Localization on Unbounded Domains

Why Localization Is Needed

On a bounded domain, stopping a diffusion before it leaves the domain makes the coefficients and derivatives in an Ito verification argument manageable. On an unbounded domain, one can first stop on bounded pieces and then enlarge those pieces. The second step is not automatic: although the stopped random variables may converge almost surely, their expectations need not converge. A Feynman–Kac representation on an unbounded domain therefore has two parts: a valid identity at each bounded stopping level, and conditions that justify taking the limit.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$. Consider a scalar diffusion started at $X_0=x$, satisfying $dX_s=\mu(X_s)\,ds+\sigma(X_s)\,dB_s$. Assume $\mu$ and $\sigma$ are globally Lipschitz and have linear growth, so that a global continuous strong solution exists. Here a strong solution is adapted to the given filtration and Brownian motion. The moment assumption used below will be stated separately. Every Ito integral is understood to have a progressively measurable integrand $H_s$ with $\mathbb{E}\int_0^T H_s^2\,ds<\infty$, or to be stopped so that this condition holds.

Let $D\subseteq\mathbb{R}$ be open, and choose bounded open sets $D_n$ with compact closures such that $\overline{D_n}\subset D_{n+1}$ and $\bigcup_n D_n=D$. Define $\tau_n=\inf\{s\ge0:X_s\notin D_n\}$, with $\inf\varnothing=\infty$, and $\tau_D=\lim_{n\to\infty}\tau_n$. The times $\tau_n$ increase. For a fixed finite horizon $T$, all stopping below is before time $T$.

Definition: Localization on an unbounded domain means applying an Ito verification argument up to $\rho_n=(T-t)\wedge\tau_n$, where $D_n$ is bounded, and then passing to the limit as $n\to\infty$. The limiting exit time $\tau_D$ may be infinite; in that case the horizon $T-t$ stops the representation.

The Stopped Feynman–Kac Identity

Let $q(t,x)\ge0$ be a continuous killing or discount rate, $f(t,x)$ a running reward, and $u\in C^{1,2}([0,T]\times D)$. Write $\mathcal{L}u=\mu(x)u_x+\tfrac12\sigma(x)^2u_{xx}$. The equation to verify is $u_t+\mathcal{L}u-qu+f=0$. The terminal and boundary values are not needed for the stopped identity itself; they enter when identifying the limit.

Proposition: Suppose $u$, its derivatives, and the coefficients are continuous on each compact subset relevant to the stopped process. For every $n$ such that $x\in D_n$, and $0\le t<T$, $$ u(t,x)=\mathbb{E}\left[e^{-\int_0^{\rho_n}q(t+s,X_s)\,ds}u(t+\rho_n,X_{\rho_n})+\int_0^{\rho_n}e^{-\int_0^s q(t+r,X_r)\,dr}f(t+s,X_s)\,ds\right]. $$

Proof. Set $A_s=\int_0^s q(t+r,X_r)\,dr$, and apply Ito's formula to $F(s,z)=e^{-A_s}u(t+s,z)$, with the finite-variation process $A_s$ included in the time derivative. The derivatives of $u$ appearing in the formula are $u_t$, $u_x$, and $u_{xx}$, and the quadratic-variation correction is $\tfrac12\sigma(X_s)^2u_{xx}\,ds$, since $dB_s\,dB_s=ds$, $dB_s\,ds=0$, and $ds\,ds=0$. Thus $$ d\big(e^{-A_s}u(t+s,X_s)\big)=e^{-A_s}(u_t+\mathcal{L}u-qu)(t+s,X_s)\,ds+e^{-A_s}\sigma(X_s)u_x(t+s,X_s)\,dB_s. $$ The PDE makes the drift $-e^{-A_s}f(t+s,X_s)\,ds$. Integrate to $\rho_n$ and rearrange. Before $\tau_n$, the stopped state lies in a bounded set; on the finite time interval the continuous derivatives and coefficients are bounded on the relevant compact set. Consequently the stopped stochastic-integral integrand is progressively measurable and square-integrable, so its Ito integral has expectation zero. Taking expectations gives the displayed identity. $\square$

Worked Example: Brownian Motion and a Quadratic Terminal Payoff

Take $D=\mathbb{R}$, $\mu=0$, $\sigma=1$, $q=f=0$, and terminal payoff $h(x)=x^2$. The candidate is $u(t,x)=x^2+T-t$. Its derivatives are $u_t=-1$, $u_x=2x$, and $u_{xx}=2$, so $u_t+\tfrac12u_{xx}=-1+1=0$, and $u(T,x)=x^2$.

