Gronwall Bounds for Mean-Square SDE Estimates

From Local Error Estimates to Finite-Time Bounds

For solutions of a stochastic differential equation, coefficient estimates usually control differences at a single time in terms of the current difference between the solutions. Gronwall's inequality converts that local control into a bound over a whole time interval. In mean-square estimates, the drift contributes through an ordinary time integral, while the diffusion contributes through the Ito isometry. Their combined effect is an integral inequality for the expected squared difference.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$. For a finite horizon $T>0$, an Ito integrand $H$ below is progressively measurable and belongs to $\mathcal{L}^2$, meaning $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. The Ito isometry then gives $\mathbb{E}[(\int_0^t H_s\,dB_s)^2]=\mathbb{E}\int_0^t H_s^2\,ds$ for $0\le t\le T$.

Definition: A strong solution of $dX_t=b(t,X_t)\,dt+\sigma(t,X_t)\,dB_t$, with initial value $\xi$, is an adapted continuous process on the given filtered probability space that satisfies $$ X_t=\xi+\int_0^t b(s,X_s)\,ds+\int_0^t \sigma(s,X_s)\,dB_s $$ almost surely for each $t$, with the stochastic integrand progressively measurable and square-integrable on each finite horizon. The function $b$ is the drift coefficient; $\sigma$ is the diffusion coefficient. A strong solution is driven by the specified Brownian motion. Pathwise uniqueness means that any two such solutions on the same space, with the same Brownian motion and equal initial values almost surely, are indistinguishable. Uniqueness in law instead means equality of the probability distributions of solutions; it is a different assertion.

The estimate below is conditional on solutions existing with the stated integrability. It does not prove existence. We first establish the deterministic inequality that will be applied to their mean-square difference.

The Integral Gronwall Inequality

Theorem: Let $u:[0,T]\to[0,\infty)$ be continuous, and let $a,c\ge0$. If $$ u(t)\le a+c\int_0^t u(s)\,ds\qquad(0\le t\le T), $$ then $u(t)\le ae^{ct}$ for every $t\in[0,T]$. More generally, if $\beta:[0,T]\to[0,\infty)$ is integrable and $u(t)\le a+\int_0^t\beta(s)u(s)\,ds$, then $u(t)\le a\exp(\int_0^t\beta(s)\,ds)$.

Proof. First suppose $c\ge0$ is constant. Set $v(t)=a+c\int_0^t u(s)\,ds$. Then $u(t)\le v(t)$, $v(0)=a$, and $v'(t)=cu(t)\le cv(t)$. The absolutely continuous function $e^{-ct}v(t)$ therefore has derivative $e^{-ct}(v'(t)-cv(t))\le0$ almost everywhere. Hence $v(t)\le ae^{ct}$, and $u(t)\le ae^{ct}$. This reasoning includes $a=0$: $v$ is nonnegative, nonincreasing after multiplication by $e^{-ct}$, and starts at zero, so $v=u=0$. For the variable coefficient statement, set $v(t)=a+\int_0^t\beta(s)u(s)\,ds$. Then $v\ge u$, $v(0)=a$, and $v'\le\beta v$ almost everywhere. Multiplying by $\exp(-\int_0^t\beta(s)\,ds)$ gives an absolutely continuous function with nonpositive derivative almost everywhere. Thus $v(t)\le a\exp(\int_0^t\beta(s)\,ds)$, which proves the assertion, including $a=0$. $\square$

Worked Example: Solving a Constant-Coefficient Inequality

Suppose a continuous nonnegative function satisfies $u(t)\le2+3\int_0^t u(s)\,ds$ on $[0,1]$. The theorem gives $u(t)\le2e^{3t}$, and in particular $u(1)\le2e^3$.

Check: The proposed bound $U(t)=2e^{3t}$ satisfies the associated equality, since $2+3\int_0^t2e^{3s}\,ds=2+2(e^{3t}-1)=2e^{3t}=U(t)$. This verifies the constants and exponent in the bound; the theorem applies to $u$ because it satisfies the inequality, not because $u$ is known to equal $U$.

