Proof Sketch of the Feynman-Kac Formula¶
The Feynman-Kac formula connects PDE solutions to stochastic expectations. This page presents a detailed proof sketch of both directions: starting from the PDE and arriving at the expectation formula, and starting from the expectation and deriving the PDE. The core technique is Ito's lemma applied to a carefully constructed process that becomes a martingale.
Related Content
- Feynman-Kac Formula -- statement and applications
- Discounted Feynman-Kac -- extension with discounting
- Kolmogorov Backward Equation -- the special case \(r = 0\)
Setup¶
The SDE¶
on \([t, T]\), where \((t, x)\) is the initial condition.
The PDE¶
with terminal condition \(u(T, x) = g(x)\).
In compact notation:
where \(\mathcal{L} = \mu\,\partial_x + \frac{1}{2}\sigma^2\,\partial_{xx}\) is the infinitesimal generator.
The Claimed Representation¶
Direction 1: PDE Solution Implies Expectation Formula¶
Given: \(u\) is a smooth (\(C^{1,2}\)) solution of the PDE.
Goal: Show \(u(t, x) = \mathbb{E}\!\left[e^{-\int_t^T r(s, X_s)\,ds}\,g(X_T) \mid X_t = x\right]\).
Step 1: Define the Discount Factor¶
Let:
This is the stochastic discount factor from time \(t\) to time \(s\). Note that:
Why Stochastic Discounting?
The discount factor \(D(t,s)\) depends on the path of \(X\) through the rate function \(r(s, X_s)\). When \(r\) is constant, \(D(t,s) = e^{-r(s-t)}\) and the discounting is deterministic.
Step 2: Apply Ito's Lemma to the Discounted Solution¶
Define the process:
By the product rule for Ito processes:
(The cross-variation \(d[D, u]\) is zero because \(D\) has no Brownian component -- it is adapted but driven only by \(ds\).)
Expanding \(du\) via Ito's lemma:
Expanding \(dD\):
Step 3: Substitute and Simplify¶
Combine the \(ds\) terms:
Step 4: Apply the PDE¶
Since \(u\) solves \(\partial_t u + \mathcal{L}u - ru = 0\), the \(ds\) coefficient vanishes:
The discounted solution process \(Y_s\) is a local martingale (it has no drift term -- only a stochastic integral against \(dW\)).
This Is the Key Step
The PDE is precisely the condition that eliminates the drift from \(Y_s\). This is not a coincidence: the Feynman-Kac PDE is designed so that the discounted solution is a martingale. This is the analytical expression of the fundamental theorem of asset pricing.
Step 5: From Local Martingale to Martingale¶
Under suitable integrability conditions (e.g., \(u\) and its derivatives have polynomial growth, and \(\sigma\) satisfies linear growth), the stochastic integral \(\int_t^s D\,\sigma\,u_x\,dW\) is a true martingale, not merely a local martingale.
Where Regularity Matters
This step is where the regularity assumptions are essential. If \(\sigma u_x\) grows too fast, the stochastic integral may only be a local martingale, and the expectation argument below fails. Sufficient conditions include:
- \(u\) has at most polynomial growth in \(x\)
- \(\sigma\) satisfies a linear growth condition
- \(r\) is bounded below
Step 6: Take Expectations¶
Since \(Y_s\) is a martingale on \([t, T]\):
At \(s = t\): \(Y_t = D(t, t)\,u(t, X_t) = 1 \cdot u(t, x) = u(t, x)\)
At \(s = T\): \(Y_T = D(t, T)\,u(T, X_T) = e^{-\int_t^T r\,d\tau}\,g(X_T)\)
Therefore:
\(\square\)
Direction 2: Expectation Defines a PDE Solution¶
Given: Define \(u(t, x) = \mathbb{E}\!\left[e^{-\int_t^T r(s, X_s)\,ds}\,g(X_T) \mid X_t = x\right]\).
