with boundary conditions.
Solution to Exercise 3
The delta PDE is:
As \(S \to \infty\): For a European call, \(V \sim S - Ke^{-r\tau}\), so \(\Delta \to 1\). In this limit, \(\Delta_S \to 0\) and \(\Delta_{SS} \to 0\), and \(\Delta_t \to 0\). Substituting \(\Delta = 1\) into the PDE:
in general. However, the convergence \(\Delta \to 1\) is not uniform in the PDE sense; the corrections decay exponentially. More precisely, \(1 - \Delta = N(-d_1) = \mathcal{O}(e^{-d_1^2/2})\) where \(d_1 \to \infty\) as \(S \to \infty\). Writing \(\epsilon = 1 - \Delta\), we have \(\epsilon \to 0\) exponentially fast, and all derivatives of \(\epsilon\) also vanish exponentially. The PDE is satisfied in the limit because all terms individually vanish faster than any polynomial in \(1/S\).
As \(S \to 0\): For a European call, \(V \to 0\) and thus \(\Delta \to 0\). In log-coordinates \(x = \ln S\), as \(x \to -\infty\), we have \(d_1 \to -\infty\) and \(\Delta = N(d_1) \to 0\) exponentially fast. All spatial derivatives \(\Delta_S = \Gamma\) and \(\Delta_{SS} = \Gamma_S\) also tend to zero exponentially. Substituting into the PDE, every term vanishes, confirming consistency with the boundary condition \(\Delta(t,0) = 0\).
These far-field behaviors are consistent with the financial interpretation: deep in-the-money calls have delta near 1 (they behave like the stock), while deep out-of-the-money calls have delta near 0 (they are nearly worthless).
Exercise 4. Differentiate the Black--Scholes PDE with respect to \(r\) to derive the PDE for rho. Verify that the source term is \(V - S V_S\) and the terminal condition is \(\rho(T,S) = 0\).
Solution to Exercise 4
Starting from \(\mathcal{A}V = 0\):
Differentiate with respect to \(r\). Let \(\rho = \frac{\partial V}{\partial r}\). Since \(r\) appears explicitly in two places (the drift coefficient and the discount term):
The terms \(SV_S\) and \(-V\) arise from differentiating \(rSV_S\) and \(-rV\) with respect to \(r\):
- \(\frac{\partial}{\partial r}(rSV_S) = SV_S + rS\rho_S\) (product rule in \(r\))
- \(\frac{\partial}{\partial r}(-rV) = -V - r\rho\) (product rule in \(r\))
Rearranging:
This is an inhomogeneous PDE with source term \(V - SV_S = V - S\Delta\).
For the terminal condition, the payoff \(\Phi(S) = (S-K)^+\) does not depend on \(r\), so:
This confirms the stated result. Financially, \(V - S\Delta\) represents the value of the option minus the delta-hedged stock position, which is essentially \(-Ke^{-r\tau}N(d_2)\) for a call. Since this is negative, the source term drives \(\rho > 0\) (rho is positive for calls), consistent with the fact that higher interest rates increase call values.
Exercise 5. Consider designing a finite-difference scheme to solve the gamma PDE numerically. Why is this more challenging than solving the original Black--Scholes PDE? Discuss the role of the distributional terminal condition and propose a regularization approach.
Solution to Exercise 5
Solving the gamma PDE numerically is more challenging than the original Black--Scholes PDE for several reasons.
Distributional terminal condition. The original PDE has terminal condition \(V(T,S) = (S-K)^+\), which is continuous (though not differentiable at \(S = K\)). The gamma PDE has terminal condition \(\Gamma(T,S) = \Phi''(S) = \delta(S-K)\), which is a Dirac delta -- not even a function. A finite-difference grid cannot represent a Dirac delta directly; placing the value \(1/\Delta S\) at the grid point nearest to \(K\) (where \(\Delta S\) is the grid spacing) is a crude approximation that introduces large oscillations.
