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I have a PDE which involves the predictable finite-variation parts of some semi-martingales and a quadratic-covariation process and I tried to derive a Feynman-Kac style expectation from the PDE. However, the result looks somewhat strange: it is in the form $$ f(t, x) = \mathbb E \left[f(T, X_T)+\int^T_t d[Z, X]_s \,\bigg|\, X_t = x\right] $$ where $Z_t = f(t, X_t)$ and $X_t$ is a semi-martingale and $[.,.]_t$ is quadratic-covariation. The reason I have doubts about this is that, evidently, the expectation contains itself. Typically, Feynman-Kac formulas take forms like $$ f(t, x) = \mathbb E \left[ f(T, X_t) \,\bigg|\, X_t = x\right] $$ where $f(T, X_T)$ corresponds to some terminal condition, but in my case, the integral inside the expectation includes the solution itself at every point between time $t$ and $T$.

Might it then be inferred that differential operators corresponding to the drift-terms of a quadratic-variation process do not lead to meaningful Feynman-Kac representations?


Derivation of the expression: Assume $X$ is a semi-martingale with the Markov property. Also assume $[Z, X]_t$ can be decomposed as $[Z, X]_t = M_t + A_t$ where $M_t$ is a local martingale and $$A_t = \int^t_0 Lf(s, X_s) ds$$ is its predictable finite-variation part where $L$ is some differential operator. Likewise assume $Z_t=M^Z_t + A^Z_t $ where $$ A^Z_t = \int^t_0 Kf(s, X_s)ds $$ where $K$ is another operator. Now take the PDE $$ Lf(t, x)=Kf(t, x) $$ Define $Y_s = Z_s - \int^s_t d[Z, X]$. Assume $f(t, x)$ solves the PDE. Then clearly $dA_t-dA^Z_t=0$ so we get $$ Y_T-Y_t = \int^T_t d(M_s - M^Z_s) $$ and taking expectations and conditioning on $X_t = x$, the expectation of the integral is zero so we get $$ E[Y_T \mid X_t = x] = E[Y_t \mid X_t = x] = f(t, x) $$

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  • $\begingroup$ it could help to add the derivation steps $\endgroup$ – Nikos M. Sep 8 '15 at 21:18

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