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Stability, Invariant Measures and Long-Time Behavior · Tutorial 278 of 600

Almost-Sure Stability of Scalar Linear SDEs

Solve the scalar linear SDE explicitly, then identify the precise drift–diffusion condition for almost-sure convergence to zero.

Advanced 11 min read

What You'll Learn

  • Derive and verify the exponential solution of a scalar linear SDE.
  • Distinguish the drift coefficient from the solution’s almost-sure exponential growth rate.
  • Apply the Brownian strong law to determine long-time behavior away from the critical case.
  • Handle the critical case using Brownian oscillation.
  • Separate stability of the origin from convergence for the zero initial condition.
  • Recognize why mean behavior alone does not determine almost-sure stability.

From the SDE to Its Long-Time Rate

Consider a scalar state whose drift is proportional to its current value and whose noise is also proportional to that value. Random fluctuations may cause individual paths to grow or shrink even when the mean follows a different trend. To decide whether a path converges to zero, the key quantity will be its almost-sure exponential rate, not just the drift coefficient \(a\).

Throughout, 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. For constants \(a,\sigma\in\mathbb{R}\) and deterministic initial value \(x_0\in\mathbb{R}\), study the Itô SDE

$$ dX_t=aX_t\,dt+\sigma X_t\,dB_t,\qquad X_0=x_0. $$

Here \(aX_t\) is the drift coefficient and \(\sigma X_t\) is the diffusion coefficient. The integral form uses the Itô integral \(\int_0^t \sigma X_s\,dB_s\), with progressively measurable integrand in \(\mathcal{L}^2\): \(\mathbb{E}\int_0^T \sigma^2X_s^2\,ds<\infty\) for each finite \(T\). A strong solution is an adapted process on the given space driven by its given Brownian motion. Pathwise uniqueness means two such solutions with the same Brownian motion and initial value agree almost surely at all times. Uniqueness in law, by contrast, means that any solutions have the same distribution; it does not itself assert that two solutions on the same space agree. All probabilities and expectations below are under \(\mathbb{P}\).

The coefficients \(x\mapsto ax\) and \(x\mapsto \sigma x\) are globally Lipschitz and have linear growth: their differences are bounded by \((|a|+|\sigma|)|x-y|\), and their absolute values are bounded by a constant times \(1+|x|\). These are standard sufficient conditions for a unique global strong solution. Here the solution and uniqueness can also be checked directly.

Key rate: Define \(\lambda=a-\tfrac12\sigma^2\). For a nonzero initial value the candidate solution is \(X_t=x_0\exp(\lambda t+\sigma B_t)\). Its logarithmic rate is \(\lambda+\sigma B_t/t\), so the Brownian strong law identifies \(\lambda\) as the almost-sure exponential growth rate.

Deriving and Verifying the Explicit Solution

Set \(f(t,b)=x_0\exp(\lambda t+\sigma b)\). Its partial derivatives are \(f_t=\lambda f\), \(f_b=\sigma f\), and \(f_{bb}=\sigma^2f\). Itô’s formula, including its second-derivative correction, gives

$$ df(t,B_t)=f_t\,dt+f_b\,dB_t+\tfrac12 f_{bb}\,dt =\left(\lambda+\tfrac12\sigma^2\right)f\,dt+\sigma f\,dB_t =a f\,dt+\sigma f\,dB_t. $$

Thus the candidate satisfies the SDE and starts at \(x_0\). To check the required integrability, its square has expectation \(\mathbb{E}[X_t^2]=x_0^2e^{(2a+\sigma^2)t}\), using the normal moment-generating function for \(B_t\sim N(0,t)\). This is finite, and its integral over any finite interval is finite. Therefore the stochastic integral with integrand \(\sigma X_s\) is an \(\mathcal{L}^2\) Itô integral.

