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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 $(2)$?

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Since for $s>t$, $\mathsf{E}(B_s-B_t)^2=\mathsf{E}B_{s-t}^2=s-t$, $$ \mathsf{E}\left[\sum_{j}\int_{t_j}^{t_{j+1}}(B_s-B_j)^2\,ds\right]=\sum_{j}\int_{t_j}^{t_{j+1}}\mathsf{E}(B_s-B_j)^2\,ds=\sum_{j}\int_{t_j}^{t_{j+1}}(s-t_j)\,ds. $$

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  • $\begingroup$ The first equality $\mathsf{E}\left[\sum_{j}\int_{t_j}^{t_{j+1}}(B_s-B_j)^2\,ds\right]=\sum_{j}\int_{t_j}^{t_{j+1}}\mathsf{E}(B_s-B_j)^2\,ds$ follows from linearity of expectation $\mathsf{E}$, doesn'it? $\endgroup$ – Strictly_increasing Oct 15 '20 at 10:02
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    $\begingroup$ $$ \mathsf{E}\left[\int_{t_j}^{t_{j+1}}(B_s-B_j)^2\,ds\right]=\int_{\Omega}\left[\int_{t_j}^{t_{j+1}}(B_s(\omega)-B_j(\omega))^2\,ds\right]\mathsf{P}(d\omega) $$ $\endgroup$ – d.k.o. Oct 15 '20 at 11:31
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    $\begingroup$ The product measure here is $\mu \times \mathsf{P}$, where $\mu$ is the Lebesgue measure on $\mathbb{R}$. So the "joint differential" is $d(\mu\times \mathsf{P})(s,\omega)$. $\endgroup$ – d.k.o. Oct 15 '20 at 13:55
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    $\begingroup$ Pardon, just one very last observation: so, just for confirmation, in the reference it is meant that $$\int_{t_j}^{t_{j+1}}(B_s(\omega)-B_j(\omega))^2\,ds=\int_{t_j}^{t_{j+1}}(B_s(\omega)-B_j(\omega))^2\,d\mu(s)$$ with $\mu$ Lebeasgue measure on $\mathbb{R}$? ($f(s,\omega)$ is a stochastic process) $\endgroup$ – Strictly_increasing Oct 15 '20 at 15:55
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    $\begingroup$ By the way, the equality there should be stated as $$ \int_X\left(\int_Y f(x,y)d\nu(y)\right)d\mu(x)=\int_Y\left(\int_X f(x,y)d\mu(x)\right)d\nu(y)=\ldots $$ $\endgroup$ – d.k.o. Oct 15 '20 at 16:10

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