Goal: Show \(u\) satisfies the PDE \(\partial_t u + \mathcal{L}u - ru = 0\).
Step 1: Markov Property¶
By the Markov property of \(X_t\):
Therefore the process:
equals \(\mathbb{E}\!\left[e^{-\int_t^T r\,d\tau}\,g(X_T) \mid \mathcal{F}_s\right]\), which is a martingale by the tower property.
Step 2: Apply Ito's Lemma¶
Assuming \(u\) is \(C^{1,2}\) (which requires regularity theory -- this is the hard part), apply Ito's lemma to \(M_s = D(t,s)\,u(s, X_s)\):
Step 3: Martingale Condition¶
Since \(M_s\) is a martingale, the \(ds\) coefficient must be zero:
at every point \((s, x)\) where \(u\) is smooth. \(\square\)
Classical vs Viscosity Solutions
Direction 2 requires \(u\) to be \(C^{1,2}\), which may fail if:
- The terminal condition \(g\) is non-smooth (e.g., \((x-K)^+\))
- The coefficients degenerate (\(\sigma(x) = 0\) at some points)
In such cases, \(u\) is a viscosity solution of the PDE -- a weaker notion that does not require differentiability but still ensures uniqueness.
Summary of the Proof Logic¶
``` Direction 1 (PDE → Expectation):
u solves PDE ──Itô's lemma──▶ D(t,s)·u(s,Xₛ) is a martingale │ E[Y_T | ℱ_t] = Y_t │ ▼ u(t,x) = E[D(t,T)·g(X_T)]
Direction 2 (Expectation → PDE):
u(t,x) = E[D·g] ──Markov──▶ D(t,s)·u(s,Xₛ) is a martingale │ Itô: drift must vanish │ ▼ ∂ₜu + ℒu - ru = 0 ```
Where Each Assumption Is Used¶
| Assumption | Used In | Purpose |
|---|---|---|
| \(\mu, \sigma\) Lipschitz | SDE existence | Ensures \(X_t\) exists and is unique |
| \(u \in C^{1,2}\) | Ito's lemma | Justifies the chain rule for \(u(s, X_s)\) |
| \(\sigma u_x\) integrability | Step 5 (Dir. 1) | Promotes local martingale to true martingale |
| \(g\) continuous, polynomial growth | Terminal condition | Ensures \(u\) is well-defined |
| \(r\) bounded below | Discounting | Prevents \(D(t,T)\) from exploding |
| Markov property | Step 1 (Dir. 2) | Allows \(u\) to depend only on \((s, X_s)\) |
The Special Case \(r = 0\) (Kolmogorov Backward)¶
When \(r \equiv 0\), the Feynman-Kac formula reduces to:
and the PDE becomes the Kolmogorov backward equation:
The proof simplifies: \(Y_s = u(s, X_s)\) is directly a martingale (no discounting needed), and Ito's lemma gives:
The PDE forces the drift to vanish, so \(u(s, X_s)\) is a martingale. Taking expectations: \(u(t, x) = \mathbb{E}[u(T, X_T)] = \mathbb{E}[g(X_T)]\).
The Martingale Interpretation¶
The proof reveals a deep structural insight:
This equivalence is the mathematical core of risk-neutral pricing:
- PDE perspective: The option price satisfies a deterministic equation
- Martingale perspective: The discounted option price is a martingale under the risk-neutral measure
- Feynman-Kac: These are the same statement
In finance, the risk-free rate \(r\) serves as both the discounting rate in the expectation and the coefficient in the PDE. The no-arbitrage condition (that discounted prices are martingales) is exactly the Feynman-Kac PDE.
Extension: Running Payoff¶
When there is a source term \(f(s, X_s)\) (running payoff), the PDE becomes:
and the representation is:
The proof proceeds identically, except the process:
is the martingale. The \(-f\) from the PDE cancels the \(+f\) from the integral term. See Feynman-Kac Formula for details.
