Verifying Feynman-Kac Solutions with Stopped Ito Processes

From a Backward PDE to a Stopped Expectation

Feynman–Kac formulas identify a solution of a partial differential equation with an expectation along a diffusion. The verification direction is especially direct: start with a sufficiently regular candidate solution, apply Ito’s formula along the diffusion, include any discount factor, stop at a bounded stopping time, and take expectations. The key justification is that the stochastic integral must be a true martingale with integrable terminal value; being a local martingale alone would not suffice.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_s)_{s\ge0},\mathbb{P})$ carrying a Brownian motion $B_s$. Fix $t<T$ and $x\in\mathbb{R}$, and let $X_s$, $t\le s\le T$, solve $dX_s=\mu(s,X_s)\,ds+\sigma(s,X_s)\,dB_s$, with $X_t=x$. Assume $\mu$ and $\sigma$ are bounded, continuous in time, and globally Lipschitz in space, uniformly in time. In particular, they satisfy a linear growth bound: for some $C<\infty$, $|\mu(s,y)|+|\sigma(s,y)|\le C(1+|y|)$. The standard existence and uniqueness theorem for Lipschitz SDEs gives a strong solution.

Let $c$ and $f$ be bounded continuous functions, and suppose $u\in C^{1,2}([t,T]\times\mathbb{R})$, with $u,u_s,u_x,u_{xx}$ bounded. Here $u_s$ denotes the time derivative. Assume $u$ satisfies the backward equation and terminal condition below. The diffusion generator is $\mathcal{L}_s u=\mu(s,x)u_x+\tfrac12\sigma(s,x)^2u_{xx}$; its second-derivative term is the Ito correction arising from $dB_s\,dB_s=ds$, not from the ordinary chain rule.

Theorem: Let $\tau$ be any $(\mathcal{F}_s)$-stopping time with $t\le\tau\le T$ almost surely. If $$ u_s(s,x)+\mu(s,x)u_x(s,x)+\tfrac12\sigma(s,x)^2u_{xx}(s,x)-c(s,x)u(s,x)+f(s,x)=0 $$ for $t\le s<T$, and $u(T,x)=g(x)$, then $$ u(t,x)=\mathbb{E}\left[D_\tau u(\tau,X_\tau)+\int_t^\tau D_s f(s,X_s)\,ds\right], \qquad D_s=\exp\left(-\int_t^s c(r,X_r)\,dr\right). $$ All expectations here are under $\mathbb{P}$.

Proof. The process $D_s$ is continuous and of finite variation, with $D_t=1$ and $dD_s=-c(s,X_s)D_s\,ds$. Its absolute value is bounded by $e^{(T-t)\|c\|_\infty}$. Apply Ito’s formula to the specified function $(s,y)\mapsto u(s,y)$. Its partial derivatives are $u_s,u_x,u_{xx}$, so the formula, including its correction term, is

du(s,X_s)= \left(u_s+\mu u_x+\tfrac12\sigma^2u_{xx}\right)(s,X_s)\,ds +u_x(s,X_s)\sigma(s,X_s)\,dB_s.

Because $D$ has finite variation, its covariation with $u(s,X_s)$ is zero. The product rule therefore gives

d\big(D_su(s,X_s)\big) =D_s\left(u_s+\mu u_x+\tfrac12\sigma^2u_{xx}-cu\right)(s,X_s)\,ds +D_su_x(s,X_s)\sigma(s,X_s)\,dB_s =-D_sf(s,X_s)\,ds+D_su_x(s,X_s)\sigma(s,X_s)\,dB_s.

Integrate from $t$ to $\tau$. The stochastic-integral integrand, stopped at $\tau$, is progressively measurable and bounded: $D,u_x,\sigma$ are bounded. Thus it belongs to $\mathcal{L}^2$, the class of progressively measurable $H$ with $\mathbb{E}\int_t^T H_s^2\,ds<\infty$. By the Ito isometry its Ito integral is square-integrable, and hence is a true martingale with conditional mean-zero increments; in particular its expectation at $\tau$ is zero. Taking expectations in the integrated identity and using $D_tu(t,X_t)=u(t,x)$ proves the formula. No passage from a local martingale to a martingale is being assumed: square integrability supplies the required integrability. $\square$

The terminal condition enters when the stopping time reaches $T$: on $\{\tau=T\}$, $u(\tau,X_\tau)=g(X_T)$. Likewise, if $\tau$ is an exit time from a domain, the boundary condition $u(s,y)=h(s,y)$ on the exit boundary converts $u(\tau,X_\tau)$ into the boundary payoff there. The theorem itself allows any bounded stopping time and retains $u(\tau,X_\tau)$ as the payoff unless such a boundary or terminal condition identifies it with another function.

