# Prove directly from the definition of the Ito's integral

I am trying to solve the exercises from the book Stochastic differential equations -An Introduction with applications by Bernt Oksendal and I am stuck on 1 question.

Prove directly from the definition of Ito's integral that

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

I have tried for a couple of hours but I simply cannot prove it . I would be grateful if somebody could help me out , even a hint would be very helpful

Note that \begin{align*}B_T^3 &=\sum_{j=1}^n B_{s_j}^3 - B_{s_{j-1}}^3 = \sum_{j=1}^n (B_{s_{j-1}}+(B_{s_j}-B_{s_{j-1}}))^3-B_{s_{j-1}}^3 \notag \\ &= \underbrace{\sum_{j=1}^n (B_{s_j}-B_{s_{j-1}})^3}_{=: I_1}+ 3 \underbrace{\sum_{j=1}^n B_{s_{j-1}} \cdot (B_{s_j}-B_{s_{j-1}})^2}_{=:I_2} + 3 \underbrace{\sum_{j=1}^n B_{s_{j-1}}^2 \cdot (B_{s_j}-B_{s_{j-1}})}_{=: I_3} \tag{1}\end{align*}

We consider the terms separately and start with $I_3$:

For \begin{equation*} g^\Pi(s,w) := \sum_{j=1}^n \underbrace{B_{s_{j-1}}^2}_{\in L^2} \cdot 1_{[s_{j-1},s_j)}(s) \end{equation*}

we have \begin{align*} \mathbb{E} \left( \int_0^T |g^\Pi(s)-B_s^2|^2 \, ds \right) &\stackrel{\text{Fub}}{=} \sum_{j=1}^n \int_{s_{j-1}}^{s_j} \mathbb{E}(|B_{s_{j-1}}^2-B_s^2|^2) \, ds \\ &\leq \sup_{|s-t| \leq |\Pi|} |B_s^2-B_t^2|^2 \sum_{j=1}^n \int_{s_{j-1}}^{s_j} 1 \ \, ds \\ &\to 0 \qquad (|\Pi| \to 0)\end{align*}

where $|\Pi| := \max (s_j-s_{j-1})$ denotes the mesh size of the partition $\Pi = \{0=s_0<\ldots<s_n=T\}$. In the last step, we have used that $[0,T] \ni t \mapsto B_t$ is uniformly continuous (almost surely). From the definition of the Itô integral, it follows that

$$g^\Pi \bullet B_T = \sum_{j=1}^n B_{s_{j-1}}^2 \cdot (B_{s_j}-B_{s_{j-1}}) \stackrel{L^2}{\to} \int_0^T B_s^2 \, dB_s$$

Hence, $$I_3 \stackrel{|\Pi| \to 0}{\to} \int_0^T B_s^2 \, dB_s. \tag{2}$$

Next we consider $I_2$. To this end, let $X_j := B_{s_{j-1}} \cdot ((B_{s_j}-B_{s_{j-1}})^2-(s_j-s_{j-1})$. Then \begin{align*} \left\| \left( \sum_{j=1}^n X_j \right)^2 \right\|_{L^2(\mathbb{P})}^2 &\stackrel{\ast}{=} \sum_{j=1}^n \mathbb{E}(X_j^2) = \sum_{j=1}^n \mathbb{E}((B_{s_{j-1}}^2 \cdot ((B_{s_j}-B_{s_{j-1}})^2-(s_j-s_{j-1})))^2) \\ &= \sum_{j=1}^n \underbrace{\mathbb{E}(B_{s_{j-1}}^2)}_{s_{j-1}} \cdot \underbrace{\mathbb{E}( ((B_{s_j}-B_{s_{j-1}})^2-(s_j-s_{j-1}))^2)}_{\stackrel{L1.6}{=} (s_j-s_{j-1})^2} \\ &= \sum_{j=1}^n s_{j-1} \cdot (s_j-s_{j-1})^2 \leq |\Pi| \cdot \underbrace{\sum_{j=1}^n s_{j-1} \cdot (s_j-s_{j-1})}_{\leq T^2} \\ &\to 0 \qquad (|\Pi| \to 0) \end{align*} For $(\ast)$ note that \begin{align*} \mathbb{E}(X_j \cdot X_k) &= \mathbb{E}(\mathbb{E}(X_j \cdot X_k|\mathcal{F}_{k-1})) \\ &= \mathbb{E}(X_j \cdot B_{s_{k-1}} \cdot \underbrace{\mathbb{E}((B_{s_j}-B_{s_{j-1}})^2)-(s_j-s_{j-1})|\mathcal{F}_{k-1})}_{= \mathbb{E}((B_{s_j}-B_{s_{j-1}})^2-(s_j-s_{j-1}) = 0}) = 0 \end{align*} for all $j>k$ where we have used that $B_{s_j}-B_{s_{j-1}}$ is independent from $\mathcal{F}_{k-1}:=\mathcal{F}_{s_{k-1}}$. Consequently, there exists a subsequence converging almost surely to $0$. Since $$\sum_{j=1}^n B_{s_{j-1}} \cdot (s_j-s_{j-1}) \to \int_0^T B_s \, ds \quad \text{a.s.} \quad (|\Pi| \to 0)$$ we conclude $$I_2 \to \int_0^T B_s \, ds \quad \text{a.s.} \quad (|\Pi| \to 0) \tag{3}$$

