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i need to use the definition of the stochastic integral via Riemann-Stieltjes sums to prove that $\int_0^t B_s^2 dB_2 = 1/3 B_t^3 - \int_0^t B_sds $. ($B_s$ Brownian motion)

I figuered out, that $1/3 B_t^3 - \int_0^t B_sds = 1/3 \sum_{k=1}^n( B_{t_k}^3-B_{t_{k-1}}^3 - 3B_{t_{k-1}}(t_k-t_{k_1})) \\ = \sum_{k=1}^n B_{t_{k-1}}^2(B_{t_k}-B_{t_{k-1}}) + \sum_{k=1}^n [B_{t_{k-1}}(B_{t_k}-B_{t_{k-1}})^2 + \frac{1}{3}(B_{t_k}-B_{t_{k-1}})(B_{t_k}-B_{t_{k-1}})^2 - B_{t_{k-1}}(t_k-t_{k-1})] = \\ \int_0^t B_s^2 dB_s + \sum_{k=1}^n [B_{t_{k-1}}(B_{t_k}-B_{t_{k-1}})^2 + \frac{1}{3}(B_{t_k}-B_{t_{k-1}})(B_{t_k}-B_{t_{k-1}})^2 - B_{t_{k-1}}(t_k-t_{k-1})]$.

But now i don't know, how i should show that $\sum_{k=1}^n [B_{t_{k-1}}(B_{t_k}-B_{t_{k-1}})^2 + \frac{1}{3}(B_{t_k}-B_{t_{k-1}})(B_{t_k}-B_{t_{k-1}})^2 - B_{t_{k-1}}(t_k-t_{k-1})]=0$. Does anyone have a hint? or do i approach it not in the right way?

Nelli0607
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  • If you're allowed to use Ito lemma, then use it. If not, then you're right in the way of proving this, but you have an calculation error. Also it might be usefull to consider this question – openspace Mar 09 '23 at 17:04
  • @openspace Thanks for the respond, unfortunately i am not allowed to use ito's lemma. I think i had some typos above. Now it should be correct. About that other question. I dont see why, $3B_j(\Delta B_j)^2 = 3\int B_s ds$. – Nelli0607 Mar 09 '23 at 17:38
  • That's the fact from the Ito calculus: $dB_s^2 = ds$. You can check it using $L_2$ limits. – openspace Mar 09 '23 at 20:30

1 Answers1

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Using Taylor

\begin{align*} f(X)=f(X_i)+f^{\prime}(X_i)(X-X_i)+\dfrac{f^{\prime \prime}(X_i)}{2}(X-X_i)^{2}+O((X-X_i)^{3}) \end{align*}

where $f(X)=X^3$, $X=B_{t_i}$ and $X_i=B_{t_{i-1}}$. Then,

\begin{align*} B_{t_i}^{3}& =B_{t_{i-1}}^{3}+3B_{t_{i-1}}^{2}(B_{t_{i}}-B_{t_{i-1}})+3B_{t_{i-1}}(B_{t_{i}}-B_{t_{i-1}})^{2}+O((B_{t_{i}}-B_{t_{i-1}})^{3}) \\ %&=B_{t_{i-1}}^{3}+3B_{t_{i-1}}^{2}(B_{\Delta t_{i}})+3B_{t_{i-1}}(B_{\Delta t_{i}}^{2})+O(B_{\Delta t_{i}}^{3}) \\ &=B_{t_{i-1}}^{3}+3B_{t_{i-1}}^{2}(\Delta B_{t_{i}})+3B_{t_{i-1}}(\Delta B_{ t_{i}})^{2}+O((\Delta B_{t_{i}})^{3}) \end{align*}

Sum on $\Pi_n$ a partition of $[0,t]$, we have \begin{align*} \sum_{\Pi_n} B_{t_i}^{3}-B_{t_{i-1}}^{3}=\sum_{\Pi_n} 3B_{t_{i-1}}^{2}(\Delta B_{t_{i}})+3B_{t_{i-1}}(\Delta B_{ t_{i}})^{2}+O((\Delta B_{t_{i}})^{3}), \end{align*}

adding and subtracting

\begin{align*} \sum_{\Pi_n} B_{t_i}^{3}-B_{t_{i-1}}^{3}=\sum_{\Pi_n} 3B_{t_{i-1}}^{2}(\Delta B_{t_{i}})+3B_{t_{i-1}}(\Delta B_{ t_{i}})^{2} - 3B_{t_i}\Delta t_i + 3B_{t_i}\Delta t_i + O((\Delta B_{t_{i}})^{3}). \end{align*}

So,

\begin{align*} \lim_{n \to \infty} \sum_{\Pi_n} B_{t_i}^{3}-B_{t_{i-1}}^{3} = & \lim_{n \to \infty} \sum_{\Pi_n} 3B_{t_{i-1}}^{2}(\Delta B_{t_{i}})+ \lim_{n \to \infty} \sum_{\Pi_n}3B_{t_{i-1}}[(\Delta B_{ t_{i}})^{2} - \Delta t_i] \\ &+ \lim_{n \to \infty} \sum_{\Pi_n}3B_{t_i}\Delta t_i + \lim_{n \to \infty} \sum_{\Pi_n}O((\Delta B_{t_{i}})^{3}). \end{align*}

Also, note that $$ \lim_{n \to \infty} \sum_{\Pi_n} 3 B_{t_{i-1}}[(\Delta B_{ t_{i}})^{2} - \Delta t_i] \displaystyle \longrightarrow 0 \ \mbox{ in } L^{2}$$

and $$ \lim_{n \to \infty} \sum_{\Pi_n}O((\Delta B_{t_{i}})^{3}) \longrightarrow 0 \ \mbox{in Probability.}$$

Therefore,

\begin{align*} B_{t}^{3} = \int_0^t 3B_{s}^{2} \ d B_{s} +\int_0^t 3B_{s}\ ds \ \Longrightarrow \ \int_0^t B_{s}^{2} \ dB_s = \dfrac{B_t^3}{3}- \int_0^t B_{s}\ ds. \end{align*}

Jesus Carreño
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