Why the PDE Is Not the Whole Representation
The Feynman–Kac formula connects a parabolic PDE to an expectation along a diffusion. The connection is powerful, but it is not automatic: solving the differential equation does not by itself establish that the expectation exists or that it equals the solution. A diffusion may explode before the terminal time, a payoff may fail to be integrable, or a boundary term created by localization may fail to disappear.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). Consider the SDE \(dX_s=\mu(s,X_s)\,ds+\sigma(s,X_s)\,dB_s\), with initial condition \(X_t=x\). Here \(\mu\) is the drift and \(\sigma\) the diffusion coefficient. Unless an integral is explicitly stopped and localized, an Itô integrand \(H_s\) is required to be adapted and progressively measurable with \(\mathbb{E}\int_t^T H_s^2\,ds<\infty\); such integrands form the class \(\mathcal{L}^2\). Locally defined SDE integrals are interpreted by stopping on compact regions where the integrands are in \(\mathcal{L}^2\).
The classical theorem gives this representation under appropriate regularity, existence, and integrability conditions. On an unbounded state space, a common sufficient condition is that the solution \(u\) is bounded, the diffusion is nonexplosive, and the coefficients are locally bounded. More general results are possible, but they must control the boundary contribution or establish suitable uniform integrability. The following localization argument makes that requirement explicit.
What Localization Proves—and What It Does Not
Proof. Set \(F(s,y)=e^{-r(s-t)}u(s,y)\). Its derivatives are \(F_s=e^{-r(s-t)}(u_s-ru)\), \(F_y=e^{-r(s-t)}u_y\), and \(F_{yy}=e^{-r(s-t)}u_{yy}\). Itô’s formula, including its quadratic-variation correction, gives $$ dF(s,X_s)=e^{-r(s-t)}\left(u_s+\mu u_x+\tfrac12\sigma^2u_{xx}-ru\right)(s,X_s)\,ds +e^{-r(s-t)}u_x(s,X_s)\sigma(s,X_s)\,dB_s. $$ The drift term is zero by the PDE. Up to \(\tau_n\), the process remains in \([-n,n]\). On the compact set \([t,T]\times[-n,n]\), local boundedness of \(\mu,\sigma\), continuity of \(u_x\), and boundedness of the exponential imply $$ \mathbb{E}\int_t^{\tau_n}\left(e^{-r(s-t)}u_x(s,X_s)\sigma(s,X_s)\right)^2 ds<\infty. $$ Thus the stopped Itô integral is a square-integrable martingale: it has finite second moment by the Itô isometry, and its conditional expectation of each future increment given the present filtration is zero. Taking expectations in the stopped identity yields the asserted formula. Finally, \(L^1\) convergence implies convergence of expectations, so passing to the limit gives the unstopped representation. \(\square\)
This argument also gives a useful special case. If \(X\) is nonexplosive, \(u\) is bounded, and \(u(T,\cdot)=g\), then the stopped values are bounded by a fixed constant when \(r\ge0\). For each continuous nonexplosive path, \(\tau_n=T\) for all sufficiently large \(n\), so the stopped values converge almost surely to the terminal value. Bounded convergence gives \(L^1\) convergence. Without boundedness or another uniform-integrability argument, the same passage is not justified.
Worked Example: A Diffusion That Explodes Before the Payoff Time
Consider \(dX_s=X_s^2\,ds+0\,dB_s\), started at time \(t=0\) from \(x=1\). Find its path and determine whether a terminal payoff \(g(X_2)\) is defined by the ordinary state-space model.
The equation is the deterministic ODE \(X'_s=X_s^2\). For \(X_s>0\), differentiating \(1/X_s\) gives \((1/X_s)'=-1\). Since \(X_0=1\), integration yields \(1/X_s=1-s\), hence \(X_s=1/(1-s)\) for \(0\le s<1\). As \(s\uparrow1\), \(X_s\to+\infty\). The solution therefore explodes at time \(1\), before the proposed terminal time \(2\).
