From Pathwise Order to Hitting-Time Bounds
Comparison is useful because an order between sample paths reverses when it is translated into a hitting time for an upper level. If a continuous path \(X\) lies below a path \(Y\), then \(Y\) cannot take longer than \(X\) to reach a level above their common starting point. This simple observation gives bounds on hitting probabilities and, in suitable settings, on explosion times.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a Brownian motion \(B_t\). All stochastic integrals below are Ito integrals. In an SDE, the integrand \(\sigma(X_t)\) is progressively measurable; the integral is defined locally by stopping the solution on compact state intervals, where \(\mathbb{E}\int_0^T \sigma(X_s)^2\,ds<\infty\). In particular, each such stopped integrand belongs to the class \(\mathcal{L}^2\). Solutions are strong solutions driven by this same \(B_t\), not merely weak solutions specified by their distributions. Pathwise uniqueness means that two solutions with the same initial value and Brownian motion agree almost surely up to their lifetimes; it is the coupling that makes pathwise comparison meaningful.
The comparison result used here is the scalar SDE comparison theorem. In the form needed below, the equations have the same Brownian motion and the same diffusion coefficient \(\sigma\), their coefficients are locally Lipschitz, and their initial values are ordered. If their drift coefficients satisfy \(b_1(x)\le b_2(x)\) throughout the common state interval, then their strong solutions satisfy \(X^1_t\le X^2_t\) almost surely for all times before either solution exits that interval. We use this standard result from the earlier material on scalar SDE comparison; its hypotheses matter, in particular the common diffusion coefficient and common Brownian motion.
Proof. Fix a sample path on which the stated order holds. If \(X\) reaches \(R\) at a finite time \(s\), then \(Y^+_s\ge X_s\ge R\), so \(Y^+\) has reached \(R\) no later than \(s\). This proves \(\tau_R(Y^+)\le\tau_R(X)\); if \(X\) never reaches \(R\), the inequality still holds because its right side is \(\infty\). Likewise, if \(Y^-\) reaches \(R\) at time \(s\), then \(X_s\ge Y^-_s\ge R\), so \(\tau_R(X)\le s\). If \(Y^-\) never reaches \(R\), the right side of this latter inequality is \(\infty\). Thus the hitting-time inequalities hold pathwise, including the cases of infinite hitting times. For each fixed \(t\), they imply the event inclusions \(\{\tau_R(Y^-)\le t\}\subseteq\{\tau_R(X)\le t\}\subseteq\{\tau_R(Y^+)\le t\}\). Taking probabilities proves the claim. \(\square\)
Worked Example: A Bounded Drift and a Level-Hitting Probability
Let \(X_0=0\) and let \(X\) solve \(dX_t=(1+\tfrac12\sin X_t)\,dt+dB_t\). Bound the probability that \(X\) reaches level \(1\) by time \(1\).
The drift satisfies \(\tfrac12\le 1+\tfrac12\sin x\le\tfrac32\) for every real \(x\). The lower and upper comparison processes, driven by the same Brownian motion, are therefore \(Y^-_t=\tfrac12t+B_t\) and \(Y^+_t=\tfrac32t+B_t\). Indeed, integrating the three equations gives \(Y^-_t\le X_t\le Y^+_t\) pathwise. All three equations have constant diffusion coefficient \(1\); their coefficients are globally Lipschitz, so the strong solutions exist for all finite times.
For \(c>0\), the reflection-principle formula for Brownian motion with constant drift gives \(\mathbb{P}(\sup_{0\le s\le t}(B_s+cs)\ge a)=\Phi((ct-a)/\sqrt{t})+e^{2ca}\Phi((-ct-a)/\sqrt{t})\), for \(a>0\) and \(t>0\). Applying it at \(a=t=1\), first with \(c=\tfrac12\) and then with \(c=\tfrac32\), yields $$ \Phi(-\tfrac12)+e\,\Phi(-\tfrac32) \ \le\ \mathbb{P}(\tau_1(X)\le1)\ \le\ \Phi(\tfrac12)+e^3\Phi(-\tfrac52). $$
Check: The lower expression uses \((ct-a)/\sqrt{t}=(\tfrac12-1)/1=-\tfrac12\) and \(2ca=2(\tfrac12)(1)=1\). The upper expression uses \((\tfrac32-1)/1=\tfrac12\) and \(2(\tfrac32)(1)=3\); its second CDF argument is \(-\tfrac32-1=-\tfrac52\). The direction of the bounds also agrees with the path order: the faster comparison process \(Y^+\) has the larger probability of reaching the level by time \(1\).
