On the orthogonal component of BSDEs in a Markovian setting

TL;DR

This paper proves that in a Markovian setting, the orthogonal component of a quadratic BSDE driven by a continuous martingale vanishes, simplifying the solution to just (Y,Z) without the need for an additional orthogonal martingale.

Contribution

It demonstrates that the orthogonal martingale component in quadratic BSDEs disappears in Markovian settings, simplifying the solution structure.

Findings

Orthogonal component N is zero in Markovian quadratic BSDEs.

Solution reduces to (Y,Z) without orthogonal martingale.

Orthogonal component vanishes even without martingale representation property.

Abstract

In this Note we consider a quadratic backward stochastic differential equation (BSDE) driven by a continuous martingale $M$ and whose generator is a deterministic function. We prove (in Theorem \ref{theorem:main}) that if $M$ is a strong homogeneous Markov process and if the BSDE has the form \eqref{BSDE} then the unique solution $(Y,Z,N)$ of the BSDE is reduced to $(Y,Z)$, \textit{i.e.} the orthogonal martingale $N$ is equal to zero showing that in a Markovian setting the "usual" solution $(Y,Z)$ has not to be completed by a strongly orthogonal even if $M$ does not enjoy the martingale representation property.