Feynman–Kac Representation of a Jump-Adjusted Put Payoff

Summary

The document sets up a pricing problem for a value function with geometric Brownian motion, a terminal put payoff, a constant running cost, and a constant intensity term that pulls the value toward a separate put-like payoff. It applies a Feynman–Kac representation, expressing the solution as the expectation of discounted running cash flows plus the discounted terminal payoff, conditional on the current underlying level.

The author derives the lognormal form of the process and recognizes that the terminal component can be valued with the Black–Scholes put formula. The unresolved question is how to evaluate the time integral of the expected intermediate payoff. No answer or derivation is included, so the post does not establish a closed form or numerical method for that component. The setup is useful for understanding how running rewards and terminal claims enter a stochastic pricing representation, but its signs and modeling interpretation would need checking before applying it.

Key ideas

• The pricing equation combines a terminal put payoff with running cash flows and an intensity adjustment.
• Feynman–Kac converts the differential equation into a conditional expectation over the underlying process.
• The terminal put component is identified as a Black–Scholes expectation.
• The post leaves the integral of intermediate put-like payoffs unresolved.

Tags

# Feynman-Kac to derive stochastic representation


# Feynman-Kac to derive stochastic representation












$u_t + \frac{1}{2}\sigma^2x^2u_{xx} - \alpha + \lambda((K_d - x)^+ - u) = 0$ with terminal condition $u(T, X) = (K_m - X(T))^+$

$dX = \sigma X(t)dW_t$

$\alpha$ and $\lambda$ are constants

Ok so using Feynman-Kac (courtesy of Wikipedia) I came to the conclusion that I need to calculate the expectation of the following thing:

$u(t, x) = E[\int_t^T (\lambda(K_d - X(r))^+ - \alpha)e^{-\int_t^r \lambda d\tau}dr + e^{-\int_t^T \lambda d\tau}(K_m - X(T))^+ |X(t) = x]$

Using Ito's lemma I derived that $X(T) = X(t)e^{-\frac{1}{2}\sigma^2(T-t) + \sigma W_{T-t}}$. Then, since $\lambda$ is a constant, I ended up with

$u(t,x) = E[\int_t^T(\lambda(K_d - X(r))^+ - \alpha)e^{\lambda(t-r)}dr + e^{\lambda(t-T)}(K_m - X(T))^+|X(t) = x] $

Now, the second part of the equation is the standard put option calculation $E[(K - S)^+]$ which is the BS formula for put, but what about the first one? For some reason I just can't figure out how to deal with this outer integral. Or maybe I'm completely wrong and made a mistake somewhere?

I'd be glad for some help.

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