Check: differentiation gives \(X'_s=(1-s)^{-2}=X_s^2\), and \(X_0=1\). There is no finite-valued \(X_2\) in this model, so \(g(X_2)\) is not an ordinary terminal random variable. A PDE calculation cannot supply a missing terminal state. One must change the model or specify an appropriate boundary or cemetery-state payoff before asking for a representation.
Nonintegrable Payoffs Can Break the Expectation
Even for a nonexplosive process, the expectation in the formula may be infinite. The failure can occur at a critical starting time even when a smooth PDE solution exists at every later time. This example uses standard Brownian motion, whose paths do not explode.
Worked Example: A Critical Gaussian Payoff
Let \(X_s=B_s\), \(T>0\), \(r=0\), and \(g(y)=\exp(y^2/(2T))\). Compute the expectation for \(0\le t<T\), starting from \(X_t=x\), and check its behavior at \(t=0\).
For \(t>0\), \(X_T\) has the normal distribution with mean \(x\) and variance \(T-t\). Put \(q=T-t\). Its density gives $$ \mathbb{E}[g(X_T)\mid X_t=x] =\frac{1}{\sqrt{2\pi q}}\int_{\mathbb{R}} \exp\!\left(\frac{y^2}{2T}-\frac{(y-x)^2}{2q}\right)dy. $$ The coefficient of \(y^2\) in the exponent is \(\tfrac1{2T}-\tfrac1{2q}=-\tfrac{t}{2Tq}\), which is negative when \(t>0\). Completing the square and evaluating the resulting Gaussian integral gives $$ u(t,x)=\sqrt{\frac{T}{t}}\exp\!\left(\frac{x^2}{2t}\right),\qquad t>0. $$
At \(t=0\), the density has variance \(T\), so the factors involving \(y^2\) cancel: $$ \mathbb{E}[g(B_T)]=\frac{1}{\sqrt{2\pi T}}\int_{\mathbb{R}}\exp\!\left(-\frac{y^2}{2T}+\frac{y^2}{2T}\right)dy =\frac{1}{\sqrt{2\pi T}}\int_{\mathbb{R}}1\,dy=\infty. $$ Thus the payoff is not integrable at the critical starting time. The finite expression for \(t>0\) does not extend to a finite value at \(t=0\).
Check: for any fixed \(t>0\), the completed-square exponent has a strictly negative quadratic coefficient, so the integral is finite. At \(t=0\), that coefficient is exactly zero, leaving an infinite integral. The distinction is integrability, not a failure of Brownian nonexplosion.
Boundary Behavior: A PDE Solution from a Strict Local Martingale
Localization yields an identity at each finite stopping level. To recover the terminal expectation, one must still control the stopped values as the levels increase. A local martingale need not be a martingale: it may have the zero-drift Itô form after localization but fail the integrability and conditional-expectation requirements of a true martingale.
Worked Example: A Formal PDE Solution That Is Not the Expectation
Let \(R_s\) be a three-dimensional Bessel process started from \(R_0=a>0\), satisfying \(dR_s=dB_s+\frac{1}{R_s}\,ds\), and put \(X_s=1/R_s\). Show that \(u(t,x)=x\) solves the terminal-payoff PDE for \(g(x)=x\), but does not equal the corresponding expectation.
Itô’s formula for \(f(y)=1/y\) has derivatives \(f'(y)=-y^{-2}\) and \(f''(y)=2y^{-3}\). Therefore $$ dX_s=-R_s^{-2}\left(dB_s+\frac1{R_s}ds\right) +\frac12(2R_s^{-3})\,ds=-X_s^2\,dB_s. $$ This stochastic integral is understood locally; after stopping \(X\) in a compact subinterval of \((0,\infty)\), its integrand is bounded and belongs to \(\mathcal{L}^2\) on finite time intervals. The generator of \(X\) on \((0,\infty)\) is \(\mathcal{L}f(x)=\tfrac12x^4 f''(x)\). Since \(u(t,x)=x\) has \(u_t=0\), \(u_x=1\), and \(u_{xx}=0\), it satisfies \(u_t+\mathcal{L}u=0\) and \(u(T,x)=x=g(x)\).