Explosion at Positive Infinity
For a process that can leave its state space through more than one boundary, its full maximal lifetime need not be the limit of upper-level hitting times. It is therefore useful to define the upper explosion time separately. For levels \(R>x_0\), the times \(\tau_R(X)\) are nondecreasing as \(R\) increases. Define \(\zeta_{+\infty}(X)=\lim_{R\to\infty}\tau_R(X)\), allowing the value \(\infty\). This records escape through positive infinity. It equals the maximal lifetime only when the solution cannot terminate through another boundary and its only possible finite-time loss of existence is explosion to positive infinity.
Proof. For each \(R>x_0\), the theorem gives \(\tau_R(Y^+)\le\tau_R(X)\le\tau_R(Y^-)\). As \(R\) increases, each sequence of hitting times is nondecreasing, so each has a limit in \([0,\infty]\). Taking limits in the inequalities preserves their order and gives the displayed bounds. The assumption excluding finite-time loss at any other boundary ensures that the upper-level limits describe the full maximal lifetimes in this proposition. Without that assumption the limit still defines upper explosion, but need not be the maximal lifetime. \(\square\)
Worked Example: An Explicit Reference Diffusion for Explosion
Let \(X_0=1\) and consider $$ dX_t=(2+\sin X_t)X_t^2\,dt+X_t^2\,dB_t,\qquad X_t>0. $$ Bound its upper explosion time using explicit reference diffusions.
For every \(x>0\), \(x^2\le(2+\sin x)x^2\le3x^2\). The comparison equations have drift \(cx^2\), diffusion coefficient \(x^2\), and initial value \(1\), with \(c=1\) and \(c=3\). Denote their solutions by \(Y^{(1)}\) and \(Y^{(3)}\). The coefficients are locally Lipschitz on \((0,\infty)\), and all three equations use the same diffusion coefficient and Brownian motion. Thus, up to their lifetimes, \(Y^{(1)}_t\le X_t\le Y^{(3)}_t\).
To solve a reference equation \(dY_t=cY_t^2\,dt+Y_t^2\,dB_t\), apply Ito’s formula to \(f(y)=1/y\), up to the first time \(Y\) leaves a compact subinterval of \((0,\infty)\). The derivatives are \(f'(y)=-y^{-2}\) and \(f''(y)=2y^{-3}\); the Ito correction is \(\tfrac12 f''(Y_t)(Y_t^2)^2\,dt=Y_t\,dt\). Hence, for \(U_t=1/Y_t\), $$ dU_t=-c\,dt-dB_t+U_t\,dt=(U_t-c)\,dt-dB_t. $$ The integral \(\int_0^t e^{-s}\,dB_s\) is an Ito integral with deterministic progressively measurable integrand \(e^{-s}\), and \(\mathbb{E}\int_0^t e^{-2s}\,ds<\infty\). Solving the linear equation gives $$ U_t=e^t\left(1-c(1-e^{-t})-\int_0^t e^{-s}\,dB_s\right). $$
Before \(U\) reaches zero, \(Y=1/U\) is positive and finite. The displayed explicit formula is finite and continuous at every finite time, so \(U\) cannot diverge to positive infinity in finite time; consequently \(Y\) cannot reach zero in finite time. Its finite-time loss of existence, if any, is therefore explosion to positive infinity, exactly when \(U\) first reaches zero. The lower comparison process \(Y^{(1)}\) remains positive up to its explosion; the pathwise order then prevents \(X\) from reaching zero earlier. Thus \(X\) also has no finite-time loss at the lower boundary, and its maximal lifetime is its upper explosion time. The proposition gives $$ \zeta_{+\infty}(Y^{(3)})\le\zeta_{+\infty}(X)\le\zeta_{+\infty}(Y^{(1)}). $$
These bounds are explicit in terms of first hitting times of zero by the displayed linear processes \(U^{(3)}\) and \(U^{(1)}\). For instance, the upper bound says that whenever \(U^{(1)}\) hits zero by time \(t\), \(X\) has exploded by time \(t\). It gives a lower bound on the probability of explosion for \(X\), without requiring a closed-form distribution for the first hitting time of the linear process.