Summary¶
| Step | Direction 1 (PDE \(\to\) Expectation) | Direction 2 (Expectation \(\to\) PDE) |
|---|---|---|
| Start | \(u\) solves PDE | \(u(t,x) = \mathbb{E}[D \cdot g]\) |
| Key tool | Ito's lemma on \(D \cdot u\) | Markov property + Ito's lemma |
| Core argument | PDE \(\Rightarrow\) drift vanishes \(\Rightarrow\) martingale | Martingale \(\Rightarrow\) drift vanishes \(\Rightarrow\) PDE |
| Conclusion | \(u(t,x) = \mathbb{E}[D \cdot g]\) | \(\partial_t u + \mathcal{L}u - ru = 0\) |
| Regularity | Needs integrability for true martingale | Needs \(C^{1,2}\) for Ito |
The Feynman-Kac proof is fundamentally a martingale argument: the PDE is the condition that makes the discounted solution process driftless, and driftless processes are martingales, whose values at different times are related by conditional expectations.
See Also¶
- Feynman-Kac Formula -- full statement with applications
- Discounted Feynman-Kac -- the discounting/killing extension
- Kolmogorov Backward Equation -- the \(r = 0\) special case
- Applications to Option Pricing -- financial applications
Exercises¶
Exercise 1. In Direction 1 (PDE to Expectation), the key step is showing that \(Y_s = D(t,s)\,u(s, X_s)\) is a martingale when \(u\) solves the PDE. Apply Ito's lemma to \(D(t,s)\,u(s, X_s)\) for constant \(r\) and verify that the drift vanishes when \(\partial_t u + \mathcal{L}u - ru = 0\).
Solution to Exercise 1
For constant \(r\), we have \(D(t, s) = e^{-r(s-t)}\) and \(Y_s = e^{-r(s-t)}u(s, X_s)\).
Applying Ito's lemma to \(u(s, X_s)\):
By the product rule (the cross-variation is zero since \(e^{-r(s-t)}\) is deterministic in \(W\)):
Substituting:
The drift coefficient is \(e^{-r(s-t)}[\partial_s u + \mathcal{L}u - r\,u]\). When \(u\) solves \(\partial_t u + \mathcal{L}u - r\,u = 0\), this drift vanishes identically, leaving:
which is a pure stochastic integral against \(dW\) with no drift -- confirming \(Y_s\) is a (local) martingale. \(\checkmark\)
Exercise 2. In Direction 2 (Expectation to PDE), the starting point is \(u(t,x) = \mathbb{E}[D(t,T)g(X_T) | X_t = x]\). Using the Markov property, show that \(u(t, X_t)\) can be expressed as a conditional expectation and hence \(D(t,s)u(s, X_s)\) is a martingale. Conclude that the drift of this process must be zero.
Solution to Exercise 2
Define \(u(t, x) = \mathbb{E}[D(t, T)\,g(X_T) \mid X_t = x]\). By the Markov property of \(X_t\), the conditional expectation at time \(s > t\) can be expressed as:
Now consider the process \(M_s = D(t, s)\,u(s, X_s)\):
The last expression is a conditional expectation of the \(\mathcal{F}_T\)-measurable random variable \(D(t,T)\,g(X_T)\) given \(\mathcal{F}_s\). By the tower property, this is a martingale: for \(s_1 < s_2\),
Since \(M_s\) is a martingale and (assuming \(u \in C^{1,2}\)) Ito's lemma gives \(dM_s = D(t,s)[\partial_s u + \mathcal{L}u - r\,u]\,ds + (\text{martingale})\), the drift must be zero. Therefore \(\partial_s u + \mathcal{L}u - r\,u = 0\).
Exercise 3. In the proof, the integrability condition ensures that the local martingale \(Y_s\) is a true martingale. Give an example where the stochastic integral \(\int_t^s D\,\sigma\,u_x\,dW\) might fail to be a true martingale, and explain the consequence for the Feynman-Kac representation.
