Tutorials › Stochastic Calculus › A Priori Second-Moment Bounds for SDE Solutions

Existence and Uniqueness Theorems · Tutorial 227 of 600

A Priori Second-Moment Bounds for SDE Solutions

Learn how to localize a solution and turn linear growth of its drift and diffusion into a finite bound on its expected squared maximum.

Advanced 10 min read

What You'll Learn

  • State separate linear-growth bounds for the drift and diffusion coefficients.
  • Localize a continuous SDE solution so its stochastic integral is square-integrable.
  • Use the drift estimate, Ito isometry and Doob’s inequality to control a running supremum.
  • Apply Gronwall’s inequality and remove the stopping by Fatou’s lemma.
  • Interpret the bound in constant-coefficient, deterministic and time-dependent examples.

Why Bound the Running Maximum?

For an SDE, controlling the second moment at a single time does not by itself control the largest value reached over a whole time interval. A useful a priori estimate instead bounds the expected squared supremum of the solution. Such a bound quantifies how large the solution can become on a finite horizon and supplies an integrability guarantee for later arguments. “A priori” means that the estimate uses the equation’s coefficients and initial value, rather than an explicit formula for the solution.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). Fix a finite horizon \(T>0\) and a deterministic initial value \(x_0\). Consider a continuous adapted solution of \(dX_t=\mu(t,X_t)\,dt+\sigma(t,X_t)\,dB_t\), with \(X_0=x_0\). The drift integral is a Lebesgue integral. The stochastic integral is the Ito integral; its integrand is progressively measurable. For a process \(H\), write \(H\in\mathcal{L}^2([0,T])\) when it is progressively measurable and \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). Before localization, the solution’s stochastic integral may only be defined locally; the proof below constructs square-integrable stopped integrands.

Definition: The coefficients have separate linear-growth bounds on \([0,T]\) if there are constants \(K_\mu,K_\sigma\ge0\) such that, for all \(t\in[0,T]\) and \(x\in\mathbb{R}\), \(|\mu(t,x)|^2\le K_\mu(1+|x|^2)\) and \(|\sigma(t,x)|^2\le K_\sigma(1+|x|^2)\).

The constants may be zero: for example, \(K_\sigma=0\) means the diffusion coefficient vanishes. These are squared-coefficient versions of linear growth. The estimate does not require the solution itself to be a martingale. Its stochastic-integral term will be a square-integrable martingale after stopping; the drift term generally prevents \(X\) from being a martingale.

The Finite-Horizon Bound

Theorem: Suppose \(X\) is a continuous adapted solution on \([0,T]\) with deterministic \(X_0=x_0\), and its coefficients satisfy the separate linear-growth bounds above. Set \(b=3TK_\mu+12K_\sigma\). Then \(\mathbb{E}\big[\sup_{0\le t\le T}|X_t|^2\big]\le(3|x_0|^2+bT)e^{bT}<\infty\).

Proof. We first stop the solution before it can become large. For each integer \(n>|x_0|\), let \(\tau_n=\inf\{t\ge0:|X_t|\ge n\}\), with the infimum of the empty set defined as \(+\infty\), and let \(X^n_t=X_{t\wedge\tau_n}\). For \(s\le\tau_n\), continuity gives \(|X_s|\le n\). The process \(H^n_s=\mathbf{1}_{\{s\le\tau_n\}}\sigma(s,X_s)\) is progressively measurable. Its squared integral has finite expectation, since the growth bound gives \(|H^n_s|^2\le K_\sigma(1+n^2)\) on \([0,T]\). Thus \(H^n\in\mathcal{L}^2([0,T])\). The Ito integral \(M^n_t=\int_0^t H^n_s\,dB_s\) is a square-integrable martingale: in particular it is integrable and \(\mathbb{E}[M^n_t\mid\mathcal{F}_u]=M^n_u\) for \(u\le t\). By the Ito isometry, \(\mathbb{E}|M^n_t|^2=\mathbb{E}\int_0^t|H^n_s|^2\,ds\). These facts concern the stopped integral and do not assert that an unstopped local martingale is a true martingale.

Define \(A_n(t)=\mathbb{E}[\sup_{0\le u\le t}|X^n_u|^2]\). The stopped equation, the inequality \(|a+c+d|^2\le3(|a|^2+|c|^2+|d|^2)\), and Cauchy–Schwarz for the drift integral give \(\displaystyle A_n(t)\le3|x_0|^2+3t\int_0^t\mathbb{E}\big[\mathbf{1}_{\{s\le\tau_n\}}|\mu(s,X_s)|^2\big]\,ds+3\mathbb{E}\sup_{0\le u\le t}|M^n_u|^2.\) The standard Doob \(L^2\) maximal inequality for a square-integrable martingale states that \(\mathbb{E}\sup_{u\le t}|M^n_u|^2\le4\mathbb{E}|M^n_t|^2\); this is the standard maximal-inequality result used here. Substituting this inequality and the Ito isometry yields \(\displaystyle A_n(t)\le3|x_0|^2+3t\int_0^t\mathbb{E}\big[\mathbf{1}_{\{s\le\tau_n\}}|\mu(s,X_s)|^2\big]\,ds+12\int_0^t\mathbb{E}\big[\mathbf{1}_{\{s\le\tau_n\}}|\sigma(s,X_s)|^2\big]\,ds.\)