It remains to determine $\lim_{|\Pi| \to 0} I_1.$ To this end, we note that \begin{align*} \mathbb{E} \left( \left| \sum_{j=1}^n (B_{s_j}-B_{s_{j-1}})^3 \right| \right) &\leq \sum_{j=1}^n \mathbb{E}(|B_{s_j}-B_{s_{j-1}}| \cdot |(B_{s_j}-B_{s_{j-1}})^2|) \\ &\stackrel{\text{CSI}}{\leq} \sum_{j=1}^n \underbrace{\|B_{s_j}-B_{s_{j-1}}\|_{L^2(\mathbb{P})}}_{\sqrt{s_j-s_{j-1}}} \cdot \underbrace{\|(B_{s_j}-B_{s_{j-1}})^2\|_{L^2(\mathbb{P})}}_{\sqrt{3 (s_j-s_{j-1})^2}} \\ &= \sqrt{3} \cdot \sum_{j=1}^n (s_j-s_{j-1})^{\frac{3}{2}} \leq |\Pi|^{\frac{1}{2}} \cdot \underbrace{\sum_{j=1}^n (s_j-s_{j-1})}_{T} \\ &\to 0 \quad (|\Pi| \to 0) \end{align*} i.e. $I_1(\Pi) \stackrel{L^1}{\to} 0$ as $|\Pi| \to 0$. Consequently, there exists a subsequence converging to $0$ almost surely.

Adding all up, we get \begin{align*} B_T^3 &\stackrel{(1)}{=} \lim_{|\Pi| \to 0} \bigg( \sum_{j=1}^n (B_{s_j}-B_{s_{j-1}})^3 + 3 \underbrace{\sum_{j=1}^n B_{s_j} \cdot (B_{s_j}-B_{s_{j-1}})^2}_{\stackrel{(3)}{\to} \int_0^T B_s \, ds} + 3 \cdot \underbrace{\sum_{j=1}^n B_{s_{j-1}}^2 \cdot (B_{s_j}-B_{s_{j-1}})}_{\stackrel{(2)}{\to} \int_0^T B_s^2 \, dB_s} \bigg) \\ &= 0 + 3 \cdot \int_0^T B_s \, ds + 3 \cdot \int_0^T B_s^2 \, dB_s \end{align*}

• Thank you. The proof was indeed long. – user3503589 Jan 19 '15 at 16:16
• What happened to the square in the LHS of $$\mathbb{E} \left( \int_0^T |g^\Pi(s)-B_s^2|^2 \, ds \right) \stackrel{\text{Fub}}{=} \sum_{j=1}^n \int_{s_{j-1}}^{s_j} \mathbb{E}(|B_{s_{j-1}}^2-B_s^2|) \, ds$$ ? – user893458 Mar 29 '17 at 22:33
• @AlexW Note that $L^2(\mathbb{P})$-convergence implies $L^1(\mathbb{P})$-convergence which means that all three terms converge in $L^1(\mathbb{P})$. (Actually it would be enough to show that all the terms converge in probability.) – saz Jun 28 '18 at 13:51
• @AlexW That's correct, but that's not the point. Abstractly speaking: We have three sequences $(X_n)_n$, $(Y_n)_n$ and $(Z_n)_n$ and we know that $X_n \to X$ in $L^1$, $Y_n \to Y$ and $Z_n \to Z$ in $L^2$. All we need to conclude is that this implies $X_n+Y_n+Z_n \to X+Y+Z$ in $L^1$. (It doesn't matter whether e.g. $Y$ is a stochastic integral which is itself defined as an $L^2$-limit) – saz Jun 28 '18 at 14:04
• @AlexW Yes, exactly... but convergence in probability would be enough. (And just as a side remark: it is possible to define the stochastic integral as a limit in probability... in fact, this definition is somewhat convenient if we want to define stochastic integrals with respect to processes which are less nice than Brownian motion (e.g. processes which do not have a finite 2nd moment)) – saz Jun 28 '18 at 15:50