From a Killing Rate to a Survival Weight
A diffusion can be removed from consideration at a rate that depends on its current state. For example, a particle might be more likely to be absorbed in regions where a hazard is high. The rate \(c(x)\) is not itself a probability: over a short time interval of length \(dt\), it gives an approximate conditional killing probability \(c(X_t)dt\). Accumulating this rate along the path produces the survival factor that enters the boundary-value representation.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). Consider a diffusion, started at \(X_0=x\in(a,b)\), satisfying \(dX_t=\mu(X_t)\,dt+\sigma(X_t)\,dB_t\) until its first exit time \(\tau=\inf\{t\ge0:X_t\notin(a,b)\}\). Here \(\mu\) is the drift coefficient and \(\sigma\) is the diffusion coefficient. The Ito integral is used: its integrand is progressively measurable and belongs to \(\mathcal{L}^2\), meaning \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\) for each finite \(T\). Assume that a continuous solution exists, that the coefficients are bounded on \([a,b]\), and that \(\mathbb{E}_x[\tau]<\infty\). These assumptions ensure that the stopped Ito integrals below are true martingales. Write \(\mathcal{L}f(x)=\mu(x)f'(x)+\tfrac12\sigma(x)^2f''(x)\) for the diffusion generator.
The last assertion follows directly from the exponential tail: conditional on the path, \(A_t\) is fixed and \(\mathbb{P}(\zeta>t\mid (X_s)_{0\le s\le t})=\mathbb{P}(\xi>A_t)=e^{-A_t}\). In particular, on the event that the diffusion reaches the boundary before it is killed, a boundary reward \(g(X_\tau)\) is received. Conditional on the diffusion path up to \(\tau\), the probability of surviving to that boundary is \(e^{-A_\tau}\). Thus the expected reward, including zero reward when killing occurs first, is \(\mathbb{E}_x[e^{-A_\tau}g(X_\tau)]\). This is an expectation with a survival weight, not generally an average under a probability distribution on the two boundary points.
The Boundary-Value Representation
Proof. Define \(Y_t=e^{-A_{t\wedge\tau}}u(X_{t\wedge\tau})\). Ito's formula, applied to \(u\), uses \(u'(x)\) and \(u''(x)\) and includes the quadratic-variation correction \(\tfrac12u''(X_t)\sigma(X_t)^2\,dt\), since \(dB_t\,dB_t=dt\), \(dB_t\,dt=0\), and \(dt\,dt=0\). The product rule with the finite-variation factor \(e^{-A_t}\) gives, up to the exit time, \[ d(e^{-A_t}u(X_t)) =e^{-A_t}\sigma(X_t)u'(X_t)\,dB_t +e^{-A_t}\big(\mathcal{L}u(X_t)-c(X_t)u(X_t)\big)\,dt. \] The drift term vanishes by the assumed differential equation. Consequently, \[ Y_t=u(x)+\int_0^t\mathbf{1}_{\{s<\tau\}}e^{-A_s}\sigma(X_s)u'(X_s)\,dB_s. \] The integrand is progressively measurable. Since \(u'\) and \(\sigma\) are bounded on \([a,b]\), and \(A_s\ge0\), \[ \mathbb{E}_x\int_0^t \left(\mathbf{1}_{\{s<\tau\}}e^{-A_s}\sigma(X_s)u'(X_s)\right)^2ds \le \|\sigma u'\|_\infty^2\mathbb{E}_x[\tau]<\infty. \] It is therefore an Ito-integrable \(\mathcal{L}^2\) integrand. By the Ito isometry its integral is square-integrable; by the martingale property of square-integrable Ito integrals, its conditional expectation given \(\mathcal{F}_s\) equals its value at time \(s\). In particular, its expectation is zero, so \(\mathbb{E}_x[Y_t]=u(x)\).
Because \(\tau<\infty\) almost surely and paths are continuous, \(X_{t\wedge\tau}\to X_\tau\) almost surely as \(t\to\infty\). Also \(A_{t\wedge\tau}\to A_\tau\). The random variables \(Y_t\) are bounded in absolute value by \(\|u\|_\infty\), so bounded convergence gives \(u(x)=\mathbb{E}_x[e^{-A_\tau}u(X_\tau)]\). At exit, \(X_\tau\) is either \(a\) or \(b\), and the boundary conditions give \(u(X_\tau)=g(X_\tau)\). This proves the representation.