On \(\{s\le\tau_n\}\), \(X_s=X^n_s\). Hence, for either coefficient, the growth bound implies \(\mathbb{E}\big[\mathbf{1}_{\{s\le\tau_n\}}|\mu(s,X_s)|^2\big]\le K_\mu(1+A_n(s))\) and \(\mathbb{E}\big[\mathbf{1}_{\{s\le\tau_n\}}|\sigma(s,X_s)|^2\big]\le K_\sigma(1+A_n(s))\). Using \(t\le T\), we obtain \(\displaystyle A_n(t)\le3|x_0|^2+b\int_0^t(1+A_n(s))\,ds=3|x_0|^2+bt+b\int_0^t A_n(s)\,ds.\) In particular \(A_n(t)\le3|x_0|^2+bT+b\int_0^t A_n(s)\,ds\). Integral Gronwall gives \(A_n(t)\le(3|x_0|^2+bT)e^{bt}\). For completeness, if \(Y(t)=3|x_0|^2+bT+b\int_0^t A_n(s)\,ds\), then \(A_n\le Y\) and \(Y'(t)=bA_n(t)\le bY(t)\) almost everywhere; integrating the derivative of \(e^{-bt}Y(t)\) gives the stated inequality. This also covers \(b=0\).

Each continuous solution path is bounded on \([0,T]\). Therefore, for almost every outcome, \(\tau_n>T\) for all sufficiently large \(n\), and \(\sup_{u\le T}|X^n_u|^2\to\sup_{u\le T}|X_u|^2\). Fatou’s lemma now gives \(\displaystyle \mathbb{E}\sup_{u\le T}|X_u|^2\le\liminf_{n\to\infty}A_n(T)\le(3|x_0|^2+bT)e^{bT}.\) The right-hand side is finite, proving the theorem. \(\square\)

Corollary: Under the theorem’s assumptions, \(\mathbb{E}|X_t|^2\le(3|x_0|^2+bT)e^{bT}\) for every \(t\in[0,T]\).

Proof. For each outcome and each fixed \(t\le T\), \(|X_t|^2\le\sup_{0\le u\le T}|X_u|^2\). Taking expectations and applying the theorem proves the claim. \(\square\)

Worked Applications

Worked Example: Brownian Motion on a Short Horizon

Take \(X_0=0\), \(\mu(t,x)=0\), \(\sigma(t,x)=1\), and \(T=\tfrac14\). The equation is \(dX_t=dB_t\), so \(X_t=B_t\). The growth bounds hold with \(K_\mu=0\) and \(K_\sigma=1\): the drift inequality is \(0\le0\), and \(1\le1+|x|^2\). Thus \(b=3(\tfrac14)(0)+12(1)=12\), and \(bT=12(\tfrac14)=3\). The theorem gives \(\displaystyle \mathbb{E}\sup_{0\le t\le1/4}|B_t|^2\le(3(0)^2+12(\tfrac14))e^3=3e^3.\)

Check: At the terminal time, \(\mathbb{E}|B_{1/4}|^2=1/4\), since \(B_{1/4}\) has variance \(1/4\). The supremum includes that terminal value, so its expected square is at least \(1/4\); the finite upper bound \(3e^3\) is consistent. This check uses a terminal-time moment, not an assertion that it equals the expected squared supremum.

Worked Example: A Deterministic Decaying Solution

Take \(X_0=2\), \(\mu(t,x)=-x\), \(\sigma(t,x)=0\), and \(T=1\). The solution is \(X_t=2e^{-t}\): differentiating gives \(dX_t/dt=-2e^{-t}=-X_t\), and \(X_0=2\). The growth constants \(K_\mu=1\) and \(K_\sigma=0\) work because \(|-x|^2=|x|^2\le1+|x|^2\), while \(|0|^2=0\). Hence \(b=3(1)(1)+12(0)=3\), and \(bT=3\). The theorem’s bound is \(\displaystyle \mathbb{E}\sup_{0\le t\le1}|X_t|^2\le(3(2)^2+3)e^3=15e^3.\)

Check: Since \(e^{-t}\) decreases on \([0,1]\), the actual supremum of \(|X_t|^2=4e^{-2t}\) is \(4\), attained at \(t=0\). Thus the exact expected supremum is \(4\), which is no larger than \(15e^3\). This example also shows that the bound is an estimate, not usually an exact value.

