I quote Øksendal (2003)
Statement. Start from a 1-dimensional Brownian motion $B_t$. Assume $B_0=0$. Then $$\displaystyle{\int_0^t}B_sdB_s=\displaystyle{\frac{1}{2}B_t^2}-\displaystyle{\frac{1}{2}t}$$
Proof. Put $\phi_n(s,\omega)=\sum B_j(\omega)\cdot\chi_{[t_j, t_{j+1}]}(s)$, where $B_j=B_{t_j}$ and $\chi$ denotes the indicator function on the subset $[t_j,t_{j+1}]$. Then:
\begin{align}\mathbb{E}\bigg[\int_0^t(\phi_n-B_s)^2ds)\bigg]&=\mathbb{E}\bigg[\sum_j\int_{t_j}^{t_{j+1}}(B_j-B_s)^2ds\bigg]\\&\color{red}{=}\sum_{j}\int_{t_j}^{t_{j+1}}(s-t_j)ds\\&=\cdots\end{align}
What I cannot understand is the $\color{red}{\text{red}}$ equality above. How can one pass from $$\mathbb{E}\bigg[\sum_j\int_{t_j}^{t_{j+1}}(B_j-B_s)^2ds\bigg]\tag{1}$$ to $$\sum_{j}\int_{t_j}^{t_{j+1}}(s-t_j)ds\tag{2}$$?
Possibly, which is the role of the outer expected value $\mathbb{E}$ (with respect to a probability measure $\mathbb{P}$, I guess) in this passage from $(1)$ to $(2)$?
Best Answer
Since for $s>t$, $\mathsf{E}(B_s-B_t)^2=\mathsf{E}B_{s-t}^2=s-t$, $$ \mathsf{E}\left[\sum_{j}\int_{t_j}^{t_{j+1}}(B_s-B_j)^2\,ds\right]=\sum_{j}\int_{t_j}^{t_{j+1}}\mathsf{E}(B_s-B_j)^2\,ds=\sum_{j}\int_{t_j}^{t_{j+1}}(s-t_j)\,ds. $$