Moment Bounds from Itô’s Formula and Lyapunov Functions

From a Generator Inequality to a Moment Bound

A Lyapunov function converts information about the local drift and noise of a diffusion into a bound on its moments. The essential calculation is Itô’s formula applied not immediately to the whole process, but first to a stopped process. Stopping keeps the coefficients and stochastic integral under control; localization then allows the stopping times to tend to infinity.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$. Consider a scalar diffusion $dX_t=\mu(X_t)\,dt+\sigma(X_t)\,dB_t$, with deterministic initial value $X_0=x_0$. Assume the coefficients are locally Lipschitz, so there is a unique maximal strong solution up to its explosion time. A solution is strong when it is constructed on the given filtered space using its Brownian motion; here the moment estimates concern this solution under $\mathbb{P}$, not under a risk-neutral measure $\mathbb{Q}$.

For $f\in C^2(\mathbb{R})$, its generator is $\mathcal{L}f(x)=\mu(x)f'(x)+\tfrac12\sigma(x)^2f''(x)$. The second term is the quadratic-variation correction: $[B]_t=t$, so $dB_t\,dB_t=dt$, while $dB_t\,dt=0$ and $dt\,dt=0$. In every Itô integral below, the integrand is progressively measurable; stopped integrands are in $\mathcal{L}^2$, the class satisfying $\mathbb{E}\int_0^T H_s^2\,ds<\infty$ for each finite $T$.

Definition: A Lyapunov function for a diffusion is a nonnegative $C^2$ function $V$ used to control the state or its moments through a bound on $\mathcal{L}V$. It is coercive if $V(x)\to\infty$ as $|x|\to\infty$.

The Localization Argument

For $n>|x_0|$, let $\tau_n=\inf\{t\ge0:|X_t|\ge n\}$, with the infimum of the empty set equal to infinity. Up to $\tau_n$, the state stays in a compact interval. On each fixed time interval, locally Lipschitz coefficients and $C^2$ functions have bounded values and derivatives on that interval. Consequently, the stopped Itô integrals used below have square-integrable integrands.

Theorem: Suppose the solution is non-explosive, $V\in C^2(\mathbb{R})$ is nonnegative, and constants $a,b\ge0$ satisfy $\mathcal{L}V(x)\le a+bV(x)$ for every $x$. Then, for every $t\ge0$, $$ \mathbb{E}[V(X_t)]\le \begin{cases} e^{bt}V(x_0)+\dfrac{a}{b}(e^{bt}-1),&b>0,\\[6pt] V(x_0)+at,&b=0. \end{cases} $$

Proof. Apply Itô’s formula to $V(X_{t\wedge\tau_n})$. Writing the formula with its derivatives and correction term gives $dV(X_s)=V'(X_s)\mu(X_s)\,ds+V'(X_s)\sigma(X_s)\,dB_s+\tfrac12V''(X_s)\sigma(X_s)^2\,ds$ before stopping. Thus $$ V(X_{t\wedge\tau_n})=V(x_0)+\int_0^{t\wedge\tau_n}\mathcal{L}V(X_s)\,ds +\int_0^t\mathbf{1}_{\{s<\tau_n\}}V'(X_s)\sigma(X_s)\,dB_s. $$ The integrand in the last term is progressively measurable and bounded on $[0,T]$ for each fixed $T$, because it is stopped in $[-n,n]$. It therefore belongs to $\mathcal{L}^2$. Its Itô integral is a true martingale: it is integrable by the Itô isometry, and its conditional expectation at an earlier time equals its earlier value. In particular, its expectation at time $t$ is zero.

Using $\mathcal{L}V\le a+bV$, taking expectations, and using $V\ge0$, we obtain $\mathbb{E}V(X_{t\wedge\tau_n})\le V(x_0)+at+b\int_0^t\mathbb{E}V(X_{s\wedge\tau_n})\,ds$. Here $\mathbf{1}_{\{s<\tau_n\}}V(X_s)\le V(X_{s\wedge\tau_n})$, including on paths already stopped. On a fixed interval $[0,T]$, the integral form of Gronwall’s inequality first gives the uniform bound $\mathbb{E}V(X_{t\wedge\tau_n})\le (V(x_0)+aT)e^{bT}$. Since the solution is non-explosive, $\tau_n\to\infty$ almost surely. Fatou’s lemma gives $\mathbb{E}V(X_t)<\infty$. Monotone convergence gives $\int_0^t\mathbb{E}[\mathbf{1}_{\{s<\tau_n\}}V(X_s)]\,ds\to\int_0^t\mathbb{E}V(X_s)\,ds$; Fatou’s lemma on the left side of the stopped inequality therefore yields $\mathbb{E}V(X_t)\le V(x_0)+at+b\int_0^t\mathbb{E}V(X_s)\,ds$. To obtain the stated bound, set $G(t)=V(x_0)+at+b\int_0^t\mathbb{E}V(X_s)\,ds$. Then $\mathbb{E}V(X_t)\le G(t)$, and $G'(t)=a+b\mathbb{E}V(X_t)\le a+bG(t)$ for almost every $t$. Multiplying by $e^{-bt}$ and integrating proves the formula when $b>0$; when $b=0$, the integral inequality directly gives $V(x_0)+at$. $\square$

