Let $r_t, \theta_t$ denote some stochastic processes driven by a $N$ dimensional Brownian motion $W_t$ (they are of course assumed adapted to the natural filtration $\mathcal{F}_t$ of that Brownian motion). Now, Consider the conditional expectation: $$ C_t = \Bbb{E} \left[ \int_t^T f(r_s, \theta_s, W_s) ds \,\vert\, \mathcal{F}_t \right] $$ with $f$ a sufficiently well-behaved function. The paper that I am reading (can't insert reference because of NDA) claims that the above expression satisfies an SDE of the type $$ dC_t = \alpha_t dt + \nu_t dW_t $$ where $\alpha$ and $\nu$ are possibly stochastic due to the "martingale representation theorem".

I was wondering how to show that? Do you use some kind of Leibniz rule to express $dC_t$ and conclude? Do you rather you some more elaborate tricks like Doob-Meyer decomposition and the likes?


1 Answer 1


The martingale representation theorem says that for any martingale $M$, there exists a unique stochastic process $\nu_t$ such that \begin{align*} M_t = \mathbb{E}(M_0) + \int_0^t\nu_sdW_s. \end{align*} See, for example, this book.

Since \begin{align*} C_t &= \Bbb{E} \left( \int_t^T f(r_s, \theta_s, W_s) ds \mid \mathcal{F}_t \right)\\ &= -\int_0^t f(r_s, \theta_s, W_s) ds + \Bbb{E} \left( \int_0^T f(r_s, \theta_s, W_s) ds \mid \mathcal{F}_t \right), \end{align*} and $$\Bbb{E} \left( \int_0^T f(r_s, \theta_s, W_s) ds \mid \mathcal{F}_t \right)$$ is a martingale, there exists a a unique stochastic process $\nu_t$ such that \begin{align*} \Bbb{E} \left( \int_0^T f(r_s, \theta_s, W_s) ds \mid \mathcal{F}_t \right) = a + \int_0^t \nu_s dW_s, \end{align*} where $a$ is a constant. Let $\alpha_t = -f(r_t, \theta_t, W_t)$. Then \begin{align*} dC_t = \alpha_t dt + \nu_t dW_t. \end{align*}


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