Generator and Martingales¶
The infinitesimal generator \(\mathcal{L}\) and martingales are deeply connected. This page develops the Dynkin martingale and the characterization of harmonic functions.
The Dynkin Martingale¶
For a diffusion \(dX_t = \mu(X_t)\,dt + \sigma(X_t)\,dW_t\) with generator \(\mathcal{L}\), define:
Theorem
For \(f \in C^2\), the process \(M_t\) is a local martingale.
Under the integrability condition \(\mathbb{E}\left[\int_0^t (f'(X_s)\sigma(X_s))^2\,ds\right] < \infty\), it is a true martingale.
Why It's a Martingale¶
From Itô's lemma:
Rearranging:
The Itô integral is a (local) martingale, so \(M_t\) is a (local) martingale. \(\square\)
Interpretation
- Drift: systematic, predictable change (captured by \(\mathcal{L}\))
- Martingale: unpredictable fluctuations (zero-mean increments)
The Dynkin martingale is the Itô integral — the noise that remains after subtracting the systematic drift.
Local vs True Martingale¶
| Condition | \(M_t\) is... |
|---|---|
| \(f \in C^2\) | Local martingale |
| + \(\mathbb{E}\left[\int_0^t (f'\sigma)^2\,ds\right] < \infty\) | True martingale |
| + bounded \(f'\sigma\) and finite horizon \(T\) | Uniformly integrable martingale |
True martingale means \(\mathbb{E}[|M_t|] < \infty\), \(\mathbb{E}[M_t \mid \mathcal{F}_s] = M_s\), and in particular \(\mathbb{E}[M_t] = 0\).
Local martingale means only that there exist \(\tau_n \uparrow \infty\) such that \(M_{t \wedge \tau_n}\) is a true martingale. This does not guarantee \(\mathbb{E}[M_t] = 0\).
Classic Counterexample: Strict Local Martingale
Let \(W_t\) be 3D Brownian motion with \(|W_0| = 1\). Define \(X_t = 1/|W_t|\).
- \(X_t\) is a local martingale (by Itô's lemma)
- \(X_t\) is not a true martingale: \(\mathbb{E}[X_t] < X_0\) — the expectation decreases.
- Optional stopping fails: \(\mathbb{E}[X_\tau] \neq X_0\) for some stopping times.
Optional Stopping and Dynkin's Formula¶
Dynkin's formula uses \(\mathbb{E}[M_\tau] = 0\), which requires optional stopping to hold:
| \(M_t\) is... | Dynkin's formula |
|---|---|
| True martingale + optional stopping conditions | ✓ Works directly |
| Local martingale only | ⚠ May fail without extra work |
Localization fix: If \(M_t\) is only a local martingale with localizing sequence \(\tau_n \uparrow \infty\), apply Dynkin to \(\tau \wedge \tau_n\) (which works since \(M_{t \wedge \tau_n}\) is a true martingale), then take \(n \to \infty\) using dominated or monotone convergence.
Harmonic Functions¶
Definition
A function \(f\) is \(\mathcal{L}\)-harmonic if:
Harmonic \(\Rightarrow\) Local Martingale¶
If \(\mathcal{L}f = 0\):
So \(f(X_t) - f(X_0)\) is a local martingale, i.e., \(f(X_t)\) is a local martingale.
Converse: Martingale \(\Rightarrow\) Harmonic?¶
If \(f(X_t)\) is a martingale, then \(M_t = f(X_t) - f(X_0)\) is a martingale, and also \(M_t = f(X_t) - f(X_0) - \int_0^t (\mathcal{L}f)(X_s)\,ds\) is a martingale. So \(\int_0^t (\mathcal{L}f)(X_s)\,ds\) must be a martingale. But it is also a continuous finite-variation process.
Key fact: A continuous martingale of finite variation is constant. (Proof: by the Doob–Meyer decomposition, such a process has zero quadratic variation, hence is a.s. constant.)
Therefore \(\int_0^t (\mathcal{L}f)(X_s)\,ds = 0\) for all \(t\), a.s., which implies \((\mathcal{L}f)(X_s) = 0\) for a.e.\ \(s\) along sample paths.
The Subtlety¶
This says \(\mathcal{L}f(x) = 0\) for \(x\) visited by the process — not automatically for all \(x\).
| Condition | Conclusion |
|---|---|
| \(f(X_t)\) martingale | \((\mathcal{L}f)(X_t) = 0\) a.e.\ along paths |
| + \(X_t\) irreducible | \(\mathcal{L}f(x) = 0\) for all \(x\) in state space |
| + \(f \in C^2\), \(\mathcal{L}f\) continuous | \(\mathcal{L}f(x) = 0\) everywhere |
If \(X_t\) never visits some region \(A\), we learn nothing about \(\mathcal{L}f\) on \(A\) from the martingale property alone.