Proof. It remains to verify uniqueness. For any solution \(X\), define \(G_t=\exp(-\lambda t-\sigma B_t)\). Itô’s formula gives \(dG_t=(-a+\sigma^2)G_t\,dt-\sigma G_t\,dB_t\). The product rule for continuous Itô processes includes the covariation term, so

$$ d(G_tX_t)=G_t\,dX_t+X_t\,dG_t+d[G,X]_t. $$

In this expression, the drift contributions are \(aG_tX_t\,dt\), \((-a+\sigma^2)G_tX_t\,dt\), and \(d[G,X]_t=-\sigma^2G_tX_t\,dt\); they sum to zero. The diffusion contributions \(\sigma G_tX_t\,dB_t\) and \(-\sigma G_tX_t\,dB_t\) also cancel. The calculation is valid after stopping when \(|X|\) or \(|G|\) first reaches a fixed bound: the resulting integrands are progressively measurable and bounded on each finite time interval, hence in \(\mathcal{L}^2\). Removing the stops gives \(G_tX_t=x_0\) for all \(t\), almost surely, because both processes are continuous and finite on compact time intervals. Since \(G_t>0\), \(X_t=x_0/G_t=x_0\exp(\lambda t+\sigma B_t)\). Any solution must equal the candidate, proving pathwise uniqueness and the asserted formula. \(\square\)

In particular, the formula is a strong solution. With deterministic \(x_0\), it also specifies the distribution of the solution for any Brownian driver, so the solution’s law is unique. These are distinct statements: uniqueness in law compares distributions, whereas pathwise uniqueness compares solutions driven by the same Brownian motion.

The Exact Almost-Sure Convergence Criterion

Theorem: For every deterministic \(x_0\ne0\), \(X_t\to0\) almost surely as \(t\to\infty\) if and only if \(a-\tfrac12\sigma^2<0\). Consequently, the origin attracts every deterministic initial value almost surely exactly under this strict inequality. If \(x_0=0\), the solution is identically zero for all \(a,\sigma\).

The Brownian strong law states that \(B_t/t\to0\) almost surely as \(t\to\infty\). We use this standard result without proof here. At the boundary we also use the standard Brownian oscillation property \(\limsup_{t\to\infty}B_t=+\infty\) and \(\liminf_{t\to\infty}B_t=-\infty\), whose proof is not needed for the criterion and is not given here.

Proof. First take \(x_0\ne0\). The explicit solution never changes sign, and \[ \frac{1}{t}\log|X_t|=\frac{1}{t}\log|x_0|+\lambda+\sigma\frac{B_t}{t}. \] On the probability-one event where the Brownian strong law holds, this expression tends to \(\lambda\). If \(\lambda<0\), then \(\log|X_t|\to-\infty\), so \(|X_t|\to0\). If \(\lambda>0\), then \(\log|X_t|\to+\infty\), so \(|X_t|\to+\infty\), which rules out convergence to zero.

It remains to consider \(\lambda=0\). If \(\sigma=0\), then \(a=0\) and \(X_t=x_0\), which does not converge to zero. If \(\sigma\ne0\), the Brownian oscillation property implies that \(\sigma B_t\) has arbitrarily large positive values as \(t\to\infty\): use \(\limsup B_t=+\infty\) when \(\sigma>0\), and \(\liminf B_t=-\infty\) when \(\sigma<0\). Hence \(|X_t|=|x_0|e^{\sigma B_t}\) is unbounded along times tending to infinity and cannot converge to zero. This proves both directions, including the boundary, for nonzero \(x_0\). When \(x_0=0\), the explicit formula gives \(X_t=0\) for every \(t\). \(\square\)

Worked Examples

Worked Example: Almost-Sure Decay with Positive Drift

Take \(x_0=2\), \(a=\tfrac14\), and \(\sigma=1\). The rate is \(\lambda=a-\tfrac12\sigma^2=\tfrac14-\tfrac12=-\tfrac14\). The explicit solution is therefore

$$ X_t=2\exp\left(-\tfrac14t+B_t\right). $$

By the Brownian strong law, \(B_t/t\to0\) almost surely, so \(t^{-1}\log|X_t|=t^{-1}\log2-\tfrac14+B_t/t\to-\tfrac14\). Thus \(X_t\to0\) almost surely, despite the positive drift coefficient \(a=\tfrac14\).

Check: Since \(B_t\sim N(0,t)\), \(\mathbb{E}[e^{B_t}]=e^{t/2}\), and hence \(\mathbb{E}[X_t]=2e^{-t/4}e^{t/2}=2e^{t/4}\). This agrees with the expectation equation \(d\,\mathbb{E}[X_t]/dt=a\mathbb{E}[X_t]\), whose solution from \(2\) is \(2e^{t/4}\). The mean grows while almost every path decays; a mean calculation alone does not decide almost-sure stability.

