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I am busy trying to teach myself some stochastic calculus and have come across a statement that I am trying to prove.

How can I prove that \begin{align} \int_0^t W_s^2dW_s = \frac{1}{3} W_t^3 - \int_0^t W_s d_s \end{align} where $W \in \mathbb{R} $ is Brownian motion, using Ito's rule?

user4933
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Use the other author's hint, $f(w,t)=w^3$. Then, compute the partial derivatives needed for Ito's lemma: $$\partial_{w}f=3w^2$$ $$\partial_{w^2}f=6w$$ $$\partial_{t}f=0$$ Plug in Ito's formula, $$d(W_t^3)=3W_t^2dW_t+3W_tdt$$ Then integrate both sides from $0$ to $t$ (introduce dummy variable $s$), $$W_t^3=3\int_{0}^{t}W_s^2dW_s+3\int_{0}^{t}W_sds$$ Reaarange, done. Of course, there are other $f(w,t)$'s that you can plug in.

Mike
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Hint: apply Ito's formula to $f(x)=x^3$.

$f'(x)=3x^2$, $f''(x)=6x$ so $dW^3=3W^2dW+3Wdt$

user848721
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    I'm not really sure how to use this hint. Sorry I am very new to stochastic calculus and think its better i admit my lack of understanding on the subject. – user4933 Nov 14 '20 at 21:35
  • How do I tie the hint in to prove this result? – user4933 Nov 14 '20 at 21:37