Applying Feynman–Kac to a Geometric Brownian Motion PDE

Summary

The document applies the Feynman–Kac formula to a terminal-value partial differential equation with drift proportional to the state and diffusion proportional to the state. It identifies the associated process as geometric Brownian motion and uses its conditional distribution to express the solution as the expected logarithm of the terminal state cubed. The calculation reduces the result to a term involving the current state and a term involving the expected Brownian increment.

The key learning point is that the Brownian increment after the conditioning time has conditional mean zero, so the proposed expectation yields a closed-form solution. The discussion is a worked setup rather than a full treatment: it does not address domain conditions such as the positivity of the state required by the logarithm, nor does it discuss regularity assumptions behind Feynman–Kac. The original post asks whether the reasoning is correct and does not provide follow-up validation.

Key ideas

• A PDE with drift and diffusion proportional to the state corresponds to a geometric Brownian motion representation.
• The terminal payoff is the logarithm of the cubed terminal state.
• The future Brownian increment has zero conditional expectation given the current information.
• The logarithmic payoff requires the state to remain positive.

Tags

# Correct application of Feynman Kac formula


# Correct application of Feynman Kac formula












I have a question on Feynman-Kac formula but can I ask the community if I have done it correctly? If no, may you point out to where I went wrong? Thanks!

The original FK formula states: Assume $f(t,x)$ satisfies $$\frac{\partial f}{\partial t}(t,x) + \mu(t,x)\frac{\partial f}{\partial x} + \frac{1}{2}\frac{\partial^2 f}{\partial x^2}(t,x)\sigma^2(t,x) = 0$$, $f(T,x) = \phi(x)$. Then $f(s,x) = E_{s,x} [\phi(X_T)]$ where $dX_t = \mu(t,X_t)dt + \sigma(t,X_t)dW_t$.

Btw, $E_{s,x}[.]$ denotes the conditional expectation $E[.|\mathcal F_s]$ and $X_s = x$.

The given question I have is: Find solution $F(t,x)$ to $$\frac{\partial f}{\partial t} + \mu x\frac{\partial f}{\partial x} + \frac{1}{2}\frac{\partial^2 f}{\partial x^2}\sigma^2x^2 = 0$$ with terminal condition $F(T,x) = \ln(x^3)$

My solution is to choose an appropriate $X_t$ that satisfies the dynamics of $X_t$ in the FK formula. In this case, I believe the choice is the Geometric Brownian Motion that has solution $X_t = X_s e^{(\mu - \frac{\sigma^2}{2})(t-s) + \sigma (W_t-W_s)}$ then \begin{align}F(t,x) &= E_{t,x} [\ln(X_T^3)] \\ &= E_{t,x} \left[ 3\ln(x) + 3(\mu - \frac{\sigma^2}{2})(T-t) + 3\sigma (W_T-W_t)\right] \end{align} So far, am I right? But the last term equals $0$ since $E[W_T - W_t | \mathcal F_t] = E[W_T - W_t]$.

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