Worked Example: Time-Dependent Drift and Diffusion

On \(0\le t\le T=\tfrac12\), take \(X_0=0\), \(\mu(t,x)=t\), and \(\sigma(t,x)=t\). Since \(0\le t\le\tfrac12\), \(t^2\le\tfrac14\); therefore both growth inequalities hold with \(K_\mu=K_\sigma=\tfrac14\), because \(\tfrac14\le\tfrac14(1+|x|^2)\). Compute \(b=3(\tfrac12)(\tfrac14)+12(\tfrac14)=\tfrac38+3=\tfrac{27}{8}\), and \(bT=(\tfrac{27}{8})(\tfrac12)=\tfrac{27}{16}\). The resulting bound is \(\displaystyle \mathbb{E}\sup_{0\le t\le1/2}|X_t|^2\le\frac{27}{16}e^{27/16}.\)

For an independent terminal-time check, the solution at \(T=\tfrac12\) is \(X_T=\int_0^T s\,ds+\int_0^T s\,dB_s\). The integrand \(s\) is progressively measurable and belongs to \(\mathcal{L}^2([0,T])\), since \(\int_0^{1/2}s^2\,ds=1/24<\infty\). The stochastic integral has mean zero, and Ito isometry gives its second moment as \(1/24\). The drift integral is \(\int_0^{1/2}s\,ds=1/8\), so \(\displaystyle \mathbb{E}|X_T|^2=(\tfrac18)^2+\tfrac1{24}=\tfrac1{64}+\tfrac1{24}=\tfrac{3}{192}+\tfrac{8}{192}=\tfrac{11}{192}.\) The cross term vanishes because the stochastic integral has mean zero. The supremum includes \(T\), so its expected square is at least \(11/192\); the theorem’s upper bound is finite and exceeds this terminal-time moment.

Why the Stopping and Maximal Inequality Matter

The localization is not cosmetic. Before the estimate is proved, one cannot simply assume that \(\mathbb{E}\int_0^T|\sigma(s,X_s)|^2\,ds\) is finite: linear growth bounds the integrand in terms of \(X\), and the desired moment bound is what will control \(X\). Stopping at \(\tau_n\) makes the integrand bounded on the finite horizon, so the Ito isometry and the square-integrable martingale estimates apply without circular reasoning. Fatou’s lemma then transfers the uniform bound back to the unstopped solution.

Common pitfall: using Ito isometry for a supremum. A wrong calculation replaces \(\mathbb{E}\sup_{u\le t}|M_u|^2\) by \(\mathbb{E}|M_t|^2=\mathbb{E}\int_0^tH_s^2\,ds\). Ito isometry evaluates the terminal-time second moment; it does not evaluate the maximum over earlier times. For the concrete choice \(H_s=1\) on \([0,1]\), the integral is \(M_t=B_t\). Its supremum includes both \(B_{1/2}^2\) and \(B_1^2\). The event \(\{|B_{1/2}|>|B_1|\}\) has positive probability, since the joint Gaussian vector \((B_{1/2},B_1)\) has a nonsingular covariance matrix. Consequently \(\mathbb{E}\sup_{0\le t\le1}|B_t|^2\ge\mathbb{E}\max(B_{1/2}^2,B_1^2)>\mathbb{E}B_1^2=1.\) The proposed equality with the Ito-isometry value \(1\) is false. The valid estimate used in the proof is Doob’s inequality, \(\mathbb{E}\sup_{u\le t}|M_u|^2\le4\mathbb{E}|M_t|^2\), which explains the factor \(12=3\cdot4\) in the theorem.

The bound is a finite-horizon estimate for an already specified solution. It does not by itself prove existence, uniqueness, or stability with respect to changes in initial conditions. Under the global Lipschitz setting, the existence-and-uniqueness theorem supplies a solution; this estimate gives a quantitative moment guarantee for it. Its main inputs are distinct: linear growth controls the coefficients, the Ito isometry and maximal inequality control the stochastic term, and Gronwall controls the feedback of the solution’s size into those coefficients.

Check Your Understanding

Use the theorem and keep terminal-time moments distinct from moments of the running supremum.

  1. For \(K_\mu=2\), \(K_\sigma=1\), and \(T=\tfrac12\), compute \(b\) and \(bT\).
  2. Why is the stopped integrand \(\mathbf{1}_{\{s\le\tau_n\}}\sigma(s,X_s)\) square-integrable on a finite horizon?
  3. For \(X_0=0\), \(\mu=0\), \(\sigma=2\), and \(T=1\), identify valid growth constants and compute the theorem’s bound.
  4. Which property of the stopped stochastic integral permits the use of Doob’s \(L^2\) maximal inequality?
  5. Why does Fatou’s lemma yield an upper bound for the unstopped expected supremum?