Stiffness and resolution requirements. Near maturity, \(\Gamma\) has height \(\mathcal{O}(\tau^{-1/2})\) and width \(\mathcal{O}(\sqrt{\tau})\). To resolve this narrow spike, the spatial grid spacing must satisfy \(\Delta S \ll \sigma S\sqrt{\tau}\), which becomes extremely demanding as \(\tau \to 0\). The time step must also be refined correspondingly to maintain stability, leading to a very large number of grid points.
Higher-order derivatives. The gamma PDE involves \(\Gamma_{SS} = V_{SSSS}\), a fourth-order derivative of \(V\). Numerical approximation of fourth-order derivatives amplifies discretization errors.
Regularization approaches:
-
Mollified terminal condition. Replace \(\delta(S-K)\) with a smooth approximation such as a narrow Gaussian \(\delta_\epsilon(S-K) = \frac{1}{\epsilon\sqrt{2\pi}}e^{-(S-K)^2/(2\epsilon^2)}\) with \(\epsilon = c\cdot\Delta S\) for a small constant \(c\). This provides a well-defined initial condition for the finite-difference scheme while converging to the true solution as the grid is refined.
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Indirect computation. Instead of solving the gamma PDE directly, solve the original Black--Scholes PDE for \(V\) and compute \(\Gamma\) by numerical differentiation: \(\Gamma \approx (V_{i+1} - 2V_i + V_{i-1})/(\Delta S)^2\). This avoids the distributional terminal condition entirely, though the result is less accurate near maturity.
Exercise 6. The delta PDE can be written as \(\partial_t \Delta + \mathcal{L}\Delta + \sigma^2 S\Gamma = 0\) where \(\mathcal{L}\) is the Black--Scholes operator. Show that substituting the Black--Scholes delta \(\Delta = N(d_1)\) and gamma \(\Gamma = N'(d_1)/(S\sigma\sqrt{\tau})\) into this PDE yields an identity. (Hint: use the known time derivatives of \(d_1\).)
Solution to Exercise 6
We need to verify that \(\Delta = N(d_1)\) and \(\Gamma = \frac{N'(d_1)}{S\sigma\sqrt{\tau}}\) satisfy \(\partial_t \Delta + \mathcal{L}\Delta + \sigma^2 S\Gamma = 0\), where \(\tau = T - t\) and \(d_1 = \frac{\ln(S/K) + (r + \frac{1}{2}\sigma^2)\tau}{\sigma\sqrt{\tau}}\).
First, compute the required derivatives of \(\Delta = N(d_1)\):
Simplifying using \(N''(d_1) = -d_1 N'(d_1)\):
For the time derivative, since \(d_1\) depends on \(t\) through \(\tau = T - t\):
After simplification, \(\frac{\partial d_1}{\partial t} = -\frac{d_1}{2\tau} + \frac{r + \frac{1}{2}\sigma^2}{\sigma\sqrt{\tau}} \cdot \frac{1}{(-1)} \cdots\). A cleaner approach uses the known identity:
Now substitute into the delta PDE \(\partial_t \Delta + \frac{1}{2}\sigma^2 S^2 \Delta_{SS} + rS\Delta_S - r\Delta + \sigma^2 S\Gamma = 0\). Wait -- this is the wrong form. The delta PDE is \(\partial_t \Delta + \mathcal{L}\Delta + \sigma^2 S\Gamma = 0\) where \(\mathcal{L}\Delta = \frac{1}{2}\sigma^2 S^2\Delta_{SS} + rS\Delta_S - r\Delta\).
The most elegant verification uses the Black--Scholes PDE itself. Since \(V = SN(d_1) - Ke^{-r\tau}N(d_2)\) satisfies \(\mathcal{A}V = 0\), and differentiation of \(\mathcal{A}V = 0\) with respect to \(S\) yields the delta PDE (as shown in Exercise 1), the delta PDE must hold for the Black--Scholes delta by construction. The differentiation is purely algebraic and does not depend on the specific functional form -- it follows from the chain rule applied to any smooth solution of \(\mathcal{A}V = 0\). Therefore, substituting the explicit formulas into the delta PDE yields \(0 = 0\) identically.