Dynkin's formula (from Wikipedia): for an Ito diffusion $X_t$ having infinitesimal generator $A$, $$ \mathbb{E}^x[f(X_\tau)] = f(x) + \mathbb{E}^x\left[\int_0^\tau Af(X_s)ds\right].$$

Here, $f$ is a continuous, real-valued function on the state space of $X_t$, and $\mathbb{E}^x[ \cdot ] = \mathbb{E}[\cdot | X_0=x]$ is the expectation given the initial condition $x$.

I came across a different version in a paper. It said that $$ \langle \alpha(t),\phi \rangle = \langle \alpha(0), \phi\rangle + \int_0^t Af(X_s)ds + M^N_t(\phi)$$ where $\alpha(t)$ is the empirical measure of $X_t$ (well actually it's a "macroscopically scaled" empirical measure, not sure if this would make a difference) and $\phi$ is a test function so that $\langle \alpha(t),\phi \rangle$ is the integral of $\phi$ with respect to the measure $\alpha(t)$. So I think this is sort of a weak form of the above.

Furthermore $M^N_t(\phi)$ is a martingale with quadratic variation satisfying $$ \frac{d}{dt}\langle M^N(\phi)\rangle_t = \Gamma(X_t,\phi).$$

Here $\Gamma$ is the carre du champ operator associated with $X_t$.

I don't know how this form of Dynkin's formula came across - it seems like the authors didn't take expectations but instead "compensated" with the martingale term. I also don't understand why $M^N(t)$ should satisfy that differential equation with the CDC operator.


1 Answer 1


I do not know if I can tag this as an answer but. The first formula is the expectation of the second formula (which really is the original form). By this, I mean Dynkin's martingale formula yields that if $X_t$ is a diffusion, and f is a rv continuous function : $$M^f_t = f(X_t) - f(X_0) - \int_0^t A f(X_s)ds $$ is a martingale. Rewriting it gives $f(X_t) = f(X_0) + M^f_t + \int_0^t A f(X_s)ds $. Now if you take the expectation, you get exactly : $$\mathbb{E}^x[f(X_t)] = f(x) + \mathbb{E}^x\left[\int_0^t Af(X_s)ds\right].$$ Since $M^f_0=0$. I am not familiar with "champ operators" but I'm quite sure the bracket also comes from it.

  • $\begingroup$ Thank you! I see how taking the expectation of the second equation gives the first (since a martingale is "expected" to have zero contribution). But I'm not sure why we can do the opposite - when removing the expectations, why does it suffice to add a martingale? And I don't know why that particular one has to satisfy that differential equation. $\endgroup$
    – 900edges
    Apr 28, 2022 at 14:48
  • $\begingroup$ For $dX_{t}=b(X_{t})dt+a(X_{t})dB_{t}$, by Ito $df(X_{t})=(b(X_{t})f'(X_{t})+\frac{a^{2}(X_{t})}{2}f''(X_{t}))dt+b(X_{t})dt$, whence $M(f)_{t}:=\int_{0}^{t}f'(X_{s})dB_{s}=f(X_{0})+\int_{0}^{t}Lf(X_{s})ds$ is a martingale with quadratic variation $\langle M(f)\rangle_{t}=\int_{0}^{t}(f'(X_{s}))^{2}ds$. $\endgroup$
    – Tobsn
    Apr 28, 2022 at 18:07

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