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Stability, Invariant Measures and Long-Time Behavior · Tutorial 276 of 600

Moment Estimates for Ornstein-Uhlenbeck and Geometric Brownian Motion

Use Itô’s formula and explicit solutions to calculate and interpret first and second moments for two fundamental diffusion models.

Advanced 9 min read

What You'll Learn

  • Derive the Ornstein–Uhlenbeck mean and variance from its explicit solution.
  • Use Itô’s formula to verify the OU second-moment equation.
  • Compute geometric Brownian motion moments from its exponential solution.
  • Identify how drift and diffusion jointly determine second-moment growth.
  • Distinguish moment boundedness from pathwise or mean-square stability claims.

What the Coefficients Tell Us About Moments

Two familiar diffusion models illustrate different long-time moment behavior. The Ornstein–Uhlenbeck process has a restoring drift: it pulls the state toward a fixed level, while its diffusion continually adds random variation. Geometric Brownian motion has drift and diffusion proportional to its current value, so its moments evolve exponentially. Computing the moments makes the roles of these coefficients precise.

Work on a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})\) carrying a standard Brownian motion \(B_t\). All expectations below are under \(\mathbb{P}\), not under a risk-neutral measure \(\mathbb{Q}\). Every integral against \(B_t\) is an Itô integral: its integrand is progressively measurable and belongs to \(\mathcal{L}^2\) on each finite time interval, meaning \(\mathbb{E}\int_0^T H_s^2\,ds<\infty\). The stochastic integrals used below are square-integrable martingales, so they have mean zero and conditional expectation equal to their current value. We use the standard Itô isometry for such integrals.

Both SDEs are interpreted in the Itô sense. The quadratic variation is \([B]_t=t\), giving the multiplication rules \(dB_t\,dB_t=dt\), \(dB_t\,dt=0\), and \(dt\,dt=0\). In particular, the Itô correction is not part of the ordinary chain rule. For example, \(\int_0^t H_s\circ dB_s=\int_0^t H_s\,dB_s+\tfrac12[H,B]_t\) when the Stratonovich integral is defined and the covariation exists. The explicit solutions here are adapted functions of the given Brownian motion, hence strong solutions; no claims about weak uniqueness or pathwise uniqueness are needed.

Ornstein–Uhlenbeck Moments

Take constants \(k>0\), \(\eta\in\mathbb{R}\), \(m\in\mathbb{R}\), and deterministic \(X_0=x\). The Ornstein–Uhlenbeck equation is \(dX_t=-k(X_t-m)\,dt+\eta\,dB_t\). Its drift is \(-k(X_t-m)\), which points toward \(m\), and its diffusion coefficient is the constant \(\eta\).

Theorem: For every \(t\ge0\), the OU solution has \(\mathbb{E}[X_t]=m+(x-m)e^{-kt}\), \(\operatorname{Var}(X_t)=\dfrac{\eta^2}{2k}(1-e^{-2kt})\), and \(\mathbb{E}[X_t^2]=\big(m+(x-m)e^{-kt}\big)^2+\dfrac{\eta^2}{2k}(1-e^{-2kt})\).

Proof. Apply Itô’s formula to \(f(t,z)=e^{kt}(z-m)\). Its derivatives are \(f_t=k e^{kt}(z-m)\), \(f_z=e^{kt}\), and \(f_{zz}=0\); thus the \(\tfrac12 f_{zz}\,dt\) correction is zero. Substituting the SDE gives \(d(e^{kt}(X_t-m))=\eta e^{kt}\,dB_t\). Integrating from \(0\) to \(t\) yields \(X_t=m+(x-m)e^{-kt}+\eta\int_0^t e^{-k(t-s)}\,dB_s\). The integrand \(s\mapsto\eta e^{-k(t-s)}\) is deterministic and has finite squared integral on \([0,t]\). The Itô integral therefore has mean zero, and the Itô isometry gives its variance as \(\eta^2\int_0^t e^{-2k(t-s)}\,ds=\frac{\eta^2}{2k}(1-e^{-2kt})\). Adding the square of the mean to the variance gives the second moment. These computations also establish finite second moments on every finite interval. \(\square\)

The formula shows that the mean approaches \(m\), while the variance approaches \(\eta^2/(2k)\). Thus the second moment stays bounded over all \(t\ge0\). This conclusion depends on \(k>0\); without a restoring drift, the displayed limit for the variance does not apply.