Use $D_n=(-n,n)$ and $\rho_n=(T-t)\wedge\tau_n$. The stopped identity gives $u(t,x)=\mathbb{E}[u(t+\rho_n,B_{\rho_n}+x)]$. Brownian paths are continuous and hence bounded on $[0,T-t]$; therefore, for almost every path, $\tau_n>T-t$ for all sufficiently large $n$. The stopped endpoint then equals $x+B_{T-t}$, and the stopped value converges to $(x+B_{T-t})^2$.

For the expectation limit, the stopped values are bounded by $C(1+\sup_{s\le T-t}|B_s|^2)$ for a finite constant $C$ depending on $x,T,t$. Brownian motion has a finite second moment of its supremum on a finite interval, so this bound is integrable and dominated convergence applies. Independently, $\mathbb{E}[(x+B_{T-t})^2]=x^2+2x\mathbb{E}[B_{T-t}]+\mathbb{E}[B_{T-t}^2]=x^2+T-t$, agreeing with $u(t,x)$.

Passing to the Unbounded-Domain Limit

For the limit, let $R=T-t$. Suppose $u$ extends continuously to the relevant boundary and agrees there with a boundary payoff $g$, while $u(T,x)=h(x)$. Assume that when $\tau_D<R$, the stopped values converge to $g(t+\tau_D,X_{\tau_D})$, and when $\tau_D\ge R$, they converge to $h(X_R)$. If the two events meet at $\tau_D=R$, assume the terminal and boundary values agree there. This avoids assigning inconsistent values at the corner of the time-space boundary.

Theorem: Suppose the stopped identity holds for every $n$. Suppose its discounted endpoint terms converge almost surely to the discounted terminal or boundary payoff just described and form a uniformly integrable family. Suppose also that the stopped running-reward integrals converge in $L^1$. Then $$ u(t,x)=\mathbb{E}\left[e^{-\int_0^R q(t+s,X_s)\,ds}h(X_R)\mathbf{1}_{\{\tau_D\ge R\}}+e^{-\int_0^{\tau_D}q(t+s,X_s)\,ds}g(t+\tau_D,X_{\tau_D})\mathbf{1}_{\{\tau_D<R\}}+\int_0^{R\wedge\tau_D}e^{-\int_0^s q(t+r,X_r)\,dr}f(t+s,X_s)\,ds\right]. $$

Proof. In the stopped identity, the endpoint terms converge almost surely by assumption. Uniform integrability implies convergence of their expectations to the expectation of the limit. The running integrals converge in $L^1$, which also implies convergence of their expectations. The stopping levels satisfy $\rho_n\uparrow R\wedge\tau_D$; the assumed endpoint identification separates this limit into the terminal and boundary cases. Taking limits on both sides of the stopped identity gives the stated formula. $\square$

A useful sufficient condition for the integrability step is polynomial growth together with a moment bound. Suppose, uniformly over $0\le s\le R$, that $|u(t+s,z)|$, $|h(z)|$, and $|g(t+s,z)|$ are bounded by $C(1+|z|^m)$, and $|f(t+s,z)|\le C(1+|z|^m)$. If $\mathbb{E}[\sup_{s\le R}|X_s|^m]<\infty$, then the running integrals are dominated by $CR(1+\sup_{s\le R}|X_s|^m)$, since $q\ge0$ makes the discount at most one. For uniform integrability of the endpoint terms, it suffices to have $\mathbb{E}[\sup_{s\le R}|X_s|^p]<\infty$ for some $p>m$. The standard Burkholder–Davis–Gundy and Gronwall moment estimate gives these finite-horizon moments for globally Lipschitz coefficients of linear growth; that estimate is a standard SDE result. In applications with faster-than-polynomial growth, a separate integrability argument is needed.

Worked Example: Brownian Motion Killed at the Boundary

Let $X_s=x+B_s$, with $x>0$, and $D=(0,\infty)$. Take $q=f=0$, terminal payoff $h(z)=z$, and boundary payoff $g=0$ at $z=0$. The function $u(t,z)=z$ has $u_t=0$, $u_z=1$, and $u_{zz}=0$; hence $u_t+\tfrac12u_{zz}=0$, and its boundary value at zero is zero.

