Ornstein-Uhlenbeck Solutions By Integrating Factors

Why an Integrating Factor Works

The Ornstein–Uhlenbeck equation is a linear stochastic differential equation with a restoring drift: when $X_t$ lies above its long-run level $\theta$, the drift points downward, and when it lies below $\theta$, it points upward, provided $\kappa>0$. The diffusion coefficient $\sigma$ supplies constant-amplitude random fluctuations. An integrating factor removes the term proportional to $X_t-\theta$, just as it does for a deterministic linear differential equation.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a Brownian motion $B_t$ with respect to the filtration. In particular, its increments are independent of the past and $B_t-B_s$ has distribution $N(0,t-s)$ for $s<t$. Fix constants $\kappa,\theta,\sigma\in\mathbb{R}$ and a deterministic initial value $x_0\in\mathbb{R}$. The mean-reverting case has $\kappa>0$, but the solution formula below is valid for every real $\kappa$.

Definition: The Ornstein–Uhlenbeck equation considered here is $dX_t=\kappa(\theta-X_t)\,dt+\sigma\,dB_t$, with $X_0=x_0$. The coefficient $\kappa(\theta-X_t)$ is the drift, and the constant $\sigma$ is the diffusion coefficient. An Itô integrand belongs to $\mathcal{L}^2$ on $[0,T]$ when it is progressively measurable and $\mathbb{E}\int_0^T H_s^2\,ds<\infty$.

Deriving the Explicit Formula

Center the process at $\theta$, and multiply by $e^{\kappa t}$. To justify this step, apply Itô’s formula to $f(t,x)=e^{\kappa t}(x-\theta)$. Its derivatives are $f_t(t,x)=\kappa e^{\kappa t}(x-\theta)$, $f_x(t,x)=e^{\kappa t}$, and $f_{xx}(t,x)=0$. Since the diffusion coefficient of $X_t$ is $\sigma$, the Itô correction is $\tfrac12 f_{xx}(t,X_t)\sigma^2\,dt=0$. Therefore

\begin{aligned} d\big(e^{\kappa t}(X_t-\theta)\big) &=\kappa e^{\kappa t}(X_t-\theta)\,dt +e^{\kappa t}\big(\kappa(\theta-X_t)\,dt+\sigma\,dB_t\big)\\ &=\sigma e^{\kappa t}\,dB_t. \end{aligned}

The two drift terms cancel because $\theta-X_t=-(X_t-\theta)$. Integrating from $0$ to $t$ gives the transformed equation. For a fixed finite horizon $T$, the deterministic integrand $H_s=e^{\kappa s}$ is progressively measurable and satisfies $\mathbb{E}\int_0^T H_s^2\,ds=\int_0^T e^{2\kappa s}\,ds<\infty$. Thus the integral is an Itô integral of an $\mathcal{L}^2$ integrand.

e^{\kappa t}(X_t-\theta)=x_0-\theta+\sigma\int_0^t e^{\kappa s}\,dB_s.

Theorem: For every $T<\infty$, the unique continuous adapted solution on $[0,T]$ is $$ X_t=\theta+e^{-\kappa t}(x_0-\theta) +\sigma\int_0^t e^{-\kappa(t-s)}\,dB_s,\qquad 0\le t\le T. $$ For each fixed $t$, the integral is defined with the progressively measurable integrand $H_s=e^{-\kappa(t-s)}\mathbf{1}_{[0,t]}(s)$, which satisfies $\mathbb{E}\int_0^T H_s^2\,ds<\infty$.

Proof. First suppose $X$ is a continuous adapted solution. The Itô calculation above shows that $e^{\kappa t}(X_t-\theta)=x_0-\theta+\sigma\int_0^t e^{\kappa s}\,dB_s$. Multiplying by $e^{-\kappa t}$ yields the displayed formula, because multiplying the integrand by the deterministic factor $e^{-\kappa t}$ gives $e^{-\kappa(t-s)}$.

