# What is the Onsager-Machlup function for $dX(t)=f(B(t)) dt+dB(t)$?

What is the Onsager-Machlup function for $$dX(t)=f(B(t)) dt+dB(t)$$?

I know that the Onsager-Machlup function for $$dX(t)=f(X(t))dt+dB(t)$$ is $$L(x,v)=\frac12\left[v-f(x)\right]^2+\frac12f'(x)$$

But what if instead of $$f(X)$$ it is $$f(B)$$? I have a strong suspicion that it is

$$L(x,v)=\frac12 E[(v-f(B(t)+x))^2]$$

I have tried finding the SDE for $$X(t)$$ but it is quite difficult. I tried doing this through infinitesimal generator, but I can't find it.

• Definitely not. I do not have the time right now to write a full answer, so here are hints, maybe you or someone else can complete them. You need to replace $f(B_t)$ by a functional of the solution, i.e. write $f(B_t)=g(X_0^t)$ where now $g$ is a functional. (In the classical case, $f=g$ is a pointwise function. When you now do the Girsanov, you will write $g(X_0^t)=g(\phi_0^t+X_0^t-\phi_0^t)$ ($\phi$ is the function you are trying to compute the OM functional at). Expand, and you see that the term $\int_0^T|f'(\phi_t)|^2$ is replaced by something else. Jun 15, 2020 at 5:52
• The something else is simply the integral of the square of the functional derivative along the path, something like $\int_0^T \int_0^T ds ds' \int_0^{s'}\int_0^s \partial_u g(\phi_0^s) du \partial_{u'} g(\phi_0^{s'}) du'$. Writing a full proof will require a few pages... Jun 15, 2020 at 5:59
• @oferzeitouni Would it be easier to find the Onsager-Machlup function for just $dX(t)=f(B(t))dt$? I know this is a different problem. Jun 15, 2020 at 6:02
• Sorry, too hasty. The scaling of this problem is off. Indeed, if $\|B-\phi\|<\epsilon$ you have $\|X-\int_0^T f(\phi) dt\|\sim C\epsilon$ with $C\neq 1$, so I would not expect the OM functional to make sense. Jun 15, 2020 at 6:42
• What is $\mu_0$? and are you averaging on $B$ but not $X$? they are of course dependent, so I suspect this is not correct, but I am not sure. There should not be an expectation in the definition of $g$ Jun 16, 2020 at 6:39