Phase 9 - Lesson 9.5
The Feynman–Kac Theorem and the PDE Connection
How a conditional expectation solves a partial differential equation, and vice versa.
Builds on: 9.2 Itô and 9.3 SDEs give the tools; 9.4 set up \(\mathbb{Q}\).
Leads to: 9.6 identifies this PDE as the Black–Scholes equation.
Leads to: 9.6 identifies this PDE as the Black–Scholes equation.
Learning Objectives
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- State the Feynman–Kac theorem linking a parabolic PDE to a conditional expectation.
- Derive the PDE by applying Itô to the candidate value function and setting the drift to zero.
- Interpret the discounting term as a killing rate in the expectation.
- Verify Feynman–Kac on the heat equation with a known Gaussian solution.
- Explain why the drift in the PDE is the risk-neutral drift, not the physical one.
Key Vocabulary
- Feynman–Kac
- The identity: the solution of a linear parabolic PDE equals an expectation over a diffusion path.
- Value function
- The function \(v(t,x)=\E[\text{payoff}\mid X_t=x]\) (possibly discounted) that solves the PDE.
- Terminal condition
- The payoff \(v(T,x)=h(x)\) that anchors the backward PDE at maturity.
- Killing / discount rate
- The \(-rv\) term; it discounts (or ‘kills’) value at rate \(r\) along the path.
- Infinitesimal generator
- The operator \(\mathcal{L}=a\partial_x+\tfrac12 b^2\partial_{xx}\) governing the diffusion's dynamics.
- Parabolic PDE
- A PDE of heat-equation type: first order in time, second order in space.
Intuition & Motivation
Intuition
Two seemingly different questions - ‘solve this diffusion PDE’ and ‘average this payoff over random paths’ - have the same answer. The bridge is Itô: if \(v(t,X_t)\) (discounted) is a martingale, then its expected future value equals its present value, which is a conditional expectation; and ‘being a martingale’ means its Itô drift vanishes, which is precisely a PDE. Expectation and PDE are two faces of the same coin.
The theorem
Theorem - Feynman–Kac
Suppose \(dX_u=a(u,X_u)\,du+b(u,X_u)\,dW_u\) and define \(v(t,x)=\E\big[e^{-\int_t^T r\,du}\,h(X_T)\mid X_t=x\big].\) Then \(v\) solves the terminal-value PDE
\[v_t+a(t,x)\,v_x+\tfrac12 b(t,x)^2 v_{xx}-r\,v=0,\qquad v(T,x)=h(x).\] (9.7)
Read it backward in time from the payoff \(h\) at \(T\). The operator \(\mathcal{L}v=a v_x+\tfrac12 b^2 v_{xx}\) is the generator of the diffusion; the \(-rv\) term does the discounting.
Worked Example - Deriving the PDE via Itô
1
Consider the discounted value \(M_t=e^{-\int_0^t r\,du}\,v(t,X_t)\). By construction (tower property) \(M_t\) is a martingale.
2
Apply Itô to \(v(t,X_t)\): \(dv=(v_t+a v_x+\tfrac12 b^2 v_{xx})dt+b v_x\,dW.\)
3
Then \(dM=e^{-\int r}\big[(v_t+a v_x+\tfrac12 b^2 v_{xx}-r v)dt+b v_x\,dW\big].\)
4
A martingale has zero drift. Setting the \(dt\) coefficient to zero gives PDE (9.7); the surviving \(dW\) term is the martingale part.
Worked Example - Sanity check on the heat equation
1
Take \(a=0,\ b=1,\ r=0\), so \(X=W\) and (9.7) becomes \(v_t+\tfrac12 v_{xx}=0\) with \(v(T,x)=h(x)\).
2
Feynman–Kac says \(v(t,x)=\E[h(W_T)\mid W_t=x]=\E[h(x+W_{T-t})]\), an average against a \(\Normal(0,T-t)\) kernel.
3
That is the classical solution of the (backward) heat equation by Gaussian convolution - exactly matching (9.7).
Predict the drift
Predict first: In the Feynman–Kac PDE used for pricing, is the drift coefficient \(a\) the physical drift \(\mu\) or the risk-free rate \(r\)?
