What the Scalar Comparison Theorem Says
In a scalar SDE, drift pushes the state in a direction determined by its sign, while diffusion describes its response to Brownian fluctuations. The comparison theorem says that, under suitable regularity assumptions, ordered initial states and ordered drifts preserve order when both equations use the same diffusion coefficient and the same Brownian motion. The conclusion is pathwise: outside a single event of probability zero, the two solution paths are ordered at every time in the interval under consideration.
Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\) adapted to the filtration. All stochastic integrals below are Itô integrals. An integrand \(H_s\) is in \(\mathcal{L}^2\) on \([0,T]\) when it is progressively measurable and \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\); such an integral is a square-integrable martingale. In particular, for \(u\ge t\), \(\mathbb{E}[\int_t^u H_s\,dB_s\mid\mathcal{F}_t]=0\), whereas \(\mathbb{E}[\int_0^t H_s\,dB_s\mid\mathcal{F}_t]\) is generally the integral itself, not zero.
We use globally Lipschitz coefficients to ensure well-posedness without needing to stop solutions before they leave bounded regions. The comparison argument itself focuses on how the two drifts and the shared diffusion interact.
Global Lipschitz coefficients with linear growth have unique strong solutions and finite second moments on every bounded time interval; this is the standard global Lipschitz existence-and-uniqueness theorem for SDEs. These hypotheses also ensure the stochastic integrals used in the proof are square-integrable. The drift is \(b_i\), and the diffusion coefficient is \(\sigma\); the theorem requires that the diffusion coefficient be the same function in both equations, not merely that the two diffusion terms have a particular ordering.
Proof by the Positive Part
Set \(D_t=X_t-Y_t\), so order preservation is equivalent to \(D_t^+=0\), where \(z^+=\max(z,0)\). Tanaka’s formula, the positive-part version of Itô’s formula, gives \(D_t^+=D_0^++\int_0^t\mathbf{1}_{\{D_s>0\}}\,dD_s+\tfrac12 L_t^0(D)\). Here \(L_t^0(D)\) is the symmetric local time of \(D\) at zero. We first show that this local-time term vanishes under the shared Lipschitz diffusion assumption.
Proof. Write \(L_\sigma\) for a uniform Lipschitz constant of \(\sigma\) in its space variable. The continuous semimartingale \(D\) has diffusion coefficient \(\sigma(s,X_s)-\sigma(s,Y_s)\), so its quadratic variation is \([D]_t=\int_0^t(\sigma(s,X_s)-\sigma(s,Y_s))^2\,ds\). Consequently, for \(\varepsilon>0\), the occupation-density approximation to local time is \(A_\varepsilon=\frac{1}{2\varepsilon}\int_0^T\mathbf{1}_{\{|D_s|<\varepsilon\}}\,d[D]_s\); as \(\varepsilon\downarrow0\), \(A_\varepsilon\) converges in probability to \(L_T^0(D)\). Lipschitz continuity implies \(|\sigma(s,X_s)-\sigma(s,Y_s)|\le L_\sigma|D_s|\). Thus \[ \mathbb{E}[A_\varepsilon]\le\frac{L_\sigma^2}{2\varepsilon}\int_0^T\mathbb{E}[\mathbf{1}_{\{|D_s|<\varepsilon\}}D_s^2]\,ds \le \frac{L_\sigma^2 T\varepsilon}{2}. \] The last bound tends to zero. Choose a sequence \(\varepsilon_n\downarrow0\) along which \(A_{\varepsilon_n}\to L_T^0(D)\) almost surely; such a subsequence exists because convergence in probability has an almost surely convergent subsequence. Fatou’s lemma gives \(\mathbb{E}[L_T^0(D)]\le\liminf_n\mathbb{E}[A_{\varepsilon_n}]=0\). Since local time is nonnegative, \(L_T^0(D)=0\) almost surely. The same argument applies on each fixed interval \([0,t]\), so the local-time term in Tanaka’s formula vanishes on \([0,T]\) (first take rational \(t\), then use continuity). \(\square\)
Now let \(L_1\) be a global Lipschitz constant for \(b_1\). On the event \(D_s>0\), we have \(X_s>Y_s\), and the drift ordering gives \(b_1(s,X_s)-b_2(s,Y_s)\le b_1(s,X_s)-b_1(s,Y_s)\le L_1D_s\). The final inequality follows from the Lipschitz bound. The stochastic integrand in Tanaka’s formula is \(H_s=\mathbf{1}_{\{D_s>0\}}(\sigma(s,X_s)-\sigma(s,Y_s))\). It is progressively measurable, and its square-integrability follows from the linear-growth moment bounds for \(X,Y\): for a constant \(C'\), \(\mathbb{E}\int_0^T H_s^2\,ds\le C'\int_0^T(1+\mathbb{E}[X_s^2]+\mathbb{E}[Y_s^2])\,ds<\infty\). Thus \(\int_0^tH_s\,dB_s\) is a true martingale with expectation zero.
