Stability with Respect to Initial Conditions

Why Compare Solutions on the Same Noise?

An SDE solution depends both on its initial value and on the Brownian path that drives it. To isolate the effect of changing the initial value, compare two solutions using the same Brownian motion and the same coefficients. The resulting estimate measures how far apart the solutions can be in mean square. It is a quantitative form of continuous dependence, stronger than merely knowing that solutions with identical initial values are pathwise unique.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$ relative to the filtration. Consider scalar SDEs on a fixed finite horizon $[0,T]$. The coefficients $\mu,\sigma:\mathbb{R}\to\mathbb{R}$ are measurable and satisfy global Lipschitz bounds: for constants $L_\mu,L_\sigma\ge0$ and all $z,w\in\mathbb{R}$,

|\mu(z)-\mu(w)|\le L_\mu|z-w|, \qquad |\sigma(z)-\sigma(w)|\le L_\sigma|z-w|.

For each deterministic initial value $x$, let $X^x$ be the strong solution, adapted to this filtration and driven by this $B_t$, of

X^x_t=x+\int_0^t\mu(X^x_s)\,ds+\int_0^t\sigma(X^x_s)\,dB_s.

Here and below, stochastic integrals are Ito integrals. Each integrand in an Ito integral is progressively measurable and belongs to $\mathcal{L}^2$ on $[0,T]$, meaning that $\mathbb{E}\int_0^T H_s^2\,ds<\infty$. Under the stated global Lipschitz assumptions, the existence and uniqueness theorem gives these strong solutions, and the usual finite-horizon moment bounds ensure the coefficient integrands are in $\mathcal{L}^2$. The drift integral is an ordinary time integral.

Definition: A strong solution is constructed on the given filtered probability space and is adapted to its filtration, with the specified Brownian motion driving its SDE. Pathwise uniqueness means that two solutions with the same initial value, coefficients, and Brownian motion agree almost surely at all times. Uniqueness in law instead concerns equality of distributions; it does not by itself give a distance estimate for solutions on the same probability space.

The Mean-Square Stability Estimate

Theorem: Suppose $\mu$ and $\sigma$ satisfy the global Lipschitz bounds above, and $X^x,X^y$ are strong solutions driven by the same Brownian motion with deterministic initial values $x,y$. Then, for every $t\in[0,T]$, $$ \mathbb{E}\big[|X^x_t-X^y_t|^2\big] \le |x-y|^2\exp\big((2L_\mu+L_\sigma^2)t\big). $$

Proof. Set $D_t=X^x_t-X^y_t$, $\Delta\mu_t=\mu(X^x_t)-\mu(X^y_t)$, and $\Delta\sigma_t=\sigma(X^x_t)-\sigma(X^y_t)$. Subtracting the two SDEs gives $$ D_t=x-y+\int_0^t\Delta\mu_s\,ds+\int_0^t\Delta\sigma_s\,dB_s. $$ The difference integrand $\Delta\sigma$ is progressively measurable and in $\mathcal{L}^2$: the Lipschitz bound gives $|\Delta\sigma_s|^2\le L_\sigma^2|D_s|^2$, and the finite-horizon second-moment bounds for the solutions imply $\mathbb{E}\int_0^T|D_s|^2\,ds<\infty$.

Apply Ito's formula to $f(z)=z^2$, for which $f'(z)=2z$ and $f''(z)=2$. The drift correction is explicitly $\tfrac12 f''(D_s)(\Delta\sigma_s)^2\,ds=(\Delta\sigma_s)^2\,ds$. Thus $$ d(D_t^2)=\big(2D_t\Delta\mu_t+(\Delta\sigma_t)^2\big)\,dt +2D_t\Delta\sigma_t\,dB_t. $$ The multiplication rules here are $dB_t\,dB_t=dt$, $dB_t\,dt=0$, and $dt\,dt=0$. The stochastic integral in this display need not be a true martingale on the basis of second moments alone: its square-integrability would require control of $\mathbb{E}\int_0^T D_s^2(\Delta\sigma_s)^2\,ds$, which involves fourth powers. We therefore stop before taking expectations.

