Calculating Ito Integral from definition.

Question

I have to calculate the following integral using the Definition of Ito Integral: $$\int_0^TsB_sdB_s.$$ Where $B_s$ is a brownian motion. I've already shown that the sequence of simple funcions $$\varphi_n(s)=\sum_{j=0}^{n-1}t_jB_{t_j}\mathbb{1}_{[t_j,t_{j+1})}(s)$$ approximates given integral. But now I'm stuck on finding the limit of $$\sum_{j=0}^{n-1}t_jB_{t_j}\Delta B_{t_j}$$ as $\Delta t_j\to0$. I've also used Ito Fromula to see where im heading with this limit and got the following result: $$\int_0^TsB_sdB_s=\frac{B_T^2T}{2}-\frac{T^2}{4}-\frac{1}{2}\int_0^TB_s^2ds,$$ but after trying for several hours i could not find a way to get to this, to the point where I'm not even sure the equation above is correct.

Answer 1

After you fixed the typos your last formula is correct. The integration by parts formula (hence any formula that follows from it) can be shown directly using the definition of the Ito integral: Let $X,Y$ be two continuous semimartingales. Then \begin{align} &X_TY_T-X_0Y_0=\lim_{n\to\infty}\sum_{i=0}^{n-1}X_{t_{i+1}}Y_{t_{i+1}}-X_{t_{i}}Y_{t_{i}}\quad\text{(telescoping sum)}\\ &=\lim_{n\to\infty}\sum_{i=0}^{n-1}X_{t_{i}}(Y_{t_{i+1}}-Y_{t_{i}})+Y_{t_{i}}(X_{t_{i+1}}-X_{t_{i}})+(X_{t_{i+1}}-X_{t_{i}})(Y_{t_{i+1}}-Y_{t_{i}})\\ &=\int_0^TX_s\,dY_s+\int_0^TY_s\,dX_s+\langle X,Y\rangle_T\,.\tag{1} \end{align} Your case leads to an interesting consequence I was not aware of before:

Claim. $$\boxed{\quad \lim_{n\to\infty}\sum_{i=0}^{n-1} t_i\,(B_{t_{i+1}}-B_{t_i})^2=\int_0^Tt\,d\langle B\rangle_t\,\quad}\tag{2} $$ (which of course equals $\int_0^Tt\,dt=\frac{T^2}{2})\,.$

Proof of (2). The Ito integral $2\int_0^Tt\,B_t\,dB_t$ is the limit of

\begin{align} &2\sum_{i=0}^{n-1}t_i\,B_{t_i}\,(B_{t_{i+1}}-B_{t_i})=2\sum_{i=0}^{n-1}t_i\,B_{t_i}\,B_{t_{i+1}}-t_i\,B^2_{t_i}\\ &=\sum_{i=0}^{n-1}t_{i+1}B^2_{t_{i+1}}-\,t_i\,B^2_{t_i}-B^2_{t_{i+1}}(t_{i+1}-t_i)-t_i\,B^2_{t_{i+1}}+2\,t_i\,B_{t_i}\,B_{t_{i+1}}-t_i\,B_{t_i}^2\\ &=T\,B^2_T-\underbrace{\sum_{i=0}^{n-1}B^2_{t_{i+1}}(t_{i+1}-t_i)}_{(*)}+t_i\,(B_{t_{i+1}}-B_{t_i})^2\,.\tag{3} \end{align} Since (*) converges to $\int_0^TB_t^2\,dt$ it follows from your formula that (2) must hold. $$\tag*{$\Box$} \quad $$

More general Claims. Let $A_t$ be a continuous process of finite variation and $X_t$ be a continuous semi martingale. Then, \begin{align} 2\int_0^TA_tX_t\,dX_t=A_TX_T^2-\int_0^TX_t^2\,dA_t-\int_0^TA_t\,d\langle X\rangle_t\,.\tag{4} \end{align} \begin{align}\boxed{\quad \lim_{n\to\infty}\sum_{i=0}^{n-1}A_{t_i}(X_{t_{i+1}}-X_{t_i})^2=\int_0^TA_t\,d\langle X\rangle_t\,.\quad}\tag{5} \end{align} Proof. Integration by parts yields \begin{align} d(A_tX_t)&=X_t\,dA_t+A_t\,dX_t\,,\\[2mm] d(A_tX_t^2)&=A_tX_t\,dX_t+X_t\,d(A_tX_t)+d\langle A_tX_t,X_t\rangle\\[2mm] &=A_tX_t\,dA_t+X_t^2\,dA_t+A_tX_t\,dX_t+A_t\,d\langle X\rangle_t\\[2mm] &=2A_tX_t\,dX_t+X_t^2\,dA_t+A_t\,d\langle X\rangle_t\,. \end{align} This shows (4). The claim (5) is shown exactly like (3) $$\tag*{$\Box$} \quad $$