I'm studying the paper Estimation of time-dependent Hurst exponents with variational smoothing and application to forecasting foreign exchange rates in which the author presents the following definition of the multifractional Brownian motion (mBm) (section 2.1 page 4)
$$\tag1 \begin{align*} X(t) = \frac{\sigma\sqrt{\Gamma(2H(t)+1)\sin(H(t)\pi)}}{\Gamma(H(t)+1/2)} \Bigg( &\int_{-\infty}^0 ((t-s)^{H(t)-1/2}-(-s)^{H(t)-1/2}) dW(s) + {} \\ &+ \int_0^t (t-s)^{H(t)-1/2} dW(s) \Bigg), \end{align*} $$
with $H$ Hurst function, $\sigma$ volatility, $W$ standard Brownian motion. Then in section 5.1, at the step 4 of a simulation procedure (page 15), the author presents this equation to estimate $W$, saying that it is derived from the definition of the mBm
$$\tag2 W(s\delta) = W((s-1)\delta) + (X(s\delta) - X(t\delta))\Pi(s\delta) - \mathbb1_{s>t} \sum_{\color{red}{u=t}}^{s-1} [W(u\delta)-W((u-1)\delta)](s-u+1)^{H(s\delta)-1/2}, \\ \text{where}\quad \Pi(s\delta) = \frac{\Gamma(H(s\delta)+1/2}{\sigma \delta^{H(s\delta)-1/2} \sqrt{\Gamma(2H(s\delta)+1)\sin(H(s\delta)\pi)}}, \\ %\sigma = \frac1\tau \sqrt\frac{\pi}{2} \sum_{s=t}^{t+\tau-1} \frac{|X(s\delta)-X((s-1)\delta)|}{\delta^{H(s\delta)}}, \\ \delta = 1/(\tau-1) \text{ is the time step}, \\ \tau \text{ is a positive integer.} $$
I did the computations to obtain $(2)$ from $(1)$, but there is a small detail which I don't understand, why is $u=t$ the lower bound of summation in equation $(2)$?
First of all, in order to run the simulation procedure, the equations are discretized introducing the time step $\delta$, and everything is defined for the equispaced times $0, \delta, 2\delta, ..., T\delta$, with $T$ final time. The procedure requires to compute $X((t+\tau)\delta)$ for every $t\in\{0,...,T-\tau\}$ using the values of $W(s\delta)$ for $s\in\{t,...,t+\tau-1\}$, where $\tau < T$ is a positive integer of our choice.
So I guess we should start by discretizing equation $(1)$, using the same notation as in equation $(2)$, i.e. use $s$ instead of $t$ and $u$ instead of $s$. Notice that since $t \ge 0$ then the first integral in equation $(1)$ vanishes. Moreover, moving the fraction from right to left we obtain
$$ \frac{\Gamma(H(s\delta)+1/2)}{\sigma\sqrt{\Gamma(2H(s\delta)+1)\sin(H(s\delta)\pi)}} X(s\delta) = \int_0^s ((s-u)\delta)^{H(s\delta)-1/2} dW(u\delta). $$
Now, quoting the paper: "at the origin $t$ of the time interval the mBm should be $0$, we translate $X$ by $X(t\delta)$", so we write
$$ \frac{\Gamma(H(s\delta)+1/2)}{\sigma\sqrt{\Gamma(2H(s\delta)+1)\sin(H(s\delta)\pi)}} (X(s\delta)-X(t\delta)) = \int_0^s ((s-u)\delta)^{H(s\delta)-1/2} dW(u\delta). $$
The integral can be approximated using Riemann sum: let $\pi_n = \{u_0,..,u_n\}$, with $0=u_0<...<u_n=s$, be a partition of the time interval $[0,s]$, with $|\pi_n|=\max\{u_i-u_{i-1}:i=1,...,n\}$, then
$$ \begin{align*} \int_0^s ((s-u)\delta)^{H(s\delta)-1/2} dW(u\delta) &= \lim_{|\pi_n| \to 0}\ \sum_{i=1}^n\ ((s-u_{i-1})\delta)^{H(s\delta)-1/2}(W(u_i\delta)-W(u_{i-1}\delta)) \\ &\approx \sum_{i=1}^n\ ((s-u_{i-1})\color{red}{\delta})^{H(s\delta)-1/2}(W(u_i\delta)-W(u_{i-1}\delta)). \end{align*} $$
Notice that since the red term does not depend on $i$, we can move it on the other side, hence obtaining the formula of $\Pi(s\delta)$ (see above)
$$\tag3 \Pi(s\delta) (X(s\delta)-X(t\delta)) = \sum_{i=1}^n\ (s-u_{i-1})^{H(s\delta)-1/2}(W(u_i\delta)-W(u_{i-1}\delta)). $$
Now consider equation (2), notice that $(s-u+1)$ is equal to $1$ when $u=s$, so we can add the term $W(s\delta) - W((s-1)\delta)$ into the series on the right hand side. By doing so we have to increase the upper bound of summation by $1$ (from $s-1$ to $s$), hence obtaining
$$\tag4 \sum_{u=t}^\color{red}s [W(u\delta)-W((u-1)\delta)](s-u+1)^{H(s\delta)-1/2}, $$
but how can we obtain this series starting from the series on the right hand side of $(3)$ ? I do agree that since $u_n = s$, then the upper bound of summation in $(4)$ must be $u = s$, but since $u_0=0$ why is the lower bound of summation equal to $u=t$? Shouldn't it be $u=u_1=u_0+1=1$?