Correct Characterization¶
Under suitable regularity (\(f \in C^2\), diffusion non-degenerate and irreducible):
The \(\Leftarrow\) direction requires non-degeneracy so the process visits all of the state space. Without it, we can only conclude \(\mathcal{L}f = 0\) along visited paths. For the equivalence with true martingale, the additional condition \(\mathbb{E}\left[\int_0^t (f'\sigma)^2\,ds\right] < \infty\) is needed.
Examples¶
Example 1: Brownian Motion¶
Generator: \(\mathcal{L}f = \frac{1}{2}f''\)
Harmonic functions: \(\mathcal{L}f = 0 \Leftrightarrow f'' = 0 \Leftrightarrow f(x) = ax + b\)
| Function | \(\mathcal{L}f\) | \(f(X_t)\) |
|---|---|---|
| \(f(x) = x\) | \(0\) | Martingale ✓ |
| \(f(x) = x^2\) | \(1 \neq 0\) | Not a martingale |
| \(f(x,t) = x^2 - t\) | \(\tilde{\mathcal{L}}f = 0\) | Martingale (time-dependent; uses extended generator \(\tilde{\mathcal{L}} = \partial_t + \mathcal{L}\)) |
Example 2: Brownian Motion in \(\mathbb{R}^d\)¶
Generator: \(\mathcal{L} = \frac{1}{2}\Delta\)
Harmonic functions: Solutions to \(\Delta f = 0\) (classical harmonic functions).
Examples in \(\mathbb{R}^2\): \(\log|x|\), \(\text{Re}(z^n)\), \(\text{Im}(z^n)\) (all away from the origin where \(f \notin C^2\)). Examples in \(\mathbb{R}^3\): \(1/|x|\) (away from origin), spherical harmonics.
Example 3: Ornstein–Uhlenbeck¶
Generator: \(\mathcal{L}f = -\kappa x f'(x) + \frac{\sigma^2}{2}f''(x)\)
Harmonic functions: Solve \(-\kappa x f' + \frac{\sigma^2}{2}f'' = 0\). The general solution involves parabolic cylinder functions; among bounded solutions, only \(f(x) = \text{constant}\) qualifies. This is consistent with the OU process being ergodic — a non-constant bounded harmonic function would contradict mean-reversion to the stationary distribution.
Example 4: GBM¶
Generator: \(\mathcal{L}f = \mu s f'(s) + \frac{\sigma^2 s^2}{2}f''(s)\)
Is \(f(s) = s\) harmonic? \(\mathcal{L}(s) = \mu s \neq 0\) (unless \(\mu = 0\)). So \(S_t\) is not a martingale (it is a submartingale if \(\mu > 0\)).
Harmonic functions: We solve \(\mu s f' + \frac{\sigma^2 s^2}{2}f'' = 0\) on \((0,\infty)\). Testing \(f(s) = s^\alpha\):
Setting this to zero for \(\alpha \neq 0\) (the case \(\alpha = 0\) gives the trivial constant solution \(f = 1\)):
When \(\mu = \sigma^2/2\), the nontrivial exponent degenerates to zero and only constant solutions exist.
Connection to Martingale Problem¶
The Stroock–Varadhan martingale problem uses this characterization as its foundation:
Martingale Problem (Stroock–Varadhan)
A probability measure \(\mathbb{P}\) on path space solves the martingale problem for \(\mathcal{L}\) if:
is a \(\mathbb{P}\)-martingale for all \(f \in C_c^\infty\).
This gives a probabilistic characterization of diffusions without requiring pathwise SDE solutions. See Stroock–Varadhan Martingale Problem for the full development, including uniqueness theory and applications to degenerate diffusions.
Summary¶
| Statement | Meaning |
|---|---|
| \(M_t = f(X_t) - f(X_0) - \int_0^t (\mathcal{L}f)(X_s)\,ds\) | Dynkin martingale = Itô integral |
| \(\mathcal{L}f = 0\) | \(f\) is \(\mathcal{L}\)-harmonic |
| \(\mathcal{L}f = 0 \Rightarrow f(X_t)\) local martingale | Harmonic functions give local martingales |
| \(f(X_t)\) martingale \(\Rightarrow \mathcal{L}f = 0\) | Converse holds under non-degeneracy and irreducibility |
See Also¶
Exercises¶
Exercise 1. For the Ornstein--Uhlenbeck process \(dX_t = -\kappa X_t\,dt + \sigma\,dW_t\), write down the Dynkin martingale \(M_t^f\) for \(f(x) = x^2\). Express it explicitly as an Ito integral and verify that \(\mathbb{E}[M_t^f] = 0\).