For uniqueness, let \(u_1\) and \(u_2\) be two such solutions and set \(v=u_1-u_2\). Then \((\mathcal{L}-c)v=0\) on \((a,b)\), while \(v(a)=v(b)=0\). Applying the proved representation to \(v\) yields \(v(x)=\mathbb{E}_x[e^{-A_\tau}v(X_\tau)]=0\), since \(X_\tau\) is a boundary point. Thus \(u_1=u_2\) on \((a,b)\), and they agree at the endpoints by their boundary conditions. \(\square\)
The representation theorem is a verification result: if a sufficiently regular candidate solves the differential equation and matches the boundary data, then it equals the survival-weighted expectation. It does not require the candidate to be guessed from the expectation. One can instead propose \(u\), check the equation and boundary values, and invoke the theorem.
Worked Calculations
Worked Example: A State-Dependent Rate and Its Hazard
Let \(X\) be Brownian motion on \((0,1)\), stopped at its first exit, and take \(c(x)=1/(1+x^2)\). Find the rate at \(x=\tfrac12\), and write the conditional probability of surviving until exit given the path.
At \(x=\tfrac12\), the rate is \(c(\tfrac12)=1/(1+(\tfrac12)^2)=1/(1+\tfrac14)=\tfrac45\). This is the instantaneous hazard at that state; it is not the probability of surviving to the boundary. For a path that exits at time \(\tau\), the accumulated rate and conditional survival probability are respectively \[ A_\tau=\int_0^\tau\frac{1}{1+X_s^2}\,ds, \qquad \mathbb{P}(\zeta>\tau\mid (X_s)_{0\le s\le\tau}) =e^{-A_\tau}. \] The full path matters because the rate changes as \(X_s\) moves. Since \(c\ge0\), \(A_\tau\ge0\), so \(0<e^{-A_\tau}\le1\) whenever \(\tau\) is finite. Check: if the rate were constant \(c\equiv\tfrac45\) along a path of duration \(\tau\), this formula would reduce to \(e^{-4\tau/5}\); the state-dependent formula instead integrates the actual rate along the path.
Worked Example: Verifying a State-Dependent Boundary Problem
For Brownian motion on \((0,1)\) with \(c(x)=1/(1+x^2)\), set boundary rewards \(g(0)=1\) and \(g(1)=2\). Verify a candidate for the survival-weighted expectation and evaluate it at \(x=\tfrac12\).
For this diffusion, \(\mu=0\), \(\sigma=1\), and \(\mathcal{L}u=\tfrac12u''\). Try \(u(x)=1+x^2\). Its derivatives are \(u'(x)=2x\) and \(u''(x)=2\). Therefore, at every \(x\in(0,1)\), \[ \mathcal{L}u(x)-c(x)u(x) =\tfrac12(2)-\frac{1}{1+x^2}(1+x^2) =1-1=0. \] The boundary values are \(u(0)=1+0^2=1\) and \(u(1)=1+1^2=2\), as required. The representation theorem now gives \[ \mathbb{E}_x\!\left[ e^{-\int_0^\tau(1+X_s^2)^{-1}\,ds}g(X_\tau) \right]=1+x^2. \] At \(x=\tfrac12\), this value is \(1+(\tfrac12)^2=1+\tfrac14=\tfrac54\).
Check: the value \(\tfrac54\) lies between \(1\) and \(2\) here because the verified function \(1+x^2\) takes values in \([1,2]\) for \(x\in[0,1]\). This interval claim follows from the explicit formula for this example, not from treating the survival weights as a probability average.
Worked Example: Why Survival Weighting Is Not an Ordinary Average
Take Brownian motion on \((0,1)\), constant killing rate \(c(x)=\tfrac12\), and boundary reward \(g(0)=g(1)=1\). Compute the survival-weighted expectation at \(x=\tfrac12\) and compare it with the unweighted boundary reward.
