Solving a Geometric-Diffusion Terminal-Value PDE with Feynman–Kac

Summary

The example applies Feynman–Kac to a terminal-value partial differential equation with a fourth-power terminal payoff. The PDE contains a second derivative term proportional to the square of the asset price, and the proposed solution represents the value as the conditional expectation of the terminal fourth power under a geometric diffusion with zero drift. The question itself includes a positive drift in its asset process, while the answer uses a driftless process consistent with the displayed PDE.

Applying Itô’s lemma to the logarithm makes the diffusion’s distribution explicit, allowing the conditional moment to be evaluated with the moment-generating function of a normal variable. The answer gives a closed-form expression proportional to the current price raised to the fourth power, with exponential growth determined by volatility and remaining time. This illustrates the method for a simple payoff, but the answer flags possible algebraic error and does not resolve the drift discrepancy in the original question.

Key ideas

• Feynman–Kac represents the PDE solution as a conditional expectation of the terminal payoff.
• The displayed PDE corresponds to a diffusion without a drift term in its generator.
• Taking the logarithm of the geometric diffusion turns it into a process with normally distributed increments.
• The fourth conditional moment can be computed using the normal moment-generating function.
• The answer notes possible algebraic error and uses a different drift from the process stated in the question.

Tags

# An example of Feynman-Kac


# An example of Feynman-Kac












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, assume that $S_t$ is the price of an asset with SDE $dS_t = rS_tdt+ \sigma S_tdW_t$, where $r$ and $\sigma$ are positive numbers, and $W_t$ is a standard Brownian motion under some measure. Consider the function $f(t, S_t)$, dependent on time $t$ and on the price $S_t$. How to solve the following boundary problem where the domain is $[0,T]\times \mathbb{R}$: $$ f_t +\dfrac{1}{2}\sigma^2 S^2 f_{SS}=0$$ with terminal condition $f(T,S)=S^4$?

## Answer by Nap D. Lover (score 5)

https://quant.stackexchange.com/a/74224

Hope it is okay to attempt an answer to this slightly old question. The PDE $$f_t +\frac12 \sigma^2 s^2 f_{ss}=0$$ with terminal condition $f(T,s)=s^4$, is solved by $$f(t,s)=\mathbb{E}(S_T^4|S_t=s),$$ where $(S_t)_{t\geq 0}$ solves the SDE $$dS_t = \sigma S_t dB_t.$$

This is by Feynman-Kac. Applying Ito's lemma to $X_t = \log S_t$, gives $$d X_t = -\frac12 \sigma^2 dt + \sigma dB_t,$$ which in turn implies $$S_t = S_0 e^{-\frac12 \sigma^2 t + \sigma B_t}.$$ It follows that $$S_T = S_t e^{-\frac12 \sigma^2 (T-t) + \sigma B_{T-t}},$$ so that upon taking the $4$-th power and conditional expectation $$\mathbb{E}(S_T^4|S_t=s) = s^4e^{-2 \sigma^2 (T-t)} \mathbb{E}( e^{4\sigma B_{T-t}})$$ $$=s^4e^{-2 \sigma^2 (T-t)} M_Z(4\sigma\sqrt{T-t}),$$ where $M_Z(u)=\mathbb{E}(e^{uZ})$ is the MGF of a standard normal RV. Thus we may finally conclude $$f(t,s) = s^4 e^{6\sigma^2 (T-t)},$$ provided I have not made arithmetical/algebraic errors.

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