Worked Example: An OU Process with Exact Parameters

Let \(k=2\), \(m=3\), \(\eta=1\), and \(X_0=1\). Find the mean, variance, and second moment at \(t=1\).

First, \(x-m=1-3=-2\), so the mean formula gives \(\mathbb{E}[X_1]=3+(-2)e^{-2}=3-2e^{-2}\). For the variance, substitute \(k=2\), \(\eta=1\), and \(t=1\): \(\operatorname{Var}(X_1)=\frac{1}{2\cdot2}(1-e^{-2\cdot2\cdot1})=\frac{1-e^{-4}}4\). Therefore \(\mathbb{E}[X_1^2]=(3-2e^{-2})^2+\frac{1-e^{-4}}4\). Expanding the square gives \(9-12e^{-2}+4e^{-4}+\frac14-\frac14e^{-4}=\frac{37}{4}-12e^{-2}+\frac{15}{4}e^{-4}\).

Check: subtracting the squared mean from the second moment returns \(\frac{1-e^{-4}}4\), the variance just computed. At \(t=0\), the general formulas give mean \(1\), variance \(0\), and second moment \(1\), agreeing with the deterministic initial value.

Checking the OU Second-Moment Equation Directly

Itô’s formula also provides a differential check on the second-moment expression. For \(g(z)=z^2\), the derivatives are \(g'(z)=2z\) and \(g''(z)=2\), so the correction term is \(\tfrac12g''(X_t)\eta^2dt=\eta^2dt\). Consequently,

$$ d(X_t^2)=\big(-2kX_t(X_t-m)+\eta^2\big)\,dt+2\eta X_t\,dB_t. $$

Because the explicit solution has finite second moments on finite intervals, \(\mathbb{E}\int_0^T(2\eta X_s)^2\,ds<\infty\). Hence the stochastic integral in this identity is a square-integrable martingale and its expectation is zero. Taking expectations gives \(\frac{d}{dt}\mathbb{E}[X_t^2]=-2k\mathbb{E}[X_t^2]+2km\mathbb{E}[X_t]+\eta^2\). This equation is a check, not a replacement for the explicit derivation: inserting the already calculated mean and second moment satisfies it and the initial condition \(\mathbb{E}[X_0^2]=x^2\).

Worked Example: Verify the OU Moment Equation

For the parameters in the previous example, verify the differential equation for \(M(t)=\mathbb{E}[X_t^2]\). Here \(\mathbb{E}[X_t]=3-2e^{-2t}\), and the second-moment formula expands to \(M(t)=\frac{37}{4}-12e^{-2t}+\frac{15}{4}e^{-4t}\). Differentiating gives \(M'(t)=24e^{-2t}-15e^{-4t}\).

For these parameters the moment equation is \(M'=-4M+12\mathbb{E}[X_t]+1\). Substitution gives \(-4(\frac{37}{4}-12e^{-2t}+\frac{15}{4}e^{-4t})+12(3-2e^{-2t})+1\). The constant terms are \(-37+36+1=0\), the \(e^{-2t}\) terms are \(48e^{-2t}-24e^{-2t}=24e^{-2t}\), and the remaining term is \(-15e^{-4t}\). This equals \(M'(t)\), as required.

Check: \(M(0)=\frac{37}{4}-12+\frac{15}{4}=1=X_0^2\). The derivative and initial value both agree with the moment equation.

