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One version of Girsanov says that, that if $\mu_0$ is the law of a Brownian motion as a Borel measure on the space of continuous functions and we define the density

$$\frac{d\mu}{d\mu_0}:=\exp\left(\int_0^Tb(W(t)) dW(t)-\frac12\int_0^T b^2(W(t)) dt\right)$$

then under certain conditions the measure $\mu$ is the law of the solution to

$$dX=b(X) dt+ dW.$$

We can convert the Ito integral in the density to a Stratonovich integral and get that

$$\frac{d\mu}{d\mu_0}:=\exp\left(\int_0^Tb(W(t)) \circ dW(t)-\frac{1}{2}\int_0^T b'(W(t))dt-\frac12\int_0^T b^2(W(t)) dt\right).$$

However, I am curious if we define the new density

$$\frac{d\hat\mu}{d\mu_0}:=\frac{1}{Z}\exp\left(\int_0^Tb(W(t)) \circ dW(t)-\frac12\int_0^T b^2(W(t)) dt\right),$$

where $Z$ is a normalizing constant, what is the process associated to $\hat \mu$?

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  • $\begingroup$ Note that $\hat \mu$ is not a probability measure, since there is no reason for the expectation of $d\hat \mu/d\mu_0$ to equal $1$. $\endgroup$
    – ofer zeitouni
    Nov 20, 2022 at 1:30
  • $\begingroup$ @oferzeitouni What do you mean? Its expectation is defined to equal 1 - that's what the Z is $\endgroup$
    – user479223
    Nov 20, 2022 at 4:26
  • $\begingroup$ I missed that you normalized, sorry. This would give you a measure for fixed $T$, but there is no reason this measure defines diffusion process! In fact, what it defines (without the $Z$) corresponds (for $b'>0$) to killing at rate $b'(X_t)$; when you renormalize, it is the process with killing conditioned on not being killed. $\endgroup$
    – ofer zeitouni
    Nov 21, 2022 at 2:05

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