It is [well known](https://i.stack.imgur.com/EJ9gG.png) that for the diffusions


\begin{align\*}
dX&=f(X)dt+&cdB\\
dY&=&cdB
\end{align\*}


the density of the law of $X$ with respect to the law of $Y$ is


\begin{align\*}
\frac{d\mu\_X}{d\mu\_Y}(c B)&=\exp\left(\int\_0^T\frac{f(cB(t))}{c}dB(t)-\frac12\int\_0^T\frac{f^2(cB(t))}{c^2}dt\right)\\
&=\exp\left(\frac{1}{c^2}\left(\int\_0^Tf(cB(t))d(cB(t))-\frac12\int\_0^Tf^2(cB(t))dt\right)\right)\\
&:=\exp\left(\frac{1}{c^2}L(cB(t))\right),\end{align\*}


where $L(cB)=\int\_0^Tf(cB(t))d(cB(t))-\frac12\int\_0^Tf^2(cB(t))dt$. My interest now is in making $\frac{1}{c^2}L(cB(t))$ defined pathwise. Applying integration by parts yields that


$$\frac{1}{c^2}L(cB)=\frac{1}{c^2}\left(F(cB(T))-\frac{c^2}{2}\int\_0^T f'(cB(t))dt-\frac12\int\_0^T f^2(cB(t))dt\right):=\frac{1}{c^2}L^c(cB).$$


I guess my question is - why does the "pathwise" functional depend on $c$ where the nonpathwise functional doesn't? What is the term $\int\_0^T f'(B(t))dt$ in relation to the original $L$? Is there a nice expression?


If we define $L\_1(B)=\int\_0^Tf(B(t))d(B(t))-\frac12\int\_0^Tf^2(B(t))dt$ and $L\_2(B)= F(B(T))-\frac{1}{2}\int\_0^T f'(B(t))dt-\frac12\int\_0^T f^2(B(t))dt\ $. Then $L\_1(B)=L\_2(B)$ however seemingly $L\_1(cB)\neq L\_2(cB)$. Edit:actually this comes from the fact that Ito's lemma only holds a.s. and cB is singular with B for $c\neq \pm 1$.



Where does the extra term in the density of a diffusion with respect to $c B(t)$ come from?
In a nutshell you are asking the following:


When $F(x)=\int\_0^xf(y)\,dy$ we get from Ito
$$
F(cB(t))=\int\_0^Tf(cB(t))\,d(cB(t))+\frac{1}{2}\int\_0^Tf'(cB(t))\,c^2\,dt\,
$$
so that the expressions
\begin{align}
&F(cB(t))-\frac{c^2}{2}\int\_0^Tf'(cB(t))\,dt\,,\tag{1}\\
&\int\_0^Tf(cB(t))\,d(cB(t))\tag{2}
\end{align}
are equal. Then why (2) depends only on $cB(t)$ and not separately on $c^2$ and $cB(t)$ like (1) seems to do?


Good question. I think the answer is that (1) can also be written as
\begin{align}
&F(cB(t))-\frac{1}{2}\int\_0^Tf'(cB(t))\,d\big\langle cB\big\rangle\_t\,,\tag{1'}
\end{align}
and gone is the isolated $c^2$.


