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QUESTION

Show that if $W(t)$ is a Brownian motion, then It\^{o}'s lemma
implies the following results:

\begin{description}

\item[(i)]
$\int_0^tW^2\,dw=\frac{1}{3}W^3-\int_0^tW\,dt$

\item[(ii)]
$\int_0^tt\,dW=tW-\int_0^tW\,dt$

\end{description}


ANSWER

It\^{o}'s lemma for $f(x,t)$

$$df=\frac{\partial f}{\partial t}dt+\frac{\partial f}{\partial
x}dx+\frac{1}{2}\frac{\partial^2f}{\partial x^2}(dx)^2+\ldots$$

Let $x=w$ so that $dx=dw,\ (dx)^2=(dw)^2=dt$

\begin{description}

\item[(i)]
Consider $f=\frac{1}{3}w^3$

\begin{eqnarray*}
d(\frac{1}{3}w^3)&=&0.dt+\frac{\partial}{\partial
w}(\frac{1}{3}w^3)dw+\frac{1}{2}\frac{\partial^2}{\partial
w^2}(\frac{1}{3}w^3)dt\\ &=&w^2dw+wdt \end{eqnarray*}

Therefore

$$\int d(\frac{1}{3}w^3)=\int w^2dw+wdt$$

$$\Rightarrow \frac{1}{3}w^3=\int_0^tw^2\,dw+\int_0^tw\,ds$$

Since $w(0)=0$ as it's Brownian.

Or

$$\int_0^tw^2\,dw=\frac{1}{3}w^3-\int_0^tx\,ds$$

\item[(ii)]
Consider $f=tw$

\begin{eqnarray*}
df=d(tw)&=&\frac{\partial}{\partial
t}(tw)+\frac{\partial}{\partial
w}(tw)dw+\frac{1}{2}\frac{\partial^2}{\partial w}(tw)(dw)^2\\
&=&wdt+tdw+0\\ \Rightarrow\int d(tw)&=&\int w\,dt+\int t\,dw\\
\Rightarrow tw&=&\int_0^tw\,ds+\int_0^ts\,dw
\end{eqnarray*}

Since $w(0)=0$ when $t-0$,

Or

$$\int_0^ts\,dw=tw-\int_0^tw\,ds$$

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