Greeks via the Infinitesimal Generator¶

In Markov models, the infinitesimal generator provides a structural link between pricing, PDEs, and sensitivity analysis.

Setup: risk-neutral diffusion¶

Under \(\mathbb{Q}\), consider

\[ \mathrm{d}S_t = rS_t\,\mathrm{d}t + \sigma S_t\,\mathrm{d}W_t \]

so the generator acting on smooth test functions \(f\) is

\[ \boxed{ (\mathcal{L}f)(S) = rS f'(S) + \frac{1}{2}\sigma^2 S^2 f''(S) } \]

Pricing PDE¶

For a European payoff \(\Phi\), the price

\[ V(t,S)=\mathbb{E}^{t,S}\!\left[e^{-r(T-t)}\Phi(S_T)\right] \]

solves the backward equation

\[ \boxed{ \frac{\partial V}{\partial t} + \mathcal{L}V - rV = 0, \qquad V(T,S)=\Phi(S) } \]

Parameter sensitivity as generator sensitivity¶

Treat the generator as parameterized by \(\sigma\):

\[ \mathcal{L}_\sigma f = rS f' + \frac{1}{2}\sigma^2 S^2 f'' \]

Then vega \(\nu=V_\sigma\) is linked to the derivative of \(\mathcal{L}_\sigma\) with respect to \(\sigma\):

\[ \frac{\partial}{\partial \sigma}\mathcal{L}_\sigma f = \sigma S^2 f'' \]

This suggests a PDE for \(\nu\) of the form

\[ \boxed{ \frac{\partial \nu}{\partial t} + \mathcal{L}\nu - r\nu = - \left(\frac{\partial \mathcal{L}}{\partial \sigma}V\right) = -\sigma S^2 V_{SS} } \]

What to remember¶

The generator \(\mathcal{L}\) determines the pricing PDE.
Parameter Greeks can be viewed as sensitivities of \(\mathcal{L}\) itself.
This framework generalizes to multi-factor models.

Exercises¶

Exercise 1. Verify that the generator \((\mathcal{L}f)(S) = rSf'(S) + \frac{1}{2}\sigma^2 S^2 f''(S)\) applied to \(f(S) = S^n\) produces a polynomial in \(S^n\). Compute \(\mathcal{L}(S^2)\) and interpret the result in terms of the expected quadratic variation of \(S\).

Solution to Exercise 1

Apply the generator to \(f(S) = S^n\). We have \(f'(S) = nS^{n-1}\) and \(f''(S) = n(n-1)S^{n-2}\). Then:

\[ (\mathcal{L}f)(S) = rS \cdot nS^{n-1} + \frac{1}{2}\sigma^2 S^2 \cdot n(n-1)S^{n-2} \]

\[ = rnS^n + \frac{1}{2}\sigma^2 n(n-1)S^n = \left[rn + \frac{1}{2}\sigma^2 n(n-1)\right]S^n \]

This is indeed a polynomial (monomial) in \(S^n\), confirming that \(S^n\) is an eigenfunction of \(\mathcal{L}\) with eigenvalue \(rn + \frac{1}{2}\sigma^2 n(n-1)\).

For \(n = 2\):

\[ \mathcal{L}(S^2) = \left[2r + \frac{1}{2}\sigma^2 \cdot 2 \cdot 1\right]S^2 = (2r + \sigma^2)S^2 \]

Interpretation. The quantity \(\mathcal{L}(S^2)\) captures the instantaneous growth rate of \(\mathbb{E}[S_t^2]\). Indeed, by the connection between the generator and expected values, \(\frac{\mathrm{d}}{\mathrm{d}t}\mathbb{E}[f(S_t)] = \mathbb{E}[\mathcal{L}f(S_t)]\). For \(f = S^2\), this gives \(\frac{\mathrm{d}}{\mathrm{d}t}\mathbb{E}[S_t^2] = (2r + \sigma^2)\mathbb{E}[S_t^2]\). The term \(\sigma^2 S^2\) reflects the quadratic variation contribution: \(\mathrm{d}\langle S \rangle_t = \sigma^2 S_t^2\,\mathrm{d}t\).

Exercise 2. Starting from \(\mathcal{A}V = 0\) where \(\mathcal{A} = \partial_t + \mathcal{L} - r\), differentiate with respect to \(\sigma\) to derive the PDE for vega: \(\partial_t \nu + \mathcal{L}\nu - r\nu = -\sigma S^2 V_{SS}\). Explain why the terminal condition is \(\nu(T,S) = 0\).