Proof. Since \(D_0^+=(\xi_1-\xi_2)^+=0\) almost surely and \(L_t^0(D)=0\), Tanaka’s formula yields \[ D_t^+=\int_0^t\mathbf{1}_{\{D_s>0\}}\bigl(b_1(s,X_s)-b_2(s,Y_s)\bigr)\,ds+\int_0^tH_s\,dB_s. \] Take expectations. The martingale has expectation zero, and the drift bound just established implies \[ 0\le\mathbb{E}[D_t^+]\le L_1\int_0^t\mathbb{E}[D_s^+]\,ds. \] The function \(t\mapsto\mathbb{E}[D_t^+]\) is continuous and nonnegative, and Gronwall’s inequality therefore gives \(\mathbb{E}[D_t^+]=0\) for every \(t\in[0,T]\). A nonnegative random variable with expectation zero is zero almost surely, so \(X_t\le Y_t\) almost surely at each fixed \(t\). Apply this to the countable set of rational times in \([0,T]\). On the resulting probability-one event, the inequality holds at all rational times; continuity of \(X\) and \(Y\) extends it to every \(t\in[0,T]\). This proves the pathwise assertion. \(\square\)
Worked Examples
Worked Example: Ordered Constant Drifts
On the given filtered space, let \(B_t\) be Brownian motion and consider \(dX_t=1\,dt+dB_t,\ X_0=0\), and \(dY_t=2\,dt+dB_t,\ Y_0=1\). Determine the pathwise gap and verify the ordering.
The drift functions are \(b_1(t,x)=1\), \(b_2(t,x)=2\), and the common diffusion coefficient is \(\sigma(t,x)=1\). They are globally Lipschitz, have linear growth, satisfy \(b_1\le b_2\), and the initial values satisfy \(0\le1\). Integrating both equations gives \(X_t=t+B_t\) and \(Y_t=1+2t+B_t\). Subtracting, \[ Y_t-X_t=(1+2t+B_t)-(t+B_t)=1+t. \] For every \(t\ge0\), \(1+t\ge1>0\), so the paths are strictly ordered.
Check: Substituting the candidates into the integral equations gives \(X_t=0+\int_0^t1\,ds+\int_0^t1\,dB_s=t+B_t\) and \(Y_t=1+\int_0^t2\,ds+\int_0^t1\,dB_s=1+2t+B_t\). Their difference is indeed \(1+t\), independent of the Brownian path.
Worked Example: Ordered Initial Values with Multiplicative Noise
Consider \(dX_t=X_t\,dt+X_t\,dB_t,\ X_0=1\), and \(dY_t=Y_t\,dt+Y_t\,dB_t,\ Y_0=2\). Find the solutions and verify their order.
Both equations have drift \(b(t,x)=x\) and common diffusion coefficient \(\sigma(t,x)=x\). For initial value \(x>0\), propose \(Z_t=x\exp(\tfrac12t+B_t)\). To check this candidate, take \(f(t,z)=x\exp(\tfrac12t+z)\). Its derivatives are \(f_t=\tfrac12 f\), \(f_z=f\), and \(f_{zz}=f\). Itô’s formula, including its second-derivative correction, gives \[ df(t,B_t)=\bigl(f_t+\tfrac12 f_{zz}\bigr)\,dt+f_z\,dB_t =\bigl(\tfrac12 f+\tfrac12 f\bigr)\,dt+f\,dB_t =f\,dt+f\,dB_t. \] Also \(f(0,0)=x\), so this is the required solution. Hence \(X_t=\exp(\tfrac12t+B_t)\) and \(Y_t=2\exp(\tfrac12t+B_t)\), and \(Y_t-X_t=\exp(\tfrac12t+B_t)>0\) for every finite \(t\).
Check: For each solution, the drift coefficient obtained by Itô’s formula is exactly the state itself, and the diffusion coefficient is also exactly the state itself. Thus both SDEs are satisfied, and the ratio \(Y_t/X_t=2\) confirms the ordering directly.