Solution to Exercise 1
The OU generator is \(\mathcal{L}f = -\kappa x\,f'(x) + \frac{\sigma^2}{2}f''(x)\). For \(f(x) = x^2\):
- \(f'(x) = 2x\), \(f''(x) = 2\)
The Dynkin martingale is:
To express as an Ito integral, apply Ito's lemma to \(f(X_t) = X_t^2\):
Rearranging: \(d(X_t^2) - (-2\kappa X_t^2 + \sigma^2)\,dt = 2\sigma X_t\,dW_t\). Integrating:
This is an Ito integral with respect to \(W_t\), hence a local martingale. To verify \(\mathbb{E}[M_t^f] = 0\), note that \(\mathbb{E}\!\left[\int_0^t 4\sigma^2 X_s^2\,ds\right] < \infty\) for any finite \(t\) (since \(\mathbb{E}[X_s^2]\) is bounded on \([0,t]\) by the second moment formula), so the Ito integral is a true martingale with \(\mathbb{E}[M_t^f] = 0\).
Exercise 2. Consider geometric Brownian motion \(dS_t = \mu S_t\,dt + \sigma S_t\,dW_t\) on \((0, \infty)\). In Example 4, it was shown that \(f(s) = s^{\alpha}\) with \(\alpha = 1 - 2\mu/\sigma^2\) is \(\mathcal{L}\)-harmonic.
(a) Verify directly that \(\mathcal{L}(s^{\alpha}) = 0\) for this value of \(\alpha\).
(b) What does this imply about the process \(S_t^{\alpha}\)? Is it a true martingale or only a local martingale? (Hint: consider whether \(\mathbb{E}[S_t^{\alpha}]\) remains constant.)
Solution to Exercise 2
(a) The GBM generator is \(\mathcal{L}f = \mu s\,f'(s) + \frac{\sigma^2 s^2}{2}f''(s)\). For \(f(s) = s^\alpha\):
- \(f'(s) = \alpha s^{\alpha - 1}\), \(f''(s) = \alpha(\alpha-1)s^{\alpha-2}\)
With \(\alpha = 1 - 2\mu/\sigma^2\):
So \(\mathcal{L}(s^\alpha) = 0\), confirming \(f(s) = s^\alpha\) is \(\mathcal{L}\)-harmonic.
(b) Since \(\mathcal{L}(s^\alpha) = 0\), the process \(S_t^\alpha\) is a local martingale. To check whether it is a true martingale, compute \(\mathbb{E}[S_t^\alpha]\) directly. Since \(S_t = s_0 \exp\!\left[(\mu - \sigma^2/2)t + \sigma W_t\right]\):
The exponent is \(\left[\mu\alpha + \frac{\sigma^2}{2}\alpha(\alpha - 1)\right]t = 0\) by the calculation in part (a). Therefore \(\mathbb{E}[S_t^\alpha] = s_0^\alpha\) for all \(t\), confirming \(S_t^\alpha\) is a true martingale (not merely a local martingale).
Exercise 3. Let \(W_t\) be standard Brownian motion and define \(f(x, t) = e^{\theta x - \theta^2 t/2}\) for a constant \(\theta\). Show that \(\tilde{\mathcal{L}}f = 0\) where \(\tilde{\mathcal{L}} = \partial_t + \frac{1}{2}\partial_{xx}\) is the extended generator. Conclude that \(f(W_t, t)\) is a martingale (the exponential martingale).
Solution to Exercise 3
The extended generator for BM is \(\tilde{\mathcal{L}} = \partial_t + \frac{1}{2}\partial_{xx}\). For \(f(x,t) = e^{\theta x - \theta^2 t/2}\):
Therefore:
Since \(\tilde{\mathcal{L}}f = 0\), the process \(f(W_t, t) = e^{\theta W_t - \theta^2 t/2}\) satisfies the Dynkin martingale condition: the drift in \(df\) vanishes, and \(f(W_t, t)\) is a local martingale. It is in fact a true martingale since \(\mathbb{E}[e^{\theta W_t - \theta^2 t/2}] = e^{-\theta^2 t/2}\cdot e^{\theta^2 t/2} = 1\) for all \(t\). This is the classical exponential martingale (also known as the Wald martingale or Doleans-Dade exponential).
