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I want to calculate $\mathbb{E} \left[\left(\int_0^tB_s\text{d}B_s\right)^3\right]$ where $B_t$ is a standard Brownian motion. Using Ito's formula for $f:\mathbb{R}\rightarrow\mathbb{R}$ with $f(x)=x^2$ we can find that $\int_0^tB_s\text{d}B_s=\dfrac{B^2_t}{2}-\dfrac{t}{2}$. Then: $\mathbb{E} \left[\left(\int_0^tB_s\text{d}B_s\right)^3\right]=\frac{1}{8}\mathbb{E}\left[\left(B^2_t-t\right)^3\right]$ and $\mathbb{E}[(B^2_t-t)^3]=\mathbb{E}[B^6_t-3tB^4_t+3t^2B^2_t-t^3]=15t^3-9t^3+3t^3-t^3=8t^3$

However, I have seen in another question that $\mathbb{E} \left[\left(\int_0^tB_s\text{d}B_s\right)^3\right]=0$. Am I doing something wrong?

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2 Answers 2

up vote 1 down vote accepted

No, I think you are right. The formula

$$\int_0^t B_s \, dB_s = \frac{1}{2} (B_t^2-t)$$

can be found in almost all books about SDE's. And you could also calculate the given moments using that

$$B_t^2 -t = t \cdot \bigg( X_t^2-1 \bigg)$$

where $X_t := \frac{B_t}{\sqrt{t}}$. $X_t^2$ is $\chi_1^2$-distributed and the third centered moment of a $\chi_1^2$-random variable is equal to 8 (see here.) Hence

$$\mathbb{E}((B_t^2-t)^3)=t^3 \cdot \mathbb{E}((X_t^2-1)^3)=8t^3$$


Remark: If you consider a process $X_t := \int_0^t \sigma(s) \, dB_s$ where $\sigma$ is a deterministic(!) coefficient, then $X_t$ is a centered normal random variable and in this case $\mathbb{E}(X_t^3)=0$ holds.

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Thanks a lot Saz! I would appreciate if you could have a look in these questions since no one has answered 'till now... This one and particularly, this one –  Nick Papadopoulos Dec 21 '12 at 18:04

For example, if $\mathbb{E}[X] = 0$, we cannot conclude that $\mathbb{E}[X^2] = 0$, etc. An example is $X = Z \sim \mathcal{N}(0, 1)$, where $\mathbb{E}[X] = 0$ but $\mathbb{E}[X^2] > 0$ since $X^2$ is never negative.

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