However, the reciprocal of a three-dimensional Bessel process is a strict local martingale. To compute its mean, use the standard three-dimensional Bessel transition density $$ q_s(a,y)=\frac{y}{a}\bigl(p_s(y-a)-p_s(y+a)\bigr),\qquad p_s(z)=\frac{1}{\sqrt{2\pi s}}e^{-z^2/(2s)},\quad y>0. $$ This density formula is a standard Bessel-process result. It gives, for \(s>0\), $$ \mathbb{E}[X_s]=\int_0^\infty \frac1y q_s(a,y)\,dy =\frac1a\int_0^\infty\bigl(p_s(y-a)-p_s(y+a)\bigr)dy =\frac{2\Phi(a/\sqrt{s})-1}{a}<\frac1a=X_0. $$ The last strict inequality follows because \(a/\sqrt{s}\) is finite and the standard normal CDF satisfies \(\Phi(z)<1\) for finite \(z\).
Check: as \(s\downarrow0\), \(a/\sqrt{s}\to\infty\), so the displayed mean tends to \(1/a=X_0\). For every \(s>0\), it is strictly smaller. Thus the PDE solution \(u(t,x)=x\) is not the expectation \(\mathbb{E}[X_T\mid X_t=x]\). The discrepancy is precisely why a local martingale argument alone cannot justify removing localization.
Common Mistake: Dropping the Stopped Boundary Term
A tempting but invalid calculation is: apply Itô’s formula, observe that the PDE drift is zero, and immediately conclude \(u(t,x)=\mathbb{E}[e^{-r(T-t)}g(X_T)]\). This step silently treats the local stochastic integral as a true martingale and assumes the process reaches time \(T\) inside the state space with an integrable payoff.
The justified calculation stops first and gives $$ u(t,x)=\mathbb{E}\!\left[e^{-r(\tau_n-t)}u(\tau_n,X_{\tau_n})\right]. $$ Only after proving that these stopped terminal values converge in \(L^1\), or establishing a suitable uniform-integrability condition, may one pass to the terminal expectation. If the process explodes, the terminal state may not exist; if the payoff is nonintegrable, the expectation may be infinite; if boundary values persist, the stopped terms need not converge to the proposed terminal payoff. These are distinct failures and require distinct repairs.
In applications, check the following before treating a PDE solution as a Feynman–Kac representation:
- Existence through the horizon: verify nonexplosion, or specify how the model handles explosion.
- Payoff integrability: verify that the discounted terminal payoff has a finite expectation.
- Boundary control: justify the limit of stopped values, for example through boundedness or uniform integrability.
- Martingale status: do not replace “local martingale” by “martingale” without checking integrability and conditional expectations.
Check Your Understanding
For each question, state the hypothesis that makes the proposed expectation or limiting step meaningful.
- For \(X'_s=X_s^2\) with \(X_0=2\), compute the explosion time and decide whether \(X_1\) exists as a finite state.
- For the Gaussian payoff \(g(y)=e^{y^2/(2T)}\), identify the coefficient of \(y^2\) in the density integral when the starting time is \(t>0\), and explain why it changes at \(t=0\).
- In the localization theorem, why does local boundedness of \(\sigma\) and continuity of \(u_x\) imply square integrability of the stopped Itô integrand?
- Explain why the PDE solution \(u(t,x)=x\) for the reciprocal Bessel example does not imply that \(X\) is a true martingale.
- Give one sufficient condition that permits passage from the stopped Feynman–Kac identity to the terminal expectation.