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Applying Feynman–Kac to a Diffusion PDE with a Terminal Payoff

Article Quant Q&A · Author: Sriram Nagaraj
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Summary

The document shows how to translate a terminal-value partial differential equation into a stochastic expectation using the Feynman–Kac formula. For the operator with diffusion coefficient proportional to the state, it identifies a geometric Brownian motion with zero drift as the associated process. Its terminal value is the fourth power of the log of that process, whose distribution follows from the integrated Brownian increment.

The source term in the PDE contributes an accumulated running payoff, while the terminal condition contributes the expected terminal payoff. The answers disagree on the sign of the running contribution: the accepted response writes a positive time interval, while another response corrects it to a negative interval. Since the PDE has a positive constant source term, the latter sign is consistent with the stated equation under the usual backward Feynman–Kac convention. The discussion sets up the expectation but omits its final moment expansion.

Key ideas

  • The PDE's diffusion operator corresponds to a state process with proportional Brownian volatility and zero drift.
  • Feynman–Kac expresses the solution as an expectation of accumulated source terms and the terminal payoff.
  • The process's log terminal value is normally distributed, allowing the fourth moment to be evaluated from Gaussian moments.
  • The source term's sign in the expectation matters, and one answer explicitly corrects the other answer's sign.

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StatisticsDerivatives pricing
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# How to apply the Feynman-Kac formula?


# How to apply the Feynman-Kac formula?












I've been learning about Feynman-Kac recently and I understand the underlying ideas. I am stuck however in actually computing explicit solutions for specific problems.

For example, suppose I have the following terminal value problem:

$$F_t + \frac{1}{2}\sigma^2x^2F_{xx}=1$$

$$F(x,T) = \ln(x)^4,~x>0$$

How would I compute $F(x,t)$ in closed form, given the closed form of the right hand side $(ln(x))^4$ using Feynman-Kac?

## Answer by Gordon (score 2, accepted)

https://quant.stackexchange.com/a/22083

Based on the form of your equation, we can consider the SDE \begin{align*} dX_t = \sigma X_t dW_t, \end{align*} where $W$ is a standard Brownian motion under the measure $Q$. Since, for $0 \leq t \leq T$, \begin{align*} X_T = X_t \exp\left(-\frac{1}{2}\sigma^2 (T-t) + \sigma \int_t^T dW_s \right), \end{align*} based on Feynman–Kac formula, the solution is given by \begin{align*} F(t, x) &= E^Q\left(\int_t^T ds + (\ln X_T)^4 \mid X_t = x\right)\\ &=(T-t) + E^Q\left[\left(\ln x -\frac{1}{2}\sigma^2 (T-t) + \sigma \int_t^T dW_s\right)^4\right]. \end{align*} The remaining is now simple and is omitted.

## Answer by PTQuoc (score 3)

https://quant.stackexchange.com/a/52876

There is a little flaw in this equation: \begin{align*} F(t, x) &= E^Q\left(\int_t^T ds + (\ln X_T)^4 \mid X_t = x\right)\\ &=(T-t) + E^Q\left[\left(\ln x -\frac{1}{2}\sigma^2 (T-t) + \sigma \int_t^T dW_s\right)^4\right]. \end{align*}

The correct one should be: \begin{align*} F(t, x) &= E^Q\left( - \int_t^T ds + (\ln X_T)^4 \mid X_t = x\right)\\ &=-(T-t) + E^Q\left[\left(\ln x -\frac{1}{2}\sigma^2 (T-t) + \sigma \int_t^T dW_s\right)^4\right]. \end{align*}

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