Solution to Exercise 3
The integrability condition ensures that \(\mathbb{E}[\int_t^T (D\,\sigma\,u_x)^2\,ds] < \infty\), which guarantees that the stochastic integral \(\int_t^s D\,\sigma\,u_x\,dW\) is a true martingale (not merely a local martingale).
Example of failure: Consider the SDE \(dX_s = X_s^2\,dW_s\) and suppose \(u_x\) grows like \(e^{x^2}\). Then \(\sigma\,u_x = X_s^2\,e^{X_s^2}\) can grow extremely fast. The stochastic integral may be only a local martingale: there exist localizing stopping times \(\tau_n \uparrow T\) such that the stopped process is a martingale, but the unstopped process may not be.
Consequence: If \(Y_s\) is a local martingale but not a true martingale, then \(\mathbb{E}[Y_T] \leq Y_t\) (supermartingale) rather than \(\mathbb{E}[Y_T] = Y_t\). This means:
with strict inequality possible. The Feynman-Kac representation would give only a lower bound for the PDE solution, and the probabilistic formula would not recover the full PDE solution. This is related to the phenomenon of bubbles in mathematical finance, where the discounted price process is a strict local martingale.
Exercise 4. For the case \(r = 0\) and \(f = 0\), the Feynman-Kac proof simplifies to showing that \(u(s, X_s)\) is a martingale. Carry out this simplified proof for \(dX_s = \sigma\,dW_s\) and \(g(x) = x^2\), verifying each step explicitly.
Solution to Exercise 4
With \(r = 0\), \(f = 0\), \(dX_s = \sigma\,dW_s\), and \(g(x) = x^2\):
Step 1: The PDE is \(\partial_t u + \frac{1}{2}\sigma^2\,\partial_{xx}u = 0\) with \(u(T, x) = x^2\). The solution is \(u(t, x) = x^2 + \sigma^2(T - t)\) (computed by Feynman-Kac or by direct verification).
Step 2: Define \(Y_s = u(s, X_s)\). Apply Ito's lemma:
Compute the derivatives: \(\partial_s u = -\sigma^2\), \(\partial_x u = 2x\), \(\partial_{xx}u = 2\). Therefore:
Step 3: The PDE \(\partial_t u + \frac{1}{2}\sigma^2\,u_{xx} = -\sigma^2 + \sigma^2 = 0\) is satisfied, so the drift vanishes.
Step 4: \(Y_s = u(s, X_s)\) is a martingale (it equals \(X_s^2 + \sigma^2(T - s)\), which is the well-known result that \(W_s^2 - s\) is a martingale, rescaled).
Step 5: \(Y_t = u(t, x) = x^2 + \sigma^2(T - t)\) and \(Y_T = u(T, X_T) = X_T^2\). By the martingale property: \(u(t, x) = \mathbb{E}[X_T^2 \mid X_t = x] = x^2 + \sigma^2(T - t)\). \(\checkmark\)
Exercise 5. The running payoff extension involves the process \(Y_s = D(t,s)u(s,X_s) + \int_t^s D(t,\tau)f(\tau, X_\tau)\,d\tau\). Show that the \(+f\) from the integral and the \(-f\) from the PDE cancel in the drift of \(Y_s\), leaving \(Y_s\) as a martingale.
Solution to Exercise 5
Define \(Y_s = D(t,s)\,u(s, X_s) + \int_t^s D(t, \tau)\,f(\tau, X_\tau)\,d\tau\). We compute \(dY_s\):
From the proof of Direction 1 (without running payoff), we already know:
Adding the running payoff term:
Since \(u\) solves \(\partial_t u + \mathcal{L}u - r\,u + f = 0\), the drift coefficient is:
The cancellation occurs precisely because the PDE includes \(+f\) as a source term, and the integral \(\int_t^s D\,f\,d\tau\) contributes \(+D\,f\,ds\) to \(dY_s\). These combine with the \(-f\) implicit in the PDE (since \(\partial_t u + \mathcal{L}u - r\,u = -f\)) to make the total drift zero. Therefore \(Y_s\) is a (local) martingale.
