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Correlated Brownian Motions in a Feynman–Kac PDE

Article Quant Q&A · Author: Pefok
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Summary

The document sets up a two-dimensional Feynman–Kac problem with a mixed second derivative and a state-dependent source term. It represents the diffusion using two independent Brownian motions, linking the PDE's mixed derivative to correlation between the state processes. The question then applies the Feynman–Kac expectation formula to a terminal payoff exponential in both states and an integral of one state variable.

The author reports that the resulting candidate function fails when substituted into the PDE and asks where the calculation went wrong. The text includes the proposed correlation, diffusion coefficients, and candidate expression, but no accepted correction or final resolution. It is therefore useful as an illustration of translating covariance structure into a PDE and checking a candidate solution, while leaving the computational issue open.

Key ideas

  • A mixed second derivative in a diffusion PDE represents covariance between the underlying state processes.
  • Two correlated state processes can be expressed using independent Brownian drivers.
  • The Feynman–Kac representation combines terminal payoff expectations with an accumulated source term.
  • Substitution back into the PDE is a necessary check on a proposed solution.

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Full text
# Feymann Kac pde with correlated process


# Feymann Kac pde with correlated process












I have to solve the following PDE: \begin{equation} \begin{cases} \dfrac{\partial F}{\partial t}+\dfrac{1}{2}\dfrac{\partial^2 F}{\partial x^2}+\dfrac{1}{2}\dfrac{\partial^2 F}{\partial y^2}+\dfrac{1}{2}\dfrac{\partial^2 F}{\partial x\partial y}+y=0\\\\ F(t,x,y)=e^{x+y} \end{cases} \end{equation} In principle I assume that there exist two independent $\mathbb{P}$-Brownian motions $W^1$ and $W^2$ such that: \begin{equation} dX_t=\sigma_1dW^1_t\\\\ dY_t=\sigma_2(\rho dW^1_t+\sqrt{1-\rho^2}dW^2_t) \end{equation} Looking now at the coefficients of the PDE we have that $\sigma_1=1$, $\sigma_2\rho=\frac{1}{2}$ and $\sigma_2^2(1-\rho^2)=1$. From this we get $\rho=\frac{1}{\sqrt{5}}$ and $\sigma_2=\frac{\sqrt{5}}{2}$. Hence substituting in the previous I get: \begin{equation} dX_t=dW^1_t\\\\ dY_t=\frac{1}{2}dW^1_t+dW^2_t \end{equation} From which I obtain the dynamics: \begin{equation} X_T=X_t+(W^1_T-W^1_t)\\\\ Y_T=Y_t+\frac{1}{2}(W^1_T-W^1_t)+(W^2_T-W^2_t) \end{equation} Applying now Feymann Kac formula I get: \begin{equation} F(t,X_t,Y_t)=E(e^{X_T+Y_T})+\int^T_tE(Y_s)ds=E(\exp(X_t+\sqrt{T-t}Z_1+Y_t+\frac{1}{2}\sqrt{T-t}Z_1+\sqrt{T-t}Z_2))+Y_t(T-t) \end{equation} where $Z_1,Z_2$ are independent standard gaussian. Now solving the first expected value I get: \begin{equation} F(t,X_t,Y_t)=e^{X_t+Y_t+\frac{13}{8}(T-t)}+Y_t(T-t)\Rightarrow F(t,x,y)=e^{x+y+\frac{13}{8}(T-t)}+y(T-t) \end{equation} Now if I plug the solution into the pde I find that it doesn't work. Can someone tell me if the computation is fine or not?

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.