# Dynkin's martingale formula

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.

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.
• 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 $\langle M(f)\rangle_{t}=\int_{0}^{t}(f'(X_{s}))^{2}ds$. Apr 28, 2022 at 18:07