Geometric Brownian Motion Moments

Let \(S_0=s>0\) be deterministic, and let \(\alpha,\sigma\in\mathbb{R}\). Geometric Brownian motion satisfies \(dS_t=\alpha S_t\,dt+\sigma S_t\,dB_t\). Here \(\alpha S_t\) is the drift coefficient and \(\sigma S_t\) is the diffusion coefficient. The first moment depends on \(\alpha\); the second moment depends on both coefficients.

Theorem: The solution is \(S_t=s\exp((\alpha-\tfrac12\sigma^2)t+\sigma B_t)\). For \(t\ge0\), \(\mathbb{E}[S_t]=s e^{\alpha t}\) and \(\mathbb{E}[S_t^2]=s^2e^{(2\alpha+\sigma^2)t}\).

Proof. Define \(F(t,z)=s\exp((\alpha-\tfrac12\sigma^2)t+\sigma z)\). Its derivatives are \(F_t=(\alpha-\tfrac12\sigma^2)F\), \(F_z=\sigma F\), and \(F_{zz}=\sigma^2F\). Itô’s formula, including its correction, gives \(dF(t,B_t)=F_tdt+F_zdB_t+\tfrac12F_{zz}dt=\alpha Fdt+\sigma FdB_t\). Also \(F(0,0)=s\), so this explicit adapted process satisfies the SDE. Since \(B_t\) is normal with mean zero and variance \(t\), its moment-generating function gives \(\mathbb{E}[e^{cB_t}]=e^{c^2t/2}\) for every real \(c\). Therefore \(\mathbb{E}[S_t]=s e^{(\alpha-\sigma^2/2)t}e^{\sigma^2t/2}=se^{\alpha t}\), and \(\mathbb{E}[S_t^2]=s^2e^{(2\alpha-\sigma^2)t}e^{(2\sigma)^2t/2}=s^2e^{(2\alpha+\sigma^2)t}\). These finite moments on every finite interval ensure that the integrands \(\sigma S_t\) and \(2\sigma S_t^2\), when used in the moment checks below, are square-integrable on finite time intervals; for the latter use the same Gaussian moment formula with the fourth moment of \(S_t\). Thus the corresponding Itô integrals are square-integrable martingales with zero mean. \(\square\)

The factor \(\sigma^2\) in the second-moment exponent arises from quadratic variation. To see it directly, apply Itô’s formula to \(g(z)=z^2\): \(g'(z)=2z\), \(g''(z)=2\), and the correction is \(\tfrac12(2)\sigma^2S_t^2dt\). It follows that \(d(S_t^2)=(2\alpha+\sigma^2)S_t^2dt+2\sigma S_t^2dB_t\). Taking expectations is justified by the integrability just noted, and gives the differential equation \(M_2'=(2\alpha+\sigma^2)M_2\), with \(M_2(0)=s^2\). Its solution agrees with the formula in the theorem.

Worked Example: A Geometric Brownian Motion with Decaying Moments

Take \(s=2\), \(\alpha=-1\), and \(\sigma=\tfrac12\). At \(t=1\), the first moment is \(\mathbb{E}[S_1]=2e^{-1}\). The exponent for the second moment is \(2\alpha+\sigma^2=2(-1)+(\tfrac12)^2=-2+\tfrac14=-\tfrac74\), so \(\mathbb{E}[S_1^2]=2^2e^{-7/4}=4e^{-7/4}\). The variance is thus \(4e^{-7/4}-(2e^{-1})^2=4e^{-7/4}-4e^{-2}\).

Check: \(e^{-7/4}>e^{-2}\), so the computed variance is positive. At \(t=0\), the moment formulas give mean \(2\) and second moment \(4\), as they must for \(S_0=2\).

