Brownian Interval Problems with Boundary Values

From Boundary Rewards to an Interval Equation

Consider Brownian motion started at \(x\) inside a bounded interval. It eventually exits through one of the two endpoints. If a reward is paid at exit and discounted at a constant rate, its expected value depends on the starting point. The central question is to calculate that function and verify which boundary-value problem it solves.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\) relative to the filtration. Fix \(a<b\), let \(B_0=x\in[a,b]\) (equivalently, use Brownian motion started at zero and the process \(x+B_t\)), and define the first exit time \(\tau=\inf\{t\ge0:B_t\notin(a,b)\}\). For \(x\in(a,b)\), continuity of Brownian paths implies \(B_\tau\in\{a,b\}\) whenever \(\tau\) is finite. The rewards at the two endpoints are fixed real numbers \(A\) and \(C\). The constant killing rate \(\lambda\) will be assumed nonnegative.

This expectation is well defined: Brownian motion exits every bounded interval in finite time almost surely. For completeness, write \(L=b-a\) and choose time blocks of length \(L^2\). If the process is still inside the interval at the start of a block, an increment greater than \(L\) during that block forces it to exit by the block's end. The probability of such an increment is \(p=\mathbb{P}(N(0,L^2)>L)=\mathbb{P}(N(0,1)>1)>0\), independently of the past. Thus \(\mathbb{P}_x(\tau>(n+1)L^2)\le(1-p)\mathbb{P}_x(\tau>nL^2)\). Iterating gives \(\mathbb{P}_x(\tau>nL^2)\le(1-p)^n\), which tends to zero.

The Exit Representation and Its ODE

Proof. Fix \(x\in(a,b)\). For a finite time horizon \(T\), apply Ito's formula to \(f(t,y)=e^{-\lambda t}v(y)\) and the stopped process \(B_{t\wedge\tau}\). Its derivatives are \(f_t=-\lambda e^{-\lambda t}v(y)\), \(f_y=e^{-\lambda t}v'(y)\), and \(f_{yy}=e^{-\lambda t}v''(y)\). The Ito correction is \(\tfrac12 f_{yy}\,dt\), so the drift is \(e^{-\lambda t}(\tfrac12v''-\lambda v)\,dt=0\). In differential form, up to exit, \(d(e^{-\lambda t}v(B_t))=e^{-\lambda t}v'(B_t)\,dB_t\). More precisely, the stopped integral has integrand \(H_s=\mathbf{1}_{\{s<\tau\}}e^{-\lambda s}v'(B_s)\). For each finite \(T\), this integrand is progressively measurable and belongs to \(\mathcal{L}^2\), since \(v'\) is bounded on the compact interval after continuous extension (or, equivalently, first stop inside \((a,b)\) and pass to the boundary). In the latter formulation, one may stop at \(\tau_n=\inf\{t:B_t\notin(a+1/n,b-1/n)\}\) for sufficiently large \(n\); the stopped integrands are bounded, and continuity gives the boundary limit. The Ito integral is therefore a true martingale on each finite horizon, not merely a local martingale: its increments have conditional expectation zero by the defining \(L^2\) construction and the Ito isometry. Consequently \(\mathbb{E}_x[e^{-\lambda(T\wedge\tau)}v(B_{T\wedge\tau})]=v(x)\). As \(T\to\infty\), \(\tau<\infty\) almost surely, so the random variable inside the expectation converges to \(e^{-\lambda\tau}g(B_\tau)\). It is bounded in absolute value by \(\sup_{[a,b]}|v|\), because \(\lambda\ge0\). Dominated convergence proves the asserted representation. \(\square\)

The ODE is the one-dimensional generator equation with killing: Brownian motion has generator \(\tfrac12\,d^2/dx^2\), and discounting contributes the term \(-\lambda v\). The equation says that these two contributions balance in the interior. Its boundary values specify the rewards; they are not additional source terms.

Explicit Solution and Uniqueness

Proof. When \(\lambda=0\), the equation is \(u''=0\), so \(u(x)=m x+n\). Solving \(ma+n=A\) and \(mb+n=C\) gives the displayed linear interpolation. When \(\lambda>0\), the equation is \(u''=k^2u\). Its general solution is \(c_1e^{kx}+c_2e^{-kx}\), or equivalently a linear combination of \(\sinh(kx)\) and \(\cosh(kx)\). The displayed formula satisfies the ODE because each hyperbolic-sine term has second derivative \(k^2\) times itself. At \(x=a\), its first coefficient is \(A\) and its second is zero; at \(x=b\), its first is zero and its second is \(C\). The denominators are nonzero since \(k(b-a)>0\) and \(\sinh\) is positive on positive arguments.

To prove uniqueness, let \(w\) be the difference of two solutions. Then \(w(a)=w(b)=0\) and \(w''=2\lambda w\). If \(\lambda=0\), \(w\) is linear and its two zero endpoint values force \(w=0\). If \(\lambda>0\), a positive value of \(w\) would imply a positive interior maximum, where \(w''\le0\); but at that maximum \(w''=2\lambda w>0\), a contradiction. A negative value would imply a negative interior minimum, where \(w''\ge0\), contradicting \(w''=2\lambda w<0\). Thus \(w=0\) throughout. This proves existence and uniqueness. Applying the exit-representation theorem to this solution identifies it with the discounted boundary-reward expectation. \(\square\)

When \(\lambda\) decreases to zero, the hyperbolic formula tends to the linear one: \(\sinh(z)/z\to1\) as \(z\to0\). Thus the zero-killing case is also the continuous limit of the discounted problem.

Why the Discount Changes the Interpolation

With no killing, the expected reward is the linear interpolation of the endpoint values, because it is determined by the probabilities of exiting on each side. Positive killing weights exit rewards by the additional factor \(e^{-\lambda\tau}\). Since this factor is at most one and decreases with exit time, the discounted expectation is generally smaller than the undiscounted expected reward when the boundary rewards are nonnegative. The ODE records this change through the term \(-\lambda u\), and the hyperbolic functions replace the linear weights.

A common mistake is to keep using linear interpolation when \(\lambda>0\). In the symmetric example, that would give \(u(0)=1\) for unit endpoint rewards. But a constant candidate \(u(x)=1\) has \(u''=0\), so its ODE residual is \(\tfrac12u''-\tfrac12u=-\tfrac12\), not zero. The correct value is \(1/\cosh(1)<1\). The missing term is the killing contribution, not a change in the Brownian exit probabilities alone.

The interval setting makes the boundary representation particularly direct: exit occurs at one of two known points, the reward is bounded, and the ODE has an explicit solution. These calculations provide a useful test of the general principle that stopped, discounted diffusion expectations solve boundary-value equations. The argument relies on the bounded domain and nonnegative killing used here; other domains or additional drift and running costs require their own hypotheses and analysis.