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From Oksendal's Stochastic Differential Equations I'm to prove that, for B Brownian motion and using the definition of an Ito integral,

$$ \displaystyle{\int_0^t}B_s^2\, dB_s = \frac{1}{3}B_t^3 - \displaystyle{\int_0^t}B_s\,ds $$

However I keep coming back to this;

$$ \displaystyle{\int_0^t}B_s^2\, dB_s = B_t^3 - \displaystyle{\int_0^t}B_s\,ds $$

From the definition of the Ito integral I'm summing over all the $B_j$'s, each term in the sum is;

$$ B_{j-1}^2(B_j-B_{j-1}) = (B_j^3 -B_{j-1}^3) - B_j (B_j^2 -B_{j-1}^2) $$

As the number of terms becomes large this approaches

$$ B_{j-1}^2(B_j-B_{j-1}) = (B_j^3 -B_{j-1}^3) - B_j \, \delta t $$

Performing the sum then gives me the (incorrect) result above.
Where did the factor of ${1\over3}$ go?

W.R.P.S
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GenericUsrnme
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  • It's simply not true that the term $B_j (B_j^2-B_{j-1}^2)$ approaches $B_j \delta t$. – saz Jan 22 '17 at 18:23

1 Answers1

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I think the point is that \begin{align*} \Delta\left(B_{j}^{3}\right) & =3B_{j}^{2}\Delta B_{j}+3B_{j}\left(\Delta B_{j}\right)^{2}+\left(\Delta B_{j}\right)^{3}, \end{align*} which leads to \begin{align*} B_{t}^{3} & =3\int B_{s}^{2}dB_{s}+3\int B_{s}ds; \end{align*} assuming $B_{0}=0$.

James
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