A useful sharper estimate holds when the generator decreases at large values of $V$.

Corollary: Under the same nonnegativity, regularity, and non-explosion assumptions, if $a\ge0$, $b>0$, and $\mathcal{L}V\le a-bV$, then $$ \mathbb{E}[V(X_t)]\le e^{-bt}V(x_0)+\frac{a}{b}(1-e^{-bt}). $$

Proof. Apply the time-dependent Itô formula to $e^{bs}V(X_s)$, stopped at $\tau_n$. Its drift is $e^{bs}(bV+\mathcal{L}V)(X_s)\,ds$, which is at most $ae^{bs}\,ds$. The stochastic integral has integrand $\mathbf{1}_{\{s<\tau_n\}}e^{bs}V'(X_s)\sigma(X_s)$; it is progressively measurable and in $\mathcal{L}^2$ on each finite interval, so it is a true martingale of mean zero. Taking expectations gives $\mathbb{E}[e^{b(t\wedge\tau_n)}V(X_{t\wedge\tau_n})]\le V(x_0)+a\int_0^t e^{bs}\,ds$. Fatou’s lemma, non-explosion, and $\int_0^t e^{bs}\,ds=(e^{bt}-1)/b$ give the result after division by $e^{bt}$. $\square$

Coercivity Can Also Prevent Explosion

The non-explosion assumption can sometimes be established by the same stopped calculation. Suppose a maximal locally defined solution has exit times $\tau_n$ as above, and $V\ge0$ is coercive with $\mathcal{L}V\le a$, where $a\ge0$. Itô’s formula up to $t\wedge\tau_n$ gives $\mathbb{E}V(X_{t\wedge\tau_n})\le V(x_0)+at$, since the stopped stochastic integral is square-integrable and has mean zero. Let $m_n=\inf_{|x|\ge n}V(x)$; coercivity implies $m_n\to\infty$. On $\{\tau_n\le T\}$, continuity gives $V(X_{T\wedge\tau_n})=V(X_{\tau_n})\ge m_n$. Hence $\mathbb{P}(\tau_n\le T)\le (V(x_0)+aT)/m_n\to0$. Thus the probability of explosion by any fixed finite $T$ is zero. This argument requires that the solution is defined up to each exit time; it does not presume that the maximal solution is already global.

Worked Example: A Growth Bound Checked Against an Exact Solution

Consider $dX_t=2\,dt+dB_t$, $X_0=0$, and $V(x)=1+x^2$. This SDE has the explicit solution $X_t=2t+B_t$. Compute a Lyapunov bound for $\mathbb{E}[X_t^2]$, then compare it with the exact moment.

Here $V'(x)=2x$, $V''(x)=2$, the drift is $\mu(x)=2$, and the diffusion coefficient is $\sigma(x)=1$. The full Itô drift, including the correction, is $\mu V'+\tfrac12\sigma^2V''=4x+1$. Since $4x\le2x^2+2$, which follows from $2(x-1)^2\ge0$, we have $\mathcal{L}V(x)\le2x^2+3=1+2V(x)$. The theorem with $a=1$, $b=2$, and $V(0)=1$ gives $\mathbb{E}V(X_t)\le \tfrac32e^{2t}-\tfrac12$, and therefore $\mathbb{E}X_t^2\le\tfrac32(e^{2t}-1)$.

Check: Directly from $X_t=2t+B_t$, $\mathbb{E}X_t^2=4t^2+t$, because $\mathbb{E}B_t=0$ and $\mathbb{E}B_t^2=t$. At $t=1$, the exact moment is $5$, while the bound is $\tfrac32(e^2-1)>5$; the latter inequality follows, for example, from $e^2>1+2+2=5$. At $t=0$, both sides are zero. The bound is valid but not sharp: the generator estimate was chosen to hold for every real $x$.