Worked Checks of the Representation

Worked Example: A Stopping Rule Based on the Brownian Path

Let $X_s=B_s$, start from $X_0=0$, set $T=1$, and take $c=f=0$. Use $u(s,y)=y^2+1-s$, then verify the representation for $\tau=\tfrac12$ if $B_{1/2}\ge0$, and $\tau=1$ otherwise.

The derivatives are $u_s=-1$, $u_y=2y$, and $u_{yy}=2$. With $\mu=0$ and $\sigma=1$, the PDE expression is $-1+0+\tfrac12(2)=0$, and $u(1,y)=y^2$. The theorem says $u(0,0)=\mathbb{E}[B_\tau^2+1-\tau]$, with $u(0,0)=1$.

To compute the expectation, let $A=\{B_{1/2}\ge0\}$. Since $B_{1/2}$ has a continuous symmetric normal distribution, $\mathbb{P}(A)=\mathbb{P}(A^c)=\tfrac12$. Symmetry also gives $\mathbb{E}[B_{1/2}^2\mathbf{1}_A]=\mathbb{E}[B_{1/2}^2\mathbf{1}_{A^c}]=\tfrac14$, because their sum is $\mathbb{E}[B_{1/2}^2]=\tfrac12$. Write $B_1=B_{1/2}+(B_1-B_{1/2})$; the second term is independent of $\mathcal{F}_{1/2}$, has mean zero and variance $\tfrac12$. Consequently,

\mathbb{E}[B_\tau^2] =\mathbb{E}[B_{1/2}^2\mathbf{1}_A] +\mathbb{E}[B_1^2\mathbf{1}_{A^c}] =\tfrac14+\left(\tfrac14+\tfrac12\cdot\tfrac12\right) =\tfrac34.

Also, $1-\tau=\tfrac12$ on $A$ and is zero on $A^c$, so $\mathbb{E}[1-\tau]=\tfrac12\cdot\tfrac12=\tfrac14$. The total is $\tfrac34+\tfrac14=1=u(0,0)$. This verifies the formula by a direct computation. The event defining the earlier stopping time is measurable at time $1/2$, so the displayed rule does define a stopping time.

Worked Example: Discounted Terminal Payoff

Take $X_s=B_s$, $t=0$, $T=1$, $c=1$, and $f=0$. For terminal payoff $g(y)=\cos y$, consider $u(s,y)=e^{-\frac32(1-s)}\cos y$. Verify the formula at the deterministic stopping time $\tau=1$, starting from $x=0$.

Put $a(s)=e^{-\frac32(1-s)}$. Then $a'(s)=\tfrac32a(s)$, so the partial derivatives of $u$ are $u_s=\tfrac32a(s)\cos y$, $u_y=-a(s)\sin y$, and $u_{yy}=-a(s)\cos y$. The PDE expression is $\tfrac32a\cos y+0-\tfrac12a\cos y-a\cos y=0$, and $u(1,y)=\cos y$. Therefore the theorem gives $\mathbb{E}[e^{-1}\cos(B_1)]=u(0,0)=e^{-3/2}$.

Independently, the characteristic function of $B_1\sim N(0,1)$ gives $\mathbb{E}[\cos(B_1)]=e^{-1/2}$. Thus the discounted expectation is $e^{-1}e^{-1/2}=e^{-3/2}$, as claimed. This check also confirms that the discount factor is evaluated over the full interval from $0$ to $1$.

Worked Example: A Running Reward Before a Bounded Stop

Let $X_s=B_s$, $t=0$, $T=1$, $c=1$, and $f=1$, and define $u(s,y)=1-e^{-(1-s)}$. Verify the representation for the deterministic stopping time $\tau=\tfrac12$.