The equation is \(\tfrac12u''-\tfrac12u=0\), or \(u''=u\), with \(u(0)=u(1)=1\). The function \(u(x)=\cosh(x-\tfrac12)/\cosh(\tfrac12)\) satisfies the equation because its second derivative equals itself. At either endpoint the numerator is \(\cosh(\tfrac12)\), so both boundary values are \(1\). The representation theorem therefore gives the desired expectation. At \(x=\tfrac12\), \[ u(\tfrac12)=\frac{\cosh(0)}{\cosh(\tfrac12)} =\frac{1}{\cosh(\tfrac12)}<1, \] since \(\cosh(y)>1\) for \(y>0\). Thus the survival-weighted value is strictly less than the boundary reward \(1\).
Check: here the formula is also \(u(\tfrac12)=\mathbb{E}_{1/2}[e^{-\tau/2}]\), since the reward is always \(1\). The exit time is positive almost surely when starting in the interior, so \(e^{-\tau/2}<1\) almost surely, consistent with the strict inequality. In general, the two boundary contributions have weights \(\mathbb{E}_x[e^{-A_\tau}\mathbf{1}_{\{X_\tau=0\}}]\) and \(\mathbb{E}_x[e^{-A_\tau}\mathbf{1}_{\{X_\tau=1\}}]\). Their sum is \(\mathbb{E}_x[e^{-A_\tau}]\), which can be less than one. They do not generally define probability weights summing to one.
Reading the Representation Correctly
For nonnegative boundary rewards, the survival-weighted expectation is nonnegative and is at most the largest boundary reward: pointwise, \(0\le e^{-A_\tau}\le1\), and \(g(X_\tau)\le\max\{g(a),g(b)\}\). But there is no general lower bound by the smaller boundary reward. Killing can reduce the total weight below one, as the constant-rate example demonstrates. For rewards of mixed sign, even the simple nonnegative comparison does not apply without additional care.
A common incorrect calculation is to write the expectation as a weighted average of \(g(a)\) and \(g(b)\) merely because each survival factor is at most one, then infer that the answer lies between the two rewards. The correction is to retain the total mass of the weights: \[ \mathbb{E}_x[e^{-A_\tau}g(X_\tau)] =g(a)\mathbb{E}_x[e^{-A_\tau}\mathbf{1}_{\{X_\tau=a\}}] +g(b)\mathbb{E}_x[e^{-A_\tau}\mathbf{1}_{\{X_\tau=b\}}]. \] The coefficients are nonnegative, but their sum is generally less than one. In the state-dependent example, the value \(\tfrac54\) lies between the rewards because the independently verified solution \(1+x^2\) lies in \([1,2]\), not because the coefficients necessarily sum to one.
The killing term enters the equation as \(-c(x)u(x)\): it removes expected future boundary reward at a rate proportional to the current value. This term is distinct from the diffusion drift \(\mu\), which describes the local mean movement of \(X_t\), and the diffusion coefficient \(\sigma\), which controls its local variance. The integral used in the verification is an Ito integral, not a Stratonovich integral; the \(\tfrac12\sigma^2u''\) term comes from Brownian quadratic variation. The argument above proves the needed expectation identity under the stated bounded-domain and integrability assumptions, rather than relying on an unverified formal PDE calculation.
Check Your Understanding
Keep the survival weight in view, and distinguish the differential equation from its probabilistic interpretation.
- For \(c(x)=2+x\), write the accumulated killing rate and the conditional probability of surviving to a finite exit time \(\tau\).
- For Brownian motion on \((0,1)\) with \(c(x)=1/(1+x^2)\), verify directly that \(u(x)=1+x^2\) solves the equation with boundary values \(1\) and \(2\).
- Explain why the two survival-weighted boundary coefficients need not sum to one.
- In the stopped-Ito proof, why does boundedness of \(\sigma u'\) and finiteness of \(\mathbb{E}_x[\tau]\) ensure that the stochastic integral is a true martingale?
- For constant killing rate \(c=\lambda>0\) and unit reward at both endpoints, what equation and boundary values must the representing function satisfy?