Worked Example: A Dissipative Cubic Drift

For $dX_t=-X_t^3\,dt+dB_t$, show that $V(x)=1+x^2$ yields a uniform-in-time bound on the second moment and rules out explosion.

We have $V'(x)=2x$, $V''(x)=2$, $\mu(x)=-x^3$, and $\sigma(x)=1$. Thus the drift in Itô’s formula is $\mu V'+\tfrac12\sigma^2V''=-2x^4+1$. Put $y=x^2\ge0$. Completing the square gives $\tfrac{17}{8}-1-y-(1-2y^2)=2(y-\tfrac14)^2\ge0$, so $\mathcal{L}V(x)\le\tfrac{17}{8}-V(x)$. The stopped estimate from the coercivity argument applies with $a=17/8$; because $V$ is coercive, it proves non-explosion. The dissipative corollary then gives $\mathbb{E}V(X_t)\le e^{-t}(1+x_0^2)+\tfrac{17}{8}(1-e^{-t})$, and hence $\mathbb{E}X_t^2\le e^{-t}x_0^2+\tfrac98(1-e^{-t})$.

Check: At $t=0$, the bound equals $x_0^2$, the exact initial second moment. As $t\to\infty$, its right side tends to $9/8$, so it supplies a finite bound at all times. The generator inequality itself is tight at $x^2=1/4$, since the completed square then vanishes.

Worked Example: Bounded Noise with Restoring Drift

Consider $dX_t=-X_t\,dt+(1+X_t^2)^{-1/2}\,dB_t$. Establish a second-moment estimate using $V(x)=1+x^2$.

The drift and diffusion coefficients are locally Lipschitz and satisfy a linear growth bound, so the standard existence theorem for SDEs gives a global strong solution. With $V'(x)=2x$, $V''(x)=2$, $\mu(x)=-x$, and $\sigma(x)=(1+x^2)^{-1/2}$, Itô’s formula gives the drift $\mu V'+\tfrac12\sigma^2V''=-2x^2+(1+x^2)^{-1}$. This is at most $1-x^2=2-V(x)$, because the difference between the upper bound and this drift is $x^2+1-(1+x^2)^{-1}=x^2+x^2/(1+x^2)\ge0$. The dissipative corollary with $a=b=1$ yields $\mathbb{E}V(X_t)\le e^{-t}(1+x_0^2)+(1-e^{-t})$, or $\mathbb{E}X_t^2\le e^{-t}x_0^2+(1-e^{-t})$.

Check: If $x_0=0$, the estimate is $\mathbb{E}X_t^2\le1-e^{-t}$. Its right side is nonnegative, equals zero at $t=0$, and has derivative $1$ at zero. This matches the instantaneous second-moment drift at the initial state: $\mathcal{L}(x^2)(0)=\sigma(0)^2=1$. The estimate is a bound, not a claim that the moment has this exact formula.

Common Pitfalls and Interpretation

A generator inequality is a pointwise statement about $\mathcal{L}V(x)$; it is not itself an expectation identity. The expectation calculation becomes valid only after stopping makes the Itô integral square-integrable. An unstopped stochastic integral with an unbounded integrand may be only a local martingale, and its expectation need not be zero without an integrability argument.

A second error is to discard the stopping time without justification. The stopped estimate alone controls $X_{t\wedge\tau_n}$, not automatically $X_t$. Non-explosion and Fatou’s lemma justify passing to the terminal moment; monotone convergence justifies passing to the time integral. Finally, a bound $\mathcal{L}V\le a+bV$ gives a finite-time growth estimate, while a bound $\mathcal{L}V\le a-bV$ gives a uniform-in-time estimate. The sign of the $V$ term is what distinguishes these conclusions.

Check Your Understanding

For each question, identify the hypotheses needed before using the moment estimate.

For $V\ge0$ with $\mathcal{L}V\le a+bV$, why is the stopped Itô integral a true martingale on a finite time interval?
Which limiting arguments are used to pass from the stopped terminal value and stopped time integral to their unstopped versions?
If $V$ is coercive and $\mathcal{L}V\le a$, how does the estimate bound the probability that the diffusion exits $[-n,n]$ before time $T$?
For the cubic-drift example, what inequality follows from $2(y-\tfrac14)^2\ge0$, and how does it produce the dissipative generator bound?
Explain why $\mathcal{L}V\le a-bV$, with $b>0$, yields a different long-time conclusion from $\mathcal{L}V\le a+bV$.

Moment Bounds from Itô’s Formula and Lyapunov Functions

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