Feynman–Kac: Terminal Payoffs and Derivative Valuation

Summary

The document clarifies a common misunderstanding of the Feynman–Kac theorem in a stochastic calculus text. The theorem is not asking the reader to discover an unknown terminal function: the payoff function is specified as part of the problem. A derivative’s value at an earlier time can then be represented as the expected terminal payoff under the stated stochastic process, and the same value function satisfies a partial differential equation with a terminal condition.

The answer illustrates the idea with a call option payoff based on the underlying asset’s final value and its strike. This connects the theorem’s mathematical statement to a familiar derivatives-pricing task. The document is brief: it corrects the interpretation and gives an example, but does not discuss assumptions, numerical solution methods, or how to choose a model for the underlying process.

Key ideas

• The terminal function in the theorem represents a specified payoff, rather than an unknown function to be discovered.
• Feynman–Kac links expected terminal payoffs to solutions of a partial differential equation.
• A call option payoff is given as a concrete example of the terminal function.
• The brief explanation does not cover model assumptions or numerical methods.

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# Questions on Feynman-Kac Theorem


# Questions on Feynman-Kac Theorem












I'm self studying stochastic calculus with Shreve's book 2.

The statement of Feynman-Kac Theorem says that, for the SDE of the form

$$dX(u) = \beta(u,X(u))du + \gamma(u,X(u))dW(u),$$

there exists a Borel measurable function $h$ such that for $t \in [0,T]$, $$g(t,x) = \mathbb{E}^{t,x}h(X(T))$$ satisfies $$g_t(t,x)+\beta(t,x)g_x(t,x)+\frac{1}{2}\gamma^2(t,x)g_{xx}(t,x)=0,$$ and the terminal condition $g(T,x) = h(x)$ for all $x$.

My question is,

This statement is completely an existential statement, it's not telling us how to find such $h$ and write it explicitly. Do we care about how to find $h$ explicitly or by numerical methods?

If we are not interested in finding $h$, then what're the key points behind the theorem?

I feel lost when studying this part...

## Answer by Rylan (score 1, accepted)

https://quant.stackexchange.com/a/83706

Shreve doesn't say "there exists a Borel-measurable function $h$", he says "let $h$ be a Borel-measurable function".

In practical terms, $h$ is the payoff function of the derivative at time $T$ that two parties would agree on, eg $h(x) = (x - K)^+$ for a call option with strike $K$.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

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