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For a semilinear PDE, we usually have this FBSDE representation:

$\mathcal{X}_t=\mathcal{X}_0+\int^t_0 \mu (s,\mathcal{X}_s)\, ds\, +\int_0^t \sigma (s,\mathcal{X}_s)dW_s,\quad 0\leq t\leq T, \\ Y_t = g(\mathcal{X}_T)+ \int_t^T f(s, \mathcal{X}_s,Y_s,Z_s)ds - \int_t^TZ_sdW_s, \quad 0\leq t\leq T,$

where $W$ is a $d$-dimensional standard Brownian motion on some probability space $(\Omega,\mathbb{F},\mathbb{P})$ equipped with a filtration $\mathbb{F}=(\mathcal{F}_t)_t$, and $\mathcal{X}_0$ is an $\mathcal{F}_0$-measurable random variable valued in $\mathbb{R}^d$.

Now in this paper: https://arxiv.org/pdf/2006.01496.pdf on page 7, equation (2.11) they use this representation: $Y_t = g(\mathcal{X}_T) - \int_t^T f(s, \mathcal{X}_s,Y_s,Z_s)ds - \int_t^TZ_sdW_s, \quad 0\leq t\leq T,$

My question: How we get the first minus? I didn't find any literature/reference who stated this result. Can someone link me some reference or explain it? Any help is appreciated. Thanks.

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  • $\begingroup$ if you look at equation (1.1) in that paper, you see that $\partial u/d\partial t = f+\text{other terms}$. Since $Y=u$ you need the minus sign, $Y=-\int_t^T f(s)ds+\text{other terms}$, to ensure the $\partial Y/\partial t=f(t)+\text{other terms}$. So it all works out, doesn't it? $\endgroup$
    – Carlo Beenakker
    Commented Sep 4, 2021 at 18:53
  • $\begingroup$ Thanks Carlo, you are absolutely right. $\endgroup$
    – freshst4r
    Commented Sep 5, 2021 at 10:34

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