Worked Example: A Strictly Larger Drift with State-Dependent Diffusion
Let \(X_0=Y_0=1\), and compare \(dX_t=X_t\,dB_t\) with \(dY_t=dt+Y_t\,dB_t\). Both equations use the diffusion coefficient \(\sigma(t,x)=x\); their drifts are \(b_1(t,x)=0\) and \(b_2(t,x)=1\). Find explicit solutions and check the comparison.
Set \(E_t=\exp(B_t-\tfrac12t)\). For \(f(t,z)=\exp(z-\tfrac12t)\), the derivatives are \(f_t=-\tfrac12f\), \(f_z=f\), and \(f_{zz}=f\). Itô’s formula gives \(dE_t=(-\tfrac12f+\tfrac12f)\,dt+f\,dB_t=E_t\,dB_t\), with \(E_0=1\). Therefore \(X_t=E_t\).
Define \(I_t=\int_0^tE_s^{-1}\,ds\) and \(Y_t=E_t(1+I_t)\). The process \(I\) has finite variation and \(dI_t=E_t^{-1}dt\). The product rule gives \(d(E_tI_t)=I_t\,dE_t+E_t\,dI_t+d[E,I]_t\). Since a continuous finite-variation process has zero covariation with a continuous semimartingale, \([E,I]_t=0\). Therefore \(d(E_tI_t)=I_tE_t\,dB_t+dt\). Adding \(dE_t=E_t\,dB_t\) yields \(dY_t=dt+Y_t\,dB_t\), and \(Y_0=1\). Hence \[ Y_t-X_t=E_tI_t=E_t\int_0^tE_s^{-1}\,ds. \] The exponential \(E_t\) is positive. For every \(t>0\), the continuous function \(s\mapsto E_s^{-1}\) is positive on \([0,t]\), so its integral is strictly positive. Thus \(Y_t>X_t\) for every \(t>0\), while \(Y_0=X_0\).
Check: The candidate solutions satisfy their equations by the Itô and product-rule calculations above. Their difference is zero at \(t=0\), as required by the equal initial values, and strictly positive for \(t>0\). The stochastic integrals in these equations are square-integrable on each finite time interval: \(E_t\) has finite second moment, and the linear SDE for \(Y_t\) has finite second moments under the global Lipschitz and linear-growth bounds.
Scope, Significance, and a Common Mistake
The theorem is useful because it converts coefficient inequalities into a comparison of whole sample paths. Its conclusion is stronger than an inequality between means: it says that the two solutions, constructed with the same Brownian motion, cannot cross. The requirement of a shared diffusion function is important in the proof. It makes the difference of diffusion coefficients vanish at \(X=Y\), and the Lipschitz bound then forces the local time at zero of the difference to vanish.
Common mistake: comparing solutions driven by different Brownian motions. If the noises are not coupled through the same Brownian path, subtracting the equations does not produce the diffusion difference \(\sigma(t,X_t)-\sigma(t,Y_t)\) used in the proof. For instance, take \(dX_t=dB_t\), \(dY_t=d\widetilde B_t\), and \(X_0=Y_0=0\), where \(B\) and \(\widetilde B\) are independent Brownian motions. The drifts are equal and the diffusion functions are both constant \(1\), but the proposed shared-noise comparison setup is absent. At any fixed \(t>0\), \(X_t-Y_t\) has distribution \(N(0,2t)\), so \(\mathbb{P}(X_t\le Y_t)=\tfrac12\), not \(1\). The correction is to couple the equations using the same Brownian motion; then with equal coefficients and initial data, pathwise uniqueness gives identical solutions.
A second error is to infer pathwise order merely from ordered expected values. An expectation comparison does not rule out paths that cross. The proof instead controls the positive part \(D_t^+\) and shows its expectation is zero at each time, then uses path continuity to obtain the simultaneous pathwise statement. The stochastic integral in that argument is a true martingale because its integrand is square-integrable; no inference from a local martingale to a true martingale is being made.
Check Your Understanding
Use the hypotheses and proof above to decide what can be concluded in each case.
- Which coefficient must be identical in the two equations for the stated comparison theorem, and what other ordering assumptions are required?
- Suppose \(b_1(t,x)\le b_2(t,x)\) and \(\xi_1\le\xi_2\) almost surely, but the equations are driven by independent Brownian motions. Which step of the proof no longer applies?
- Why does the Lipschitz condition on the common diffusion coefficient imply that the difference process has zero local time at zero in the proof?
- For the first worked example, recompute \(Y_t-X_t\) directly from the two integral equations.
- For the third worked example, explain why \(\int_0^tE_s^{-1}\,ds>0\) for every \(t>0\), almost surely.