Mean-Square Difference Estimate for SDEs

Assume $b(t,x)$ and $\sigma(t,x)$ are measurable in $t$, and that for all $t\in[0,T]$, $x,y\in\mathbb{R}$, $$ |b(t,x)-b(t,y)|\le L_b|x-y|, \qquad |\sigma(t,x)-\sigma(t,y)|\le L_\sigma|x-y|, $$ where $L_b,L_\sigma\ge0$ are constants. These are global Lipschitz conditions in the state variable. Also assume linear growth: for some $K\ge0$, $|b(t,x)|+|\sigma(t,x)|\le K(1+|x|)$. Consider two strong solutions $X,Y$ driven by the same $B_t$, with square-integrable initial values and $\mathbb{E}\int_0^T (|b(s,X_s)|^2+|\sigma(s,X_s)|^2+|b(s,Y_s)|^2+|\sigma(s,Y_s)|^2)\,ds<\infty$. These assumptions ensure the time integrals are well-defined and the stochastic integrands used below belong to $\mathcal{L}^2$.

Theorem: Under these assumptions, put $u(t)=\mathbb{E}|X_t-Y_t|^2$ and $\delta^2=\mathbb{E}|\xi_X-\xi_Y|^2$. Then $$ u(t)\le 3\delta^2 e^{C_Tt},\qquad C_T=3T L_b^2+3L_\sigma^2,\qquad 0\le t\le T. $$

Proof. Subtract the integral equations for $X$ and $Y$. Write $\Delta b_s=b(s,X_s)-b(s,Y_s)$ and $\Delta\sigma_s=\sigma(s,X_s)-\sigma(s,Y_s)$. The difference is $\xi_X-\xi_Y+\int_0^t\Delta b_s\,ds+\int_0^t\Delta\sigma_s\,dB_s$. For real numbers $p,q,r$, $|p+q+r|^2\le3(|p|^2+|q|^2+|r|^2)$, by Cauchy–Schwarz. Also, $|\int_0^t\Delta b_s\,ds|^2\le t\int_0^t|\Delta b_s|^2\,ds$. Taking expectations and applying the Ito isometry to the stochastic integral gives $$ u(t)\le3\delta^2+3t\int_0^t\mathbb{E}|\Delta b_s|^2\,ds+3\int_0^t\mathbb{E}|\Delta\sigma_s|^2\,ds. $$ The Lipschitz conditions imply $\mathbb{E}|\Delta b_s|^2\le L_b^2u(s)$ and $\mathbb{E}|\Delta\sigma_s|^2\le L_\sigma^2u(s)$. Since $t\le T$, it follows that $u(t)\le3\delta^2+C_T\int_0^t u(s)\,ds$. The function $u$ is continuous under the stated moment and solution assumptions; applying the integral Gronwall inequality with $a=3\delta^2$ proves the bound. $\square$

The estimate separates the two mechanisms. The drift term is controlled by the time-integral bound and carries the factor $T L_b^2$; the diffusion term is controlled by the Ito isometry and carries $L_\sigma^2$. The numerical constants here come from the convenient three-term square inequality, so the resulting bound need not be sharp.

Worked Example: A Finite-Horizon Stability Bound

Suppose two solutions on $[0,1]$ have $\mathbb{E}|\xi_X-\xi_Y|^2=\tfrac14$, and their drift and diffusion coefficients have Lipschitz constants $L_b=\tfrac12$ and $L_\sigma=1$, respectively. The theorem gives $$ C_T=3\cdot1\cdot\left(\tfrac12\right)^2+3\cdot1^2=\tfrac34+3=\tfrac{15}{4}, \qquad u(t)\le\tfrac34 e^{15t/4}. $$ In particular, $u(1)\le\tfrac34 e^{15/4}$.