Worked Example: Diffusion Can Increase the Second Moment

Let \(s=1\), \(\alpha=-\tfrac14\), and \(\sigma=1\), and evaluate at \(t=2\). The mean is \(\mathbb{E}[S_2]=e^{(-1/4)2}=e^{-1/2}\), which is less than its initial value \(1\). The second-moment exponent is \(2\alpha+\sigma^2=2(-\tfrac14)+1=-\tfrac12+1=\tfrac12\). Therefore \(\mathbb{E}[S_2^2]=e^{(1/2)2}=e\), which is greater than its initial value \(1\).

Check: the variance is \(e-(e^{-1/2})^2=e-e^{-1}\), a positive value. The negative drift reduces the first moment, but it does not overcome the diffusion contribution to the second-moment exponent in this example.

Moment Boundedness and a Common Pitfall

Proposition: For the OU process with \(k>0\), deterministic initial value, and finite parameters, both \(|\mathbb{E}[X_t]|\) and \(\mathbb{E}[X_t^2]\) are bounded uniformly for \(t\ge0\). For geometric Brownian motion with \(s>0\), the first moment is bounded on \(t\ge0\) exactly when \(\alpha\le0\); the second moment is bounded exactly when \(2\alpha+\sigma^2\le0\). A zero exponent gives a constant moment, while a negative exponent gives decay.

Proof. For OU, \(0<e^{-kt}\le1\), so \(|\mathbb{E}[X_t]|\le |m|+|x-m|\). Also, the variance is at most \(\eta^2/(2k)\), and \(|m+(x-m)e^{-kt}|\le |m|+|x-m|\). Since the second moment is variance plus squared mean, it is bounded by \(\eta^2/(2k)+(|m|+|x-m|)^2\), uniformly in \(t\). For geometric Brownian motion, the first moment is \(s e^{\alpha t}\) with \(s>0\). If \(\alpha\le0\), this is at most \(s\); if \(\alpha>0\), it grows without bound as \(t\) increases. The second moment is \(s^2e^{(2\alpha+\sigma^2)t}\), and the same argument applied to its exponent proves the stated condition. \(\square\)

A common mistake is to use the square of the mean as the second moment of geometric Brownian motion. That calculation would give \((\mathbb{E}[S_t])^2=s^2e^{2\alpha t}\), omitting the diffusion contribution. The correction is visible by applying Itô’s formula to \(S_t^2\): the term \(\tfrac12g''(S_t)\sigma^2S_t^2dt=\sigma^2S_t^2dt\) changes the moment exponent to \(2\alpha+\sigma^2\). The square of the mean is only one part of the second moment, since \(\mathbb{E}[S_t^2]=\operatorname{Var}(S_t)+(\mathbb{E}[S_t])^2\).

These formulas are useful because moment behavior can be read directly from the coefficients. In OU, the restoring drift limits the effect of accumulated noise and produces a finite long-time variance. In geometric Brownian motion, drift and diffusion combine in the second-moment exponent, so a decreasing mean alone does not guarantee a decreasing second moment. Moment boundedness is a statement about expectations; it is not, by itself, a claim about pathwise behavior.

Check Your Understanding

Use the formulas and conditions in this tutorial to answer each question.

  1. For an OU process with parameters \(k,m,\eta,x\), what are the limiting mean and variance as \(t\) tends to infinity, assuming \(k>0\)?
  2. For \(dS_t=\alpha S_t\,dt+\sigma S_t\,dB_t\), derive the exponent governing \(\mathbb{E}[S_t^2]\) using Itô’s formula on \(S_t^2\).
  3. If \(\alpha<0\), can the second moment of geometric Brownian motion grow? State the exact condition on \(\alpha\) and \(\sigma\) for growth.
  4. Why is the stochastic integral in the OU second-moment calculation mean zero, and what integrability condition is used?
  5. For \(s=3\), \(\alpha=0\), and \(\sigma=1\), compute \(\mathbb{E}[S_1]\) and \(\mathbb{E}[S_1^2]\), then check that the variance is positive.