An exhaustion is $D_n=(1/n,n)$, for integers $n$ large enough that $x\in D_n$. On a finite time interval, the path is bounded, so it cannot exit through $n$ once $n$ is sufficiently large. If it reaches zero before the horizon, the lower exit locations $1/n$ converge to zero; otherwise the terminal location converges to $X_R$. The stopped endpoint terms are bounded in absolute value by $\sup_{s\le R}|X_s|$, which is integrable, so the limit is valid.

Thus $x=\mathbb{E}[X_R\mathbf{1}_{\{\tau_0\ge R\}}]$, where $\tau_0=\inf\{s:X_s=0\}$. To check the result directly, write $X_{R\wedge\tau_0}=x+\int_0^R\mathbf{1}_{\{s\le\tau_0\}}\,dB_s$. The integrand is progressively measurable and its squared integral has expectation at most $R$, so the Ito integral has mean zero. The stopped variable is zero on exit and $X_R$ on survival; its expectation is therefore $x$, as represented by the formula.

Worked Example: An Exponential Payoff

On $D=\mathbb{R}$, let $X_s=x+B_s$, $q=f=0$, and $h(z)=e^z$. Set $u(t,z)=e^{z+(T-t)/2}$. Its derivatives are $u_t=-\tfrac12u$, $u_z=u$, and $u_{zz}=u$, so $u_t+\tfrac12u_{zz}=0$, and $u(T,z)=e^z$.

For $D_n=(-n,n)$, localization gives the stopped identity. Since $e^{u}$ is not the relevant quantity here, bound the endpoint directly: $u(t+\rho_n,x+B_{\rho_n})\le e^{x+R/2}e^{\sup_{s\le R}|B_s|}$. The Brownian reflection principle implies that $\sup_{s\le R}|B_s|$ has finite exponential moments of every fixed order, so this is an integrable dominating variable. The limit can therefore be taken.

At $t=0,x=0,T=1$, the representation gives $u(0,0)=e^{1/2}$. The independent Gaussian calculation is $\mathbb{E}[e^{B_1}]=\int_{\mathbb{R}}e^z(2\pi)^{-1/2}e^{-z^2/2}\,dz=e^{1/2}$, by completing the square: $z-z^2/2=1/2-(z-1)^2/2$. This verifies both the PDE representation and the required integrability for this non-polynomial payoff.

Why the Limit Condition Matters

The bounded-domain identity is not the same as the unbounded-domain identity: the former is an identity for each stopped process, while the latter requires convergence of expectations. In particular, almost-sure convergence alone does not suffice. A concrete warning is $Y_n=n\mathbf{1}_{(0,1/n)}$ on the probability space $(0,1)$ with uniform probability. Then $Y_n\to0$ almost surely, but $\mathbb{E}[Y_n]=n(1/n)=1$ for every $n$. Thus replacing the stopped expectations by the expectation of their pointwise limit without uniform integrability can give a wrong answer.

For Feynman–Kac localization, polynomial growth plus a sufficiently high moment is a convenient sufficient condition, not a necessary one. Exponential payoffs can also work when the diffusion has suitable exponential moments, as in the Brownian example. The relevant question is always whether the stopped endpoint and reward terms are integrable uniformly enough to pass to the limit.

Step	What it provides	What it does not provide
Stop at $\tau_n$	A localized Ito identity	Convergence of expectations as $n$ increases
Almost-sure endpoint convergence	Identification of a candidate limiting payoff	Uniform integrability
Growth and moment bounds	A route to domination or uniform integrability	Boundary-value continuity, which must be checked separately

Check Your Understanding

Use the stopped identity first, then identify precisely which limit condition is needed.

For $u(t,x)=x^2+T-t$, compute $u_t$, $u_x$, and $u_{xx}$, and verify the Brownian backward equation.
Explain why a finite moment of order $m$ controls the running reward under polynomial growth of order $m$, while a moment of order $p>m$ is a convenient condition for uniform integrability of endpoint terms.
For the killed Brownian example, identify the terminal payoff on survival and the boundary payoff on exit.
Complete the square in the Gaussian integral for $\mathbb{E}[e^{B_T}]$ and state the resulting value.
Give an example of why almost-sure convergence of stopped endpoint terms is not, by itself, enough to pass to the limit in their expectations.

Step	What it provides	What it does not provide
Stop at \(\tau_n\)	A localized Ito identity	Convergence of expectations as \(n\) increases
Almost-sure endpoint convergence	Identification of a candidate limiting payoff	Uniform integrability
Growth and moment bounds	A route to domination or uniform integrability	Boundary-value continuity, which must be checked separately

Localization on Unbounded Domains

What You'll Learn