Check: The initial term in the integral inequality is $3\delta^2=3\cdot\tfrac14=\tfrac34$. Substituting $U(t)=\tfrac34 e^{15t/4}$ into its right-hand side gives $\tfrac34+\tfrac{15}{4}\int_0^t\tfrac34 e^{15s/4}\,ds=\tfrac34+\tfrac34(e^{15t/4}-1)=U(t)$. Thus the stated exponential bound matches the integral inequality with the computed constants.

Uniqueness from the Mean-Square Bound

When the two initial values agree almost surely, $\delta^2=0$, and the estimate gives $u(t)=0$ for every fixed $t\in[0,T]$. Since $|X_t-Y_t|^2\ge0$, expectation zero implies $X_t=Y_t$ almost surely at that time. To obtain the pathwise statement, take the intersection of the probability-one equality events over rational $t\in[0,T]$, together with the probability-one event on which both paths are continuous. Equality at the rational times then implies equality at every time by continuity. This proves pathwise uniqueness on $[0,T]$, and applying the argument on finite horizons gives it on all finite times.

Worked Example: The Uniqueness Calculation

Let $b(t,x)=-2x$ and $\sigma(t,x)=x$. These coefficients satisfy global Lipschitz bounds $L_b=2$ and $L_\sigma=1$, since $|-2x+2y|=2|x-y|$ and $|x-y|=|x-y|$. Suppose $X,Y$ are square-integrable strong solutions driven by the same Brownian motion and have the same initial value almost surely. On $[0,1]$, $\delta^2=0$ and $C_T=3\cdot1\cdot2^2+3\cdot1^2=12+3=15$. Therefore $u(t)\le15\int_0^t u(s)\,ds$. Gronwall with $a=0$ gives $u(t)\le0$; because $u(t)\ge0$, $u(t)=0$ throughout the interval. The continuity argument above yields indistinguishability.

Check: The zero function satisfies the resulting bound exactly: $0=15\int_0^t0\,ds$. The conclusion uses the zero initial difference as well as the shared Brownian motion; the calculation does not compare solutions driven by different Brownian motions.

What the Estimate Does—and Does Not—Show

The estimate also gives stability with respect to initial data: if the initial mean-square difference is small, then the difference remains bounded by $3\delta^2e^{C_Tt}$ on the finite horizon. It does not say that the bound is attained, and it does not by itself establish existence of either solution. Nor does pathwise uniqueness alone assert uniqueness in law without a further result connecting those notions under appropriate hypotheses.

A common mistake is to omit the diffusion contribution or to treat its integral as an ordinary time integral. For example, replacing $\mathbb{E}|\int_0^t\Delta\sigma_s\,dB_s|^2$ by $(\int_0^t\mathbb{E}|\Delta\sigma_s|\,ds)^2$ is not the Ito isometry and generally gives no valid estimate. The correct calculation is $$ \mathbb{E}\left|\int_0^t\Delta\sigma_s\,dB_s\right|^2 =\int_0^t\mathbb{E}|\Delta\sigma_s|^2\,ds \le L_\sigma^2\int_0^t u(s)\,ds. $$ The equality requires that $\Delta\sigma$ be progressively measurable and square-integrable. Here these conditions follow from the solution integrability assumptions; the Lipschitz estimate then supplies the final inequality.

Likewise, the Gronwall step requires a nonnegative continuous function satisfying the integral inequality on the entire interval. A bound only at one time cannot be fed into the theorem to control all later times. In applications, first derive the inequality for every $t\in[0,T]$, then apply Gronwall with constants valid uniformly on that horizon.

Check Your Understanding

For each question, identify the assumptions needed before applying the estimate.

For $u(t)\le 1+4\int_0^t u(s)\,ds$, what bound does the integral Gronwall inequality give?
Which term in the SDE difference estimate is evaluated using the Ito isometry, and what integrability condition is required?
If $L_b=0$, which part of the mean-square estimate disappears, and what coefficient remains in $C_T$?
Why does $u(t)=0$ at each fixed time imply indistinguishability only after using continuity and a countable dense set of times?
Does the mean-square estimate prove existence or uniqueness in law? Explain what it does prove when the initial values agree.