For $n>|x-y|$, let $\tau_n=\inf\{s\ge0:|D_s|\ge n\}$, with the infimum of the empty set equal to infinity. The stopped Ito integrand $2\mathbf{1}_{\{s\le\tau_n\}}D_s\Delta\sigma_s$ is bounded in absolute value by $2L_\sigma n^2$. It is progressively measurable and square-integrable on $[0,T]$, so its Ito integral is a true martingale: its conditional expectation over any future interval given the current filtration is zero. Taking expectations in the stopped Ito formula is therefore valid. The Lipschitz assumptions give $$ 2D_s\Delta\mu_s+(\Delta\sigma_s)^2 \le 2L_\mu D_s^2+L_\sigma^2D_s^2 =cD_s^2,\qquad c=2L_\mu+L_\sigma^2. $$ Consequently, for $g_n(t)=\mathbb{E}[D_{t\wedge\tau_n}^2]$, $$ g_n(t)\le |x-y|^2+c\int_0^t \mathbb{E}\big[\mathbf{1}_{\{s\le\tau_n\}}D_s^2\big]\,ds \le |x-y|^2+c\int_0^t g_n(s)\,ds. $$ The last inequality holds because $D_s=D_{s\wedge\tau_n}$ when $s\le\tau_n$, and the omitted contribution is nonnegative. Gronwall's inequality yields $g_n(t)\le |x-y|^2e^{ct}$. The paths of $D$ are continuous, so on every finite interval they are bounded; hence $D_{t\wedge\tau_n}\to D_t$ almost surely as $n\to\infty$. Fatou's lemma now gives $$ \mathbb{E}[D_t^2]\le\liminf_{n\to\infty}g_n(t) \le |x-y|^2e^{ct}. $$ This proves the estimate, including the case $x=y$, for which the right-hand side is zero. $\square$

Corollary: For each fixed $t\in[0,T]$, the map sending a deterministic initial value $x$ to the square-integrable random variable $X^x_t$ is Lipschitz in mean-square norm: $$ \big(\mathbb{E}|X^x_t-X^y_t|^2\big)^{1/2} \le e^{(2L_\mu+L_\sigma^2)t/2}|x-y|. $$

Proof. Both sides of the theorem's inequality are nonnegative. Taking square roots preserves the inequality and gives the displayed bound. It holds for every pair $x,y$, so it is precisely the Lipschitz estimate for the initial-value map under the norm $\|U\|_2=(\mathbb{E}|U|^2)^{1/2}$. $\square$

Worked Consequences

Worked Example: Constant Drift and Diffusion

Consider $dX_t=-2\,dt+3\,dB_t$, with initial values $x=4$ and $y=1$, using the same $B_t$. Find and check the mean-square distance at time $T=2$.

Both coefficients are constant, so $L_\mu=L_\sigma=0$ are valid Lipschitz constants. Direct integration gives $$ X^4_t=4-2t+3B_t,\qquad X^1_t=1-2t+3B_t. $$ Subtracting cancels both the common drift and the common Brownian term: $X^4_t-X^1_t=3$. Therefore $$ \mathbb{E}\big[|X^4_2-X^1_2|^2\big]=\mathbb{E}[3^2]=9. $$ The theorem gives $|4-1|^2e^0=9$, exactly the same value.

Check: Substituting $t=0$ into the direct difference also gives $4-1=3$, its prescribed initial difference. The equality at $T=2$ is thus consistent both with the initial condition and with the theorem's bound.

Worked Example: A Contracting Drift

Consider the deterministic equation $dX_t=-3X_t\,dt$, with initial values $1$ and $-1$. Compare the exact distance at $t=\tfrac12$ with the general stability estimate.