Here $u_s=-e^{-(1-s)}$, $u_y=0$, and $u_{yy}=0$. Hence $u_s+\tfrac12u_{yy}-u+1=-e^{-(1-s)}-(1-e^{-(1-s)})+1=0$, while $u(1,y)=0$. The right-hand side of the theorem is \[ e^{-1/2}u(1/2,B_{1/2})+\int_0^{1/2}e^{-s}\,ds =e^{-1/2}(1-e^{-1/2})+(1-e^{-1/2}) =1-e^{-1}. \] It equals $u(0,0)=1-e^{-1}$. The calculation checks both contributions: the discounted value at the stopping time and the accumulated discounted running reward.

Stopping on a Boundary and the Scope of the Formula

Corollary: Under the theorem’s assumptions, suppose a bounded stopping time $\tau$ is the earlier of $T$ and the first exit of $X$ from a domain. If $u(T,y)=g(y)$ and $u(s,y)=h(s,y)$ at every exit point with $s\le T$, then $$ u(t,x)=\mathbb{E}\left[D_\tau\,\big(\mathbf{1}_{\{\tau=T\}}g(X_T)+\mathbf{1}_{\{\tau<T\}}h(\tau,X_\tau)\big)+\int_t^\tau D_s f(s,X_s)\,ds\right]. $$ Equalities on the boundary and at terminal time are understood pointwise on the stated sets (or almost surely along the stopped process).

Proof. Apply the theorem to this $\tau$. On $\{\tau=T\}$, the terminal condition gives $u(\tau,X_\tau)=g(X_T)$. On $\{\tau<T\}$, stopping occurs at the exit boundary, where the boundary condition gives $u(\tau,X_\tau)=h(\tau,X_\tau)$. These events partition the sample space because $\tau\le T$; substituting the two values into the theorem’s expectation proves the displayed formula. $\square$

For instance, with $X_s=B_s$, $u(s,y)=y$, $c=f=0$, and $\tau$ the first exit from $(0,2)$ or time $1$, whichever comes first, the equation holds because $u_s=0$, $u_y=1$, and $u_{yy}=0$. The payoff at exit is $0$ or $2$; if there is no exit by time $1$, it is $B_1$. The theorem yields $\mathbb{E}[B_\tau]=u(0,1)=1$. The boundary payoff is not merely the terminal payoff $g(X_T)$: it depends on whether the process exits before $T$.

A Common Discounting Error

A frequent mistake is to apply Ito’s formula to $u(s,X_s)$ and then use the resulting identity as though it were already the discounted representation. In the discounted-cosine example, this would replace $\mathbb{E}[e^{-1}\cos(B_1)]$ with $\mathbb{E}[\cos(B_1)]$. The incorrect value is $e^{-1/2}$, whereas the correct value is $e^{-1}e^{-1/2}=e^{-3/2}$. The missing factor comes from differentiating $D_su(s,X_s)$: since $dD_s=-c(s,X_s)D_s\,ds$, the product rule contributes the term $-c u$ to the PDE.

The expectation identity is proved here under $\mathbb{P}$. In an application that uses a risk-neutral measure $\mathbb{Q}$, the diffusion and its coefficients must instead be specified under $\mathbb{Q}$, and the expectation must be taken under that measure. A change of measure does not follow from the stopped Ito calculation itself. Likewise, a local-martingale argument alone does not justify taking expectations; the boundedness and square-integrability assumptions above are what make the stochastic integral a true martingale.

Check Your Understanding

Use the theorem’s hypotheses and distinguish the PDE identity from the expectation step.

For $u(s,y)=y^2+T-s$ and standard Brownian motion, compute each term in $u_s+\tfrac12u_{yy}$.
In the discounted-cosine example, which product-rule term would be missing if one applied Ito’s formula to $u(s,X_s)$ without multiplying by $D_s$?
Why does boundedness of $D_su_x(s,X_s)\sigma(s,X_s)$ imply that the stopped Ito integral has expectation zero?
State the boundary condition needed to replace $u(\tau,X_\tau)$ by a boundary payoff when $\tau<T$ is an exit time.
For the running-reward example, separate the discounted stopping-time term from the integral term and verify their sum is $1-e^{-1}$.

Verifying Feynman-Kac Solutions with Stopped Ito Processes

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