It is well known 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 $c\neq \pm 1$.


1 Answer 1


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 $c^2$.


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