Solving the ordinary differential equation gives $X^1_t=e^{-3t}$ and $X^{-1}_t=-e^{-3t}$. Their difference is $2e^{-3t}$, so at $t=\tfrac12$, $$ \mathbb{E}\big[|X^1_{1/2}-X^{-1}_{1/2}|^2\big] =(2e^{-3/2})^2=4e^{-3}. $$ Here $\mu(z)=-3z$ has Lipschitz constant $L_\mu=3$, and $\sigma(z)=0$ has $L_\sigma=0$. The theorem's bound is $$ |1-(-1)|^2e^{(2\cdot3+0)(1/2)}=4e^3. $$ Since $e^{-3}\le e^3$, the exact value $4e^{-3}$ satisfies the bound.

Check: At $t=0$, the exact squared distance is $4$, and the theorem's right-hand side is also $4$. For positive time the actual distance contracts, whereas this general estimate grows; the estimate guarantees stability but need not capture the sign or strength of a particular drift.

Worked Example: Nonlinear Globally Lipschitz Coefficients

Let $\mu(z)=\sin z$ and $\sigma(z)=\cos z$. Bound the mean-square distance between solutions started at $x=0$ and $y=\pi$ at time $t=\tfrac13$.

The mean value theorem and the bounds $|\cos z|\le1$, $|\sin z|\le1$ show that both coefficients are globally Lipschitz with constants $L_\mu=L_\sigma=1$. Thus $2L_\mu+L_\sigma^2=3$, and the theorem yields $$ \mathbb{E}\big[|X^0_{1/3}-X^\pi_{1/3}|^2\big] \le |0-\pi|^2e^{3(1/3)} =\pi^2e. $$ In mean-square norm, the corresponding bound is $\big(\mathbb{E}|X^0_{1/3}-X^\pi_{1/3}|^2\big)^{1/2}\le\pi\sqrt e$.

Check: At time zero the squared distance is $\pi^2$, while the bound at time $1/3$ is $\pi^2e$; it is larger because $e>1$. The estimate is therefore consistent with its initial value and the computed exponent.

What the Estimate Does—and Does Not—Say

The exponent $2L_\mu+L_\sigma^2$ combines the effect of drift differences and diffusion differences. In the squared-distance calculation, the drift contributes $2D_t\Delta\mu_t$, while the quadratic-variation correction contributes $(\Delta\sigma_t)^2$. The drift coefficient $\mu$ and diffusion coefficient $\sigma$ therefore enter the bound in distinct ways. A large Lipschitz constant may produce a loose bound, as the contracting-drift example shows.

The shared Brownian motion is essential to this comparison. It lets us subtract the two SDEs and obtain the noise difference $\sigma(X^x_t)-\sigma(X^y_t)$. If the solutions were driven by unrelated Brownian motions, the difference would contain two separate noise terms, and this theorem would not directly give the displayed estimate. The conclusion concerns a coupled pair of strong solutions, not merely the equality or closeness of their individual distributions.

Finally, the proof does not assert that every local martingale is a true martingale. Stopping makes the stochastic integral in the squared-difference identity square-integrable, so its expectation is zero; Fatou's lemma then passes the bound to the unstopped solutions. This distinction is useful whenever applying Ito's formula to a quantity whose integrability has not yet been established.

Check Your Understanding

Use the common-noise setup and the hypotheses stated above.

If $L_\mu=2$, $L_\sigma=1$, and $|x-y|=3$, what bound does the theorem give for $\mathbb{E}|X^x_t-X^y_t|^2$?
Why does using the same Brownian motion matter when subtracting the two SDEs?
For $\mu(z)=az$ and $\sigma(z)=bz$, give valid global Lipschitz constants and write the resulting stability estimate.
Why does the proof stop the squared difference before taking expectations of its Ito formula?
What does the corollary say about the map from an initial value to the solution at a fixed time?

Stability with Respect to Initial Conditions

What You'll Learn