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In the level of generality you have stated, I have no ideas. But in some cases this can be reformulated as a type of backward SDE, as studied in section 2.2 of http://arxiv.org/pdf/0801.3505.pdf in the case that $X$ and $H$ are continuous. It is no simple problem, but a couple of special cases seem approachable.

The simplest case is if $\int H_- dX$ is a martingale. Then $y(t,T) = H_t$ is a solution. An example if $X$ is a square integrable martingale and $H$ satisfies $\mathbb{E}[\int H^2_- d[X,X]] < \infty$.

Somewhat more generally, suppose $X$ is a continuous semimartingale, so $X = M + A$ for unique continuous local martingale $M$ and continuous FV process $A$. Assume moreover that $M$ is a true martingale and that your filtration is generated by a Brownian motion $W$. Let's look for a solution $y(\cdot,T)$ among the class of continuous processes $Y$ such that $\int Y dM$ is a martingale. For such $Y$ the equation becomes

$Y_t = H_t + \mathbb{E}[\int_t^TY_sdAs | \mathcal{F}_t]$.

To make this look more like a standard BSDE, note that if such a $Y$ exists then $N_t = \mathbb{E}[\int_0^TY_sdAs | \mathcal{F}_t]$ is a martingale and

$dY_t = dH_t - Y_tdA_t + dN_t$, with $Y_T = H_T$.

Now forget about the constraint that $\int Y dM$ is a martingale, and let's look for a pair of processes $(Y,N)$ (Y,N)$, with$N$a martingale, satisfying this equation. This is a special case of the problem studied in the aforementioned paper: see equation (2.19) with$\xi = H_T$,$J = -H$,$g \equiv 0$, and$f(s,y) = y$. Theorem 2.2 of said paper will guarantee you the existence of a unique process$Y \in \mathbb{S}^p$satisfying this equation, as long as$H \in \mathbb{S}^p$and there exist two continuous BMO martingales$N_1$and$N_2$with$\langle N_1, N_2 \rangle = A$. Here$\mathbb{S}^p$is the set of continuous adapted processes whose supremum has finite$p^{th}$moment. Now if it happens that$\int Z dM$is a martingale whenever$Z \in \mathbb{S}^p$, then this$Y$uniquely solves your equation. 1 In the level of generality you have stated, I have no ideas. But in some cases this can be reformulated as a type of backward SDE, as studied in section 2.2 of http://arxiv.org/pdf/0801.3505.pdf in the case that$X$and$H$are continuous. It is no simple problem, but a couple of special cases seem approachable. The simplest case is if$\int H_- dX$is a martingale. Then$y(t,T) = H_t$is a solution. An example if$X$is a square integrable martingale and$H$satisfies$\mathbb{E}[\int H^2_- d[X,X]] < \infty$. Somewhat more generally, suppose$X$is a continuous semimartingale, so$X = M + A$for unique continuous local martingale$M$and continuous FV process$A$. Assume moreover that$M$is a true martingale and that your filtration is generated by a Brownian motion$W$. Let's look for a solution$y(\cdot,T)$among the class of continuous processes$Y$such that$\int Y dM$is a martingale. For such$Y$the equation becomes$Y_t = H_t + \mathbb{E}[\int_t^TY_sdAs | \mathcal{F}_t]$. To make this look more like a standard BSDE, note that if such a$Y$exists then$N_t = \mathbb{E}[\int_0^TY_sdAs | \mathcal{F}_t]$is a martingale and$dY_t = dH_t - Y_tdA_t + dN_t$, with$Y_T = H_T$. Now forget about the constraint that$\int Y dM$is a martingale, and let's look for a pair of processes$(Y,N)$satisfying this equation. This is a special case of the problem studied in the aforementioned paper: see equation (2.19) with$\xi = H_T$,$J = -H$,$g \equiv 0$, and$f(s,y) = y$. Theorem 2.2 of said paper will guarantee you the existence of a unique process$Y \in \mathbb{S}^p$satisfying this equation, as long as$H \in \mathbb{S}^p$and there exist two continuous BMO martingales$N_1$and$N_2$with$\langle N_1, N_2 \rangle = A$. Here$\mathbb{S}^p$is the set of continuous adapted processes whose supremum has finite$p^{th}$moment. Now if it happens that$\int Z dM$is a martingale whenever$Z \in \mathbb{S}^p$, then this$Y\$ uniquely solves your equation.