Check: The comparison directions are consistent: the larger-drift reference \(Y^{(3)}\) lies above \(X\), so it can explode no later than \(X\); the smaller-drift reference \(Y^{(1)}\) lies below \(X\), so \(X\) can explode no later than it. When \(c=1\), the formula at \(t=\log 2\) has mean \(e^t(1-(1-e^{-t}))=1\). Its variance is \(e^{2t}\int_0^t e^{-2s}\,ds=(e^{2t}-1)/2=3/2\). Thus \(U^{(1)}_{\log 2}\le0\) has probability \(\Phi(-1/\sqrt{3/2})=\Phi(-\sqrt{2/3})\). Continuity and \(U^{(1)}_0=1\) imply that this event entails a zero crossing by time \(\log 2\), so \(\mathbb{P}(\zeta_{+\infty}(X)\le\log 2)\ge\Phi(-\sqrt{2/3})\).
Worked Example: A Fixed-Level Bound from the Same Reference
For the diffusion in the preceding example, bound the probability of reaching level \(2\) by time \(\log 2\). The lower reference \(Y^{(1)}\) has reciprocal \(U^{(1)}\), which starts at \(1\). If \(U^{(1)}_{\log 2}\le\tfrac12\), continuity ensures that \(U^{(1)}\) has reached \(\tfrac12\) by that time, unless it has already reached zero. In either case \(Y^{(1)}\) has reached level \(2\) by time \(\log 2\). Since \(Y^{(1)}\le X\), this also forces \(X\) to have reached level \(2\).
At \(t=\log 2\), the calculation above gives \(U^{(1)}_t\sim N(1,\tfrac32)\). Therefore $$ \mathbb{P}(\tau_2(X)\le\log 2) \ \ge\ \mathbb{P}(U^{(1)}_{\log 2}\le\tfrac12) \ =\Phi\left(\frac{\tfrac12-1}{\sqrt{3/2}}\right) \ =\Phi\left(-\frac1{\sqrt6}\right). $$
Check: The standardized threshold is \((-\tfrac12)/\sqrt{3/2}\). Since \(\sqrt{3/2}=\sqrt6/2\), this ratio is \(-1/\sqrt6\). The event used is a subset of the reference process’s level-\(2\) hitting event; pathwise order then makes it a subset of \(X\)’s level-hitting event as well.
Common Pitfall: Upper Explosion Is Not Always the Lifetime
It is incorrect to identify a process’s maximal lifetime with \(\lim_{R\to\infty}\tau_R(X)\) without ruling out other ways the process can cease to exist. For a concrete example, take the state space \((0,\infty)\), \(X_0=1\), and \(dX_t=-dt\). The solution is \(X_t=1-t\) for \(0\le t<1\), and its maximal lifetime in that state space is \(1\), when it reaches the lower boundary. For every \(R>1\), however, it never reaches \(R\), so \(\tau_R(X)=\infty\) and \(\lim_{R\to\infty}\tau_R(X)=\infty\). The limit correctly records that there is no explosion at positive infinity; it does not record the lower-boundary termination.
Another common error is to reverse the hitting-time inequality. If \(X_t\le Y_t\), then for an upper level \(R\), it is \(Y\) that can hit first: \(\tau_R(Y)\le\tau_R(X)\). The event inclusions in the theorem are a reliable way to check both the direction and the resulting probability bounds.
In applications, the comparison processes need not give an exact distribution for the target hitting time. They can still provide useful bounds: a reference hitting event gives a one-sided probability estimate, and explicit transformations such as the reciprocal in the explosive example can turn a nonlinear diffusion into a tractable linear process. These conclusions depend on the shared-noise, shared-diffusion comparison hypotheses and on keeping upper explosion distinct from other boundary behavior.
Check Your Understanding
For each question, assume the stated hitting times are defined for continuous paths and take probability bounds under \(\mathbb{P}\).
- If \(X_t\le Y_t\) for all times and both start below \(R\), which upper-level hitting time is no larger?
- Suppose \(Y^-\le X\le Y^+\). Write the event inclusions that produce the bounds on \(\mathbb{P}(\tau_R(X)\le t)\).
- For \(dY_t=cY_t^2\,dt+Y_t^2\,dB_t\), compute the first and second derivatives of \(f(y)=1/y\) and identify the Ito correction.
- Why does \(\lim_{R\to\infty}\tau_R(X)\) fail to equal the maximal lifetime for \(dX_t=-dt\), \(X_0=1\), on \((0,\infty)\)?
- In the bounded-drift example, what drift bounds provide the lower and upper comparison processes, and why does the faster one give an upper bound on the target hitting probability?