Exercise 4. Let \(X_t\) be a continuous martingale with finite variation (i.e., \(X_t\) has paths of bounded variation on every \([0, T]\)). Prove that \(X_t = X_0\) a.s. for all \(t\). (Hint: use the fact that the quadratic variation \([X]_t = 0\) for finite-variation processes, and for a continuous local martingale, \([X]_t = 0\) implies \(X_t\) is constant.) Explain why this result is crucial for the converse direction: martingale \(\Rightarrow\) harmonic.
Solution to Exercise 4
Let \(X_t\) be a continuous martingale with paths of bounded variation on every \([0, T]\).
Step 1: For a continuous finite-variation process, the quadratic variation is zero: \([X]_t = 0\) for all \(t\). This follows because \([X]_t = \lim_{n\to\infty}\sum_{k}(X_{t_{k+1}} - X_{t_k})^2\), and for a bounded-variation process, \(\max_k |X_{t_{k+1}} - X_{t_k}| \to 0\), so:
Step 2: By the martingale representation structure, any continuous local martingale \(M_t\) satisfies \(M_t = M_0 + \int_0^t H_s\,dW_s\) for some predictable process \(H_s\) (by the Brownian martingale representation theorem in the Brownian filtration, or more generally by the Kunita-Watanabe characterization). Its quadratic variation is \([M]_t = \int_0^t H_s^2\,ds\). If \([M]_t = 0\), then \(H_s = 0\) a.e., so \(M_t = M_0\).
Alternatively, without invoking representation: \(X_t - X_0\) is a continuous local martingale with \([X - X_0]_t = [X]_t = 0\). By the identity \(\mathbb{E}[(X_t - X_s)^2 \mid \mathcal{F}_s] = \mathbb{E}[[X]_t - [X]_s \mid \mathcal{F}_s] = 0\), we get \(X_t = X_s\) a.s. for all \(s \leq t\), hence \(X_t = X_0\) a.s.
Why this matters for the converse: If \(f(X_t)\) is a martingale, then \(M_t = f(X_t) - f(X_0)\) is a martingale and also \(M_t = f(X_t) - f(X_0) - \int_0^t (\mathcal{L}f)(X_s)\,ds + \int_0^t (\mathcal{L}f)(X_s)\,ds\). The process \(A_t = \int_0^t (\mathcal{L}f)(X_s)\,ds\) is continuous and of finite variation, and \(M_t - A_t\) is a martingale by the Dynkin construction. So \(A_t = M_t - (M_t - A_t)\) is a difference of two martingales, hence a martingale. Being both a continuous finite-variation process and a martingale, \(A_t\) must be constant: \(A_t = A_0 = 0\). This forces \(\int_0^t (\mathcal{L}f)(X_s)\,ds = 0\) for all \(t\), implying \(\mathcal{L}f = 0\) along the paths of \(X\).
Exercise 5. For standard Brownian motion in \(\mathbb{R}^2\) with generator \(\mathcal{L} = \frac{1}{2}\Delta = \frac{1}{2}(\partial_{xx} + \partial_{yy})\), verify that the following functions are \(\mathcal{L}\)-harmonic:
(a) \(f(x, y) = x^2 - y^2\)
(b) \(f(x, y) = e^x \cos(y)\)
(c) \(f(x, y) = \ln(x^2 + y^2)\) for \((x, y) \neq (0, 0)\)
For each, state what the harmonic property implies about \(f(W_t^1, W_t^2)\) where \((W_t^1, W_t^2)\) is 2D Brownian motion.
Solution to Exercise 5
(a) \(f(x,y) = x^2 - y^2\):
So \(f(W_t^1, W_t^2) = (W_t^1)^2 - (W_t^2)^2\) is a local martingale.
(b) \(f(x,y) = e^x \cos(y)\):
So \(f(W_t^1, W_t^2) = e^{W_t^1}\cos(W_t^2)\) is a local martingale. (This is the real part of \(e^{W_t^1 + iW_t^2}\), connected to conformal invariance of 2D BM.)
