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I have asked a similar question involving some finance background some time ago here math.stackexchange, however no really good answer came up. I was able to find a solution at least for a special case. Removing all unnecessary information, I try to solve the following problem.

Given a given a martingale $H_t$ and a semimartingale $X_t$, let $y(t,T)$ be the process solving

$ y(t,T)=H_t + \mathbb{E}[\int_t^T y(s-,T)dX_s|\mathcal{F}_t] $

if a solution exists. I wouldn't even call this an SDE. I was able to come up with a solution in case of a deterministic $X_t$, but got stuck otherwise. It looks very similar to a OU-type SDE and looking at the solution of this kind of SDE, I thought

$y(t,T)=\mathbb{E}\left[ \mathcal{E}(X)_t\left( \frac{H_t}{\mathcal{E}(-X)_T}+\int_t^T\mathcal{E}(X)^{-1}_s(dH_s-d\langle H,X\rangle_s) \right)|\mathcal{F}_t\right]$

might work. This was some kind of educated guess, and for deterministic $X$ works fine (the integral vanishes). Also since $H$ is a martingale the integral simplifies. I'd like to solve for general $X$ (or at least an Ito-process).

Can standard SDE-Theory be applied here in any way? Is there a general method to solve such problems? Results about existence and uniqueness? I couldn't find anything in the literature. I'd be grateful for any hints.

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1 Answer 1

up vote 4 down vote accepted

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 $(Y,N)$, 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.

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Thank you! I'll check that paper out and let you know! –  Pierre Sep 13 '12 at 19:48

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