Worked Example: Growth When the Rate Is Positive

Take \(x_0=-3\), \(a=1\), and \(\sigma=1\). Now \(\lambda=1-\tfrac12=\tfrac12\), giving

$$ X_t=-3\exp\left(\tfrac12t+B_t\right). $$

The sign remains negative, and \[ \frac1t\log|X_t|=\frac1t\log3+\frac12+\frac{B_t}{t}\longrightarrow\frac12 \quad\text{almost surely}. \] Therefore \(|X_t|\to\infty\), so \(X_t\) does not converge to zero.

Check: The second moment formula gives \(\mathbb{E}[X_t^2]=9e^{(2a+\sigma^2)t}=9e^{3t}\). This is consistent with the explicit square \(9e^{t+2B_t}\), since \(\mathbb{E}[e^{2B_t}]=e^{2t}\). Both calculations give \(9e^{3t}\).

Worked Example: The Critical Rate

Take \(x_0=1\), \(a=1\), and \(\sigma=\sqrt{2}\). The rate is \(\lambda=1-\tfrac12(\sqrt2)^2=1-1=0\), and the solution is \(X_t=e^{\sqrt2 B_t}\). Brownian oscillation gives arbitrarily large positive values of \(B_t\) at times tending to infinity. Consequently \(X_t\) is unbounded along such times and cannot tend to zero. This shows why the criterion is strict: replacing \(a<\tfrac12\sigma^2\) by \(a\le\tfrac12\sigma^2\) would be incorrect.

Check: Here \(\mathbb{E}[X_t]=\mathbb{E}[e^{\sqrt2 B_t}]=e^{(\sqrt2)^2t/2}=e^t\), while \(t^{-1}\log X_t=\sqrt2 B_t/t\to0\) almost surely. The expectation’s growth does not contradict the zero pathwise exponential rate.

Interpretation and a Common Pitfall

The threshold \(a=\tfrac12\sigma^2\) is set by the Itô correction in the logarithm of the solution. For \(x_0\ne0\), the logarithm has deterministic rate \(a-\tfrac12\sigma^2\), not \(a\). The drift \(aX_t\) in the original equation is therefore not itself the pathwise growth rate. The squared diffusion coefficient contributes half its value with a negative sign to that logarithmic rate.

A common wrong calculation is to write \(\log|X_t|=\log|x_0|+at+\sigma B_t\), then infer from \(B_t/t\to0\) that the stability condition is \(a<0\). This omits the quadratic-variation correction. Applying Itô’s formula to \(g(x)=\log|x|\), for \(x\ne0\), gives \(g'(x)=1/x\) and \(g''(x)=-1/x^2\); its correction is \(\tfrac12g''(X_t)(\sigma X_t)^2dt=-\tfrac12\sigma^2dt\). The corrected expression is \(\log|X_t|=\log|x_0|+(a-\tfrac12\sigma^2)t+\sigma B_t\), matching the explicit solution.

This is an Itô equation: its stochastic integral is \(\int_0^t \sigma X_s\,dB_s\). The Stratonovich integral \(\int_0^t \sigma X_s\circ dB_s\) is a different convention; converting that equation to Itô form changes its drift. One must not silently apply the Itô stability threshold to an equation written in Stratonovich form.

The Itô integral in the solution is a true square-integrable martingale on each finite interval: its integrand is in \(\mathcal{L}^2\), so the Itô integral has finite second moment and conditional expectation \(\mathbb{E}[\int_s^t \sigma X_u\,dB_u\mid\mathcal{F}_s]=0\) for \(s\le t\). The solution \(X_t\) itself need not be a martingale; its expectation in the first example grows. This distinction matters when a mean-based test is compared with pathwise convergence.

Check Your Understanding

Use the explicit solution and distinguish the zero initial condition from stability for nonzero initial states.

  1. For \(a=\tfrac32\) and \(\sigma=2\), calculate \(\lambda\) and decide whether every nonzero initial state converges to zero almost surely.
  2. For \(a=0\) and \(\sigma=0\), determine the solution for arbitrary \(x_0\) and explain what happens at the critical rate.
  3. For \(a=\tfrac12\) and \(\sigma=1\), identify the rate and state which Brownian path property rules out convergence to zero for \(x_0\ne0\).
  4. Explain why \(x_0=0\) converges to zero regardless of the values of \(a\) and \(\sigma\).
  5. For \(a=-1\) and \(\sigma=2\), compare the sign of the mean-growth coefficient \(a\) with the sign of the almost-sure rate \(\lambda\).