(c) \(f(x,y) = \ln(x^2 + y^2)\) for \((x,y) \neq (0,0)\):
By symmetry:
So \(\ln((W_t^1)^2 + (W_t^2)^2) = \ln|B_t|^2 = 2\ln|B_t|\) (where \(B_t\) is 2D BM) is a local martingale, provided \(B_t \neq 0\). Since 2D BM is neighborhood-recurrent but point-recurrent only at the starting point, \(\ln|B_t|\) is well-defined for \(B_0 \neq 0\). This is the classical result that \(\ln|B_t|\) is a local martingale for 2D BM.
Exercise 6. Consider a diffusion \(dX_t = \mu(X_t)\,dt + \sigma(X_t)\,dW_t\) where \(\sigma(x) = 0\) for \(x \in [1, 2]\) but \(\sigma(x) > 0\) elsewhere. Explain why the equivalence \(\mathcal{L}f = 0 \Leftrightarrow f(X_t)\) is a local martingale may fail in this setting. Which specific condition (non-degeneracy or irreducibility) is violated, and what is the consequence for the converse direction?
Solution to Exercise 6
The condition violated is non-degeneracy. Since \(\sigma(x) = 0\) for \(x \in [1,2]\), the diffusion coefficient vanishes on an entire interval. Within \([1,2]\), the process follows the ODE \(dX_t = \mu(X_t)\,dt\) with no randomness — it moves deterministically and cannot explore the region \([1,2]\) freely.
The consequence for the converse direction: suppose \(f(X_t)\) is a martingale. The argument that \(\mathcal{L}f = 0\) relies on the process visiting all points in the state space. If \(X_t\) starts outside \([1,2]\) and enters this interval, it moves deterministically through it, never "sampling" all points densely. So even if \(f(X_t)\) is a martingale, we can only conclude \(\mathcal{L}f(x) = 0\) for points \(x\) actually visited by the process.
Concretely, on \([1,2]\) the generator reduces to \(\mathcal{L}f(x) = \mu(x)f'(x)\) (purely first-order). A function \(f\) could satisfy \(\mu(x)f'(x) \neq 0\) at some \(x \in [1,2]\) that the process traverses only once (deterministically), yet \(f(X_t)\) might still be a martingale globally because the non-zero generator contribution on \([1,2]\) is compensated by the behavior elsewhere. The equivalence \(\mathcal{L}f = 0 \Leftrightarrow f(X_t)\) local martingale breaks down because the degenerate region prevents the process from being irreducible in the sense needed for the converse.
Exercise 7. In the Stroock--Varadhan martingale problem, one requires \(M_t^f = f(X_t) - f(X_0) - \int_0^t (\mathcal{L}f)(X_s)\,ds\) to be a martingale for all \(f \in C_c^\infty\).
(a) Why is it sufficient to test only functions in \(C_c^\infty\) (smooth with compact support) rather than all \(C^2\) functions?
(b) Give an intuitive explanation for why the martingale problem characterization uniquely determines the law of \(X_t\) under non-degeneracy conditions.
Solution to Exercise 7
(a) The space \(C_c^\infty\) is sufficient because it is dense in the relevant function spaces (such as \(C_0^2\) or \(L^2\)) and the generator is a local operator for diffusions. More precisely:
- The martingale property \(\mathbb{E}[M_t^f \mid \mathcal{F}_s] = M_s^f\) for all \(f \in C_c^\infty\) determines all finite-dimensional distributions of \(X_t\). This is because the test functions \(f \in C_c^\infty\) separate points: for any two distinct probability measures on path space, there exists \(f \in C_c^\infty\) for which the martingale condition distinguishes them.
- Technically, \(C_c^\infty\) is a core for the generator of any non-degenerate diffusion, meaning the closure of \(\mathcal{L}|_{C_c^\infty}\) equals the full generator. Testing the martingale property on a core suffices to determine the process.
(b) Intuitively, the generator \(\mathcal{L}\) encodes both the drift \(\mu(x)\) and the diffusion coefficient \(a(x) = \sigma\sigma^\top(x)\) at every point \(x\). The martingale condition \(\mathbb{E}[M_t^f \mid \mathcal{F}_s] = M_s^f\) for all smooth \(f\) forces the conditional mean and conditional variance of infinitesimal increments to match \(\mu\) and \(a\) respectively. Under non-degeneracy (\(a(x)\) is positive definite), the second-order terms in \(\mathcal{L}\) uniquely determine the diffusion matrix, and the first-order terms uniquely determine the drift. Since the law of a Markov process is determined by its infinitesimal transition rates (equivalently, by \(\mu\) and \(a\)), uniqueness follows. This is the Stroock-Varadhan uniqueness theorem: non-degenerate \(a(x)\) with suitable regularity guarantees that the martingale problem has a unique solution.