State the Feynman-Kac representation for dX = mu(t,x) dt + sigma(t,x) dW, including the discounted version with rate r, and derive the PDE from Ito rather than quoting it.

State the Feynman-Kac representation for dX = mu(t,x) dt + sigma(t,x) dW, including the discounted version with rate r, and derive the PDE from Ito rather than quoting it.

Approach: Define the candidate function as a conditional expectation, argue it is a martingale by the tower property, then apply Ito and force the dt coefficient to vanish.

The Feynman-Kac representation says that if u(t,x) = E[g(X_T) | X_t = x] then u_t + mu u_x + (1/2) sigma^2 u_xx = 0 with the terminal condition u(T,x) = g(x), and with discounting u(t,x) = E[e^{-r(T-t)} g(X_T) | X_t = x] satisfies u_t + mu u_x + (1/2) sigma^2 u_xx - r u = 0 with the same terminal condition. The derivation runs through the martingale property: u(t, X_t) = E[g(X_T) | F_t] is a conditional expectation of a single fixed random variable, so by the tower property it is a martingale. Applying Ito's lemma to u(t, X_t) gives du = (u_t + mu u_x + (1/2) sigma^2 u_xx) dt + sigma u_x dW, and a martingale with a continuous finite variation part must have that part identically zero, which is the PDE. With discounting the martingale is e^{-rt} u(t, X_t), and the product rule adds the -r u term. The construction runs in both directions, so a solution of the PDE also has the expectation representation.

Follow-up: With a running payoff ∫_t^T f(s, X_s) ds added, where does the extra term enter the PDE?

Key concepts: Feynman-Kac, martingale property, Ito's lemma, terminal condition.