Solution to Exercise 2

Define the full pricing operator \(\mathcal{A} = \partial_t + \mathcal{L}_\sigma - r\) so that the pricing PDE is \(\mathcal{A}V = 0\).

Differentiate with respect to \(\sigma\). Since \(\partial_t\) and \(r\) do not depend on \(\sigma\):

\[ \frac{\partial}{\partial\sigma}(\mathcal{A}V) = \mathcal{A}\left(\frac{\partial V}{\partial\sigma}\right) + \frac{\partial\mathcal{L}_\sigma}{\partial\sigma}V = 0 \]

Let \(\nu = \frac{\partial V}{\partial\sigma}\). Then:

\[ \mathcal{A}\nu = -\frac{\partial\mathcal{L}_\sigma}{\partial\sigma}V \]

Computing \(\frac{\partial}{\partial\sigma}\mathcal{L}_\sigma f = \frac{\partial}{\partial\sigma}\!\left(\frac{1}{2}\sigma^2 S^2 f''\right) = \sigma S^2 f''\):

\[ \partial_t \nu + \mathcal{L}\nu - r\nu = -\sigma S^2 V_{SS} \]

Terminal condition. At \(t = T\), \(V(T,S) = \Phi(S)\), which does not depend on \(\sigma\). Therefore:

\[ \nu(T,S) = \frac{\partial}{\partial\sigma}\Phi(S) = 0 \]

This makes sense: at expiration, the payoff is determined entirely by \(S_T\) and the strike, with no sensitivity to volatility. All vega sensitivity comes from the time value of the option, which vanishes at maturity.

Exercise 3. Consider a general diffusion with generator \(\mathcal{L}_\theta f = \mu(\theta)Sf' + \frac{1}{2}\sigma(\theta)^2 S^2 f''\), where \(\theta\) is a model parameter. Compute \(\frac{\partial}{\partial \theta}\mathcal{L}_\theta f\) and use it to write down the PDE satisfied by the sensitivity \(\partial V / \partial \theta\).

Solution to Exercise 3

The general generator is:

\[ \mathcal{L}_\theta f = \mu(\theta)Sf' + \frac{1}{2}\sigma(\theta)^2 S^2 f'' \]

Differentiate with respect to \(\theta\):

\[ \frac{\partial}{\partial\theta}\mathcal{L}_\theta f = \mu'(\theta)Sf' + \sigma(\theta)\sigma'(\theta)S^2 f'' \]

The sensitivity \(w := \frac{\partial V}{\partial\theta}\) satisfies the PDE obtained by differentiating \(\mathcal{A}V = 0\) with respect to \(\theta\):

\[ \partial_t w + \mathcal{L}_\theta w - rw = -\frac{\partial\mathcal{L}_\theta}{\partial\theta}V = -\mu'(\theta)SV_S - \sigma(\theta)\sigma'(\theta)S^2 V_{SS} \]

with terminal condition \(w(T,S) = 0\) (assuming the payoff does not depend on \(\theta\)).

The source term on the right involves \(V_S\) (delta) and \(V_{SS}\) (gamma) of the original pricing problem, weighted by the parameter sensitivities \(\mu'(\theta)\) and \(\sigma'(\theta)\). In the standard Black-Scholes case where \(\theta = \sigma\), we have \(\mu(\sigma) = r\) (so \(\mu' = 0\)) and \(\sigma(\theta) = \sigma\) (so \(\sigma' = 1\)), recovering the vega PDE from Exercise 2.

Exercise 4. Using the generator approach, show that theta satisfies \(\Theta = rV - rS\Delta - \frac{1}{2}\sigma^2 S^2 \Gamma\) directly from the pricing PDE. Explain why theta does not require a separate PDE but can always be computed from other Greeks.

Solution to Exercise 4

The pricing PDE is:

\[ \frac{\partial V}{\partial t} + \mathcal{L}V - rV = 0 \]

Expanding \(\mathcal{L}V = rSV_S + \frac{1}{2}\sigma^2 S^2 V_{SS}\):

\[ \frac{\partial V}{\partial t} + rSV_S + \frac{1}{2}\sigma^2 S^2 V_{SS} - rV = 0 \]

Solving for \(\Theta = \frac{\partial V}{\partial t}\):

\[ \Theta = rV - rSV_S - \frac{1}{2}\sigma^2 S^2 V_{SS} = rV - rS\Delta - \frac{1}{2}\sigma^2 S^2 \Gamma \]

Why theta does not need a separate PDE. The pricing PDE is a constraint relating \(V_t\), \(V_S\), \(V_{SS}\), and \(V\). Since theta is simply \(V_t\), it can always be expressed algebraically in terms of the other Greeks (\(\Delta\), \(\Gamma\)) and the option price \(V\) itself. There is no need to solve an auxiliary PDE because the Black-Scholes PDE already provides the relationship directly.