Key Idea
Because pricing expectations are taken under \(\mathbb{Q}\), the \(a\) in (9.7) is the risk-neutral drift \(rx\) for a stock. This is why an option's price does not depend on the stock's real-world expected return \(\mu\).
Common Mistakes to Avoid
- Solving the PDE forward in time. It is a terminal-value problem: integrate backward from \(v(T,\cdot)=h\).
- Dropping the \(-rv\) term - that omits discounting and misprices.
- Using the physical drift \(\mu\) in the PDE when pricing; it must be the risk-neutral drift \(r\).
- Forgetting \(v_t\); the value function depends on time-to-maturity, which is the whole dynamic.
Quant Practitioner Tips
- Memorize the pattern: martingale \(\Leftrightarrow\) zero Itô drift \(\Leftrightarrow\) PDE. It recurs throughout pricing.
- Feynman–Kac lets you price by Monte Carlo (average paths) OR by PDE solvers - pick whichever is cheaper for the problem.
- The generator \(\mathcal{L}=a\partial_x+\tfrac12 b^2\partial_{xx}\) is worth recognizing on sight; Itô always produces it.
Knowledge Check
Q1 Medium
Feynman–Kac says the solution of \(v_t+av_x+\tfrac12 b^2 v_{xx}-rv=0,\ v(T,\cdot)=h\) equals:
Q2 Easy
The PDE arises from Itô because a martingale must have:
Q3 Medium
When Feynman–Kac is used for option pricing, the drift in the PDE is:
Practical Exercise
Let \(dX=\theta(m-X)dt+\sigma\,dW\) (an Ornstein–Uhlenbeck process) and \(v(t,x)=\E[X_T\mid X_t=x]\). (a) Write the Feynman–Kac PDE (here \(h(x)=x,\ r=0\)). (b) Guess \(v(t,x)=m+(x-m)e^{-\theta(T-t)}\) and verify it solves the PDE.
▶ Show full solution
(a) \(v_t+\theta(m-x)v_x+\tfrac12\sigma^2 v_{xx}=0\), \(v(T,x)=x\).
(b) With \(v=m+(x-m)e^{-\theta(T-t)}\): \(v_x=e^{-\theta(T-t)}\), \(v_{xx}=0\), \(v_t=-\theta(x-m)e^{-\theta(T-t)}\). Substitute: \(-\theta(x-m)e^{-\theta(T-t)}+\theta(m-x)e^{-\theta(T-t)}+0=0\) since \(\theta(m-x)=-\theta(x-m)\). And \(v(T,x)=m+(x-m)=x\). Verified; this is the mean-reversion formula.
After the reveal, answer for yourself: How does the \(e^{-\theta(T-t)}\) factor encode mean reversion toward \(m\)?
Lesson Summary
Feynman–Kac equates the solution of a linear parabolic PDE with a discounted expectation over diffusion paths. Apply Itô to the (discounted) value function and force zero drift: the \(dt\) term is the PDE \(v_t+av_x+\tfrac12 b^2 v_{xx}-rv=0\), the \(dW\) term is its martingale part. In pricing, the drift is the risk-neutral \(r\), so option prices ignore \(\mu\).
Retrieval Practice
Close the lesson and answer from memory before checking. This is deliberate, effortful recall - the single highest-yield study action.
▶ Show retrieval prompts & answers
Q: State the Feynman–Kac correspondence in one sentence.
A: The solution of \(v_t+av_x+\tfrac12 b^2 v_{xx}-rv=0\) with terminal data \(h\) equals the discounted expectation \(\E[e^{-\int_t^T r}h(X_T)\mid X_t=x]\).
A: The solution of \(v_t+av_x+\tfrac12 b^2 v_{xx}-rv=0\) with terminal data \(h\) equals the discounted expectation \(\E[e^{-\int_t^T r}h(X_T)\mid X_t=x]\).
Q: How does the PDE fall out of Itô?
A: Apply Itô to the discounted value function; since it is a martingale its drift must vanish, and the zero-drift condition is exactly the PDE.
A: Apply Itô to the discounted value function; since it is a martingale its drift must vanish, and the zero-drift condition is exactly the PDE.
Completion Checklist
- I can explain the core ideas in my own words
- I worked the derivations/examples by hand
- I completed the interactive workbench(es)
- I passed the knowledge check
Confidence / mastery rating
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