Exercise 6. Explain why the Feynman-Kac proof requires \(u \in C^{1,2}\) (once differentiable in \(t\), twice in \(x\)). What goes wrong with the Ito lemma application if \(u\) is only continuous? In this case, what weaker concept of solution (viscosity solution) can be used?
Solution to Exercise 6
Ito's lemma states: if \(u \in C^{1,2}\) and \(X_s\) is an Ito process, then \(u(s, X_s)\) is also an Ito process with specific drift and diffusion terms. The formula requires computing \(\partial_s u\), \(\partial_x u\), and \(\partial_{xx}u\), so \(u\) must be at least \(C^{1,2}\) (once continuously differentiable in \(t\), twice in \(x\)).
If \(u\) is only continuous: Ito's lemma cannot be applied, because the chain rule for stochastic calculus fundamentally requires second-order spatial derivatives. Without \(u_{xx}\), we cannot write \(du(s, X_s)\) and cannot identify the drift term that must vanish. The entire martingale argument breaks down at Step 2.
Viscosity solutions: When classical solutions do not exist (e.g., when \(g(x) = (x - K)^+\) has a kink, or when the diffusion coefficient degenerates), we use viscosity solutions. A continuous function \(u\) is a viscosity solution if, for every smooth test function \(\varphi\) that touches \(u\) from above (or below) at a point \((t_0, x_0)\), the PDE inequality \(\partial_t \varphi + \mathcal{L}\varphi - r\,\varphi \leq 0\) (or \(\geq 0\)) holds at that point. This definition replaces the pointwise PDE with a family of inequalities involving smooth approximations, avoiding the need for \(u\) to be differentiable. Viscosity solution theory provides existence and uniqueness under much weaker conditions than classical theory.
Exercise 7. Summarize the Feynman-Kac proof in one sentence for each of the five steps: (1) define the process, (2) apply Ito's lemma, (3) use the PDE, (4) conclude martingale property, (5) evaluate at boundary times. Then explain why step (3) is the crucial one that connects the PDE to probability.
Solution to Exercise 7
(1) Define the process: Construct \(Y_s = D(t,s)\,u(s, X_s)\), the product of the stochastic discount factor and the PDE solution evaluated along the SDE path.
(2) Apply Ito's lemma: Use the product rule and Ito's lemma to decompose \(dY_s\) into a drift term (proportional to \(ds\)) and a martingale term (proportional to \(dW_s\)).
(3) Use the PDE: Recognize that the drift coefficient equals \(D(t,s)[\partial_s u + \mathcal{L}u - r\,u]\), which vanishes because \(u\) solves the Feynman-Kac PDE.
(4) Conclude martingale property: Since the drift is zero and the stochastic integral satisfies integrability conditions, \(Y_s\) is a true martingale on \([t, T]\).
(5) Evaluate at boundary times: Apply the martingale property \(Y_t = \mathbb{E}[Y_T \mid \mathcal{F}_t]\) with \(Y_t = u(t, x)\) and \(Y_T = D(t,T)\,g(X_T)\) to obtain the representation formula.
Why step (3) is crucial: Step (3) is where the PDE and probability meet. The PDE \(\partial_t u + \mathcal{L}u - r\,u = 0\) is exactly the condition that eliminates the drift from the discounted process, and a driftless process is a martingale. Without this step, we would have a general Ito process with both drift and diffusion; the PDE is precisely the "no-drift" condition. This is also the mathematical content of the fundamental theorem of asset pricing: the PDE for option prices is equivalent to the statement that discounted prices are martingales under the risk-neutral measure.