This is in contrast to vega, which requires differentiating the PDE with respect to a parameter \(\sigma\), producing a new PDE with a source term. Theta is a state derivative (with respect to \(t\)) that is already one of the terms in the original PDE.

Exercise 5. For the CEV (constant elasticity of variance) model with \(\mathrm{d}S_t = rS_t\,\mathrm{d}t + \sigma S_t^\beta\,\mathrm{d}W_t\), write down the infinitesimal generator \(\mathcal{L}^{\text{CEV}}\). How does \(\frac{\partial}{\partial \sigma}\mathcal{L}^{\text{CEV}}\) differ from the Black--Scholes case?

Solution to Exercise 5

In the CEV model, \(\mathrm{d}S_t = rS_t\,\mathrm{d}t + \sigma S_t^\beta\,\mathrm{d}W_t\). The infinitesimal generator acts on smooth functions \(f\) as:

\[ (\mathcal{L}^{\text{CEV}}f)(S) = rSf'(S) + \frac{1}{2}\sigma^2 S^{2\beta}f''(S) \]

The key difference from Black-Scholes (where \(\beta = 1\)) is that the diffusion coefficient is \(\frac{1}{2}\sigma^2 S^{2\beta}\) instead of \(\frac{1}{2}\sigma^2 S^2\).

Differentiating with respect to \(\sigma\):

\[ \frac{\partial}{\partial\sigma}\mathcal{L}^{\text{CEV}}_\sigma f = \sigma S^{2\beta}f'' \]

Comparison with Black-Scholes. In Black-Scholes, \(\frac{\partial}{\partial\sigma}\mathcal{L}_\sigma f = \sigma S^2 f''\). In the CEV model, \(S^2\) is replaced by \(S^{2\beta}\). Consequently, the vega PDE for the CEV model has the source term \(-\sigma S^{2\beta}V_{SS}\) instead of \(-\sigma S^2 V_{SS}\).

For \(\beta < 1\) (which produces a leverage effect), the diffusion coefficient grows more slowly with \(S\), so the vega source term is relatively smaller for large \(S\) and relatively larger for small \(S\), compared to Black-Scholes. For \(\beta > 1\), the opposite holds.

Exercise 6. In a two-factor model with state variables \((S, v)\) and generator \(\mathcal{L}f = rSf_S + \kappa(\bar{v}-v)f_v + \frac{1}{2}vS^2 f_{SS} + \rho\xi vS f_{Sv} + \frac{1}{2}\xi^2 v f_{vv}\), write down the pricing PDE. How many second-order Greeks appear, and which ones are captured by cross-derivatives \(f_{Sv}\)?

Solution to Exercise 6

The pricing PDE is obtained from \(\frac{\partial V}{\partial t} + \mathcal{L}V - rV = 0\):

\[ \frac{\partial V}{\partial t} + rSV_S + \kappa(\bar{v} - v)V_v + \frac{1}{2}vS^2 V_{SS} + \rho\xi vS\,V_{Sv} + \frac{1}{2}\xi^2 v\,V_{vv} - rV = 0 \]

Counting second-order Greeks. The state variables are \((S, v)\), giving three distinct second-order partial derivatives:

\(V_{SS}\) -- gamma (sensitivity to spot squared)
\(V_{vv}\) -- volga or variance-of-variance sensitivity
\(V_{Sv}\) -- cross-derivative (sensitivity to joint spot-variance moves)

So there are three second-order Greeks.

The cross-derivative \(V_{Sv}\). This Greek captures the sensitivity of the option price to simultaneous changes in spot \(S\) and variance \(v\). It is the analogue of vanna (\(\partial^2 V / \partial S \partial \sigma\)) in the Black-Scholes setting, adapted to a stochastic variance framework. The coefficient \(\rho\xi vS\) multiplying \(V_{Sv}\) in the PDE shows that this cross-term is driven by the correlation \(\rho\) between the spot and variance Brownian motions. When \(\rho = 0\) (uncorrelated spot and variance), the cross-derivative term drops out of the PDE, and the model loses its ability to generate skew in implied volatility.