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QUESTION

If $f(t)$ is an integrable function on $[0,t]$ and $W(t)$ Brownian
show that the following integration by parts formula holds:

$$\int_0^tf(t)\,dW=f(t)W(t)-\int_0^tW\,df$$


ANSWER

Consider $g=f(t)w(t)$

\begin{eqnarray*}
\textrm{It\^{o}}\Rightarrow dg&=&\frac{\partial g}{\partial
t}dt+\frac{\partial g}{\partial
w}dw+\frac{1}{2}\frac{\partial^2g}{\partial w^2}(dw)^2\\
d(fw)&=&\frac{\partial}{\partial t}(fw)dt+\frac{\partial
g}{\partial w}(fw)dw+\frac{1}{2}\frac{\partial^2}{\partial
w^2}(fw)dt\\ d(fw)&=&w\frac{df}{dt}dt+fdw+0\\ \Rightarrow
d(fw)&=&wdf+fdw\\ \textrm{or
}\int_0^tf\,dw&=&\int_0^td(wf)-\int_0^tw\,df\\
\Rightarrow\int_0^tf(t)\,dw&=&f(t)w-\int_0^tw\,df
\end{eqnarray*}

Since $w$ is brownian and $w(0)=0$.



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