Conversely, define $N_t=\int_0^t e^{\kappa s}\,dB_s$ and set $X_t=\theta+e^{-\kappa t}(x_0-\theta)+\sigma e^{-\kappa t}N_t$. The integrand $e^{\kappa s}$ is in $\mathcal{L}^2$ on every finite interval, so $N$ has a continuous adapted version. Apply Itô’s formula to $g(t,n)=e^{-\kappa t}n$: $g_t=-\kappa e^{-\kappa t}n$, $g_n=e^{-\kappa t}$, and $g_{nn}=0$, so the correction $\tfrac12 g_{nn}\,d[N]_t$ is zero. Since $dN_t=e^{\kappa t}\,dB_t$, this gives $d(e^{-\kappa t}N_t)=-\kappa e^{-\kappa t}N_t\,dt+dB_t$. Differentiating the deterministic term gives $-\kappa e^{-\kappa t}(x_0-\theta)\,dt$. Combining these identities yields $dX_t=-\kappa(X_t-\theta)\,dt+\sigma\,dB_t=\kappa(\theta-X_t)\,dt+\sigma\,dB_t$, and $X_0=x_0$. This proves existence. The first part of the proof shows that every solution must equal this formula, which proves uniqueness. $\square$

Worked Solutions and Checks

Worked Example: Mean Reversion with a Positive Initial Displacement

Take $\kappa=2$, $\theta=3$, $\sigma=1$, and $x_0=0$. Substitute these values into the formula:

X_t=3+e^{-2t}(0-3)+\int_0^t e^{-2(t-s)}\,dB_s =3-3e^{-2t}+\int_0^t e^{-2(t-s)}\,dB_s.

At $t=1$, the deterministic part is $3-3e^{-2}$, and the random part is $\int_0^1 e^{-2(1-s)}\,dB_s$. The latter is well-defined: on $[0,1]$ its deterministic integrand is progressively measurable and its squared integral is $\int_0^1 e^{-4(1-s)}\,ds=(1-e^{-4})/4<\infty$.

Check: at $t=0$, the integral over the empty interval is zero and $3-3e^0=0=x_0$. The integrating-factor form is $e^{2t}(X_t-3)=-3+\int_0^t e^{2s}\,dB_s$, matching the transformed equation with $x_0-\theta=-3$.

Worked Example: Evaluating the Formula at a Chosen Time

Take $\kappa=1$, $\theta=-2$, $\sigma=3$, and $x_0=5$. The explicit solution is

X_t=-2+7e^{-t}+3\int_0^t e^{-(t-s)}\,dB_s.

At $t=\ln 2$, this becomes $$ X_{\ln 2}=-2+\frac{7}{2} +3\int_0^{\ln 2}e^{-(\ln 2-s)}\,dB_s =\frac{3}{2}+\frac{3}{2}\int_0^{\ln 2}e^s\,dB_s, $$ because $e^{-(\ln 2-s)}=e^s/2$. The kernel is admissible on this finite interval: it is deterministic and progressively measurable, and $\int_0^{\ln 2}e^{2s}\,ds=(e^{2\ln 2}-1)/2=3/2<\infty$.

Check: the transformed formula is $e^t(X_t+2)=7+3\int_0^t e^s\,dB_s$. At $t=\ln 2$, it gives $2(X_{\ln 2}+2)=7+3\int_0^{\ln 2}e^s\,dB_s$. Substituting the expression above gives the same right-hand side.

Worked Example: The Zero-Reversion Case

Let $\kappa=0$, $\theta=7$, $\sigma=-2$, and $x_0=1$. The drift vanishes, since $0\cdot(7-X_t)=0$, and the formula reduces to

X_t=7+(1-7)-2\int_0^t 1\,dB_s=1-2B_t.

Check: this process starts at $X_0=1$, and its differential is $dX_t=-2\,dB_t$, exactly the stated SDE when $\kappa=0$. The formula also has the correct limit as $\kappa$ tends to zero: $e^{-\kappa t}$ tends to $1$, and the kernel $e^{-\kappa(t-s)}$ tends uniformly to $1$ for $s\in[0,t]$. By the Itô isometry, the corresponding integrals converge in $L^2$, since the squared-integrand difference has integral tending to zero.

The Convolution Is Not Generally a Martingale

The transformed stochastic integral and the stochastic convolution in the explicit solution are related, but they are different processes. Define $$ N_t=\int_0^t e^{\kappa s}\,dB_s, \qquad M_t=\int_0^t e^{-\kappa(t-s)}\,dB_s. $$ For every fixed finite $T$, the integrand $e^{\kappa s}$ defining $N$ is progressively measurable and in $\mathcal{L}^2$ on $[0,T]$. The process $N$ is a true square-integrable martingale: the Itô isometry gives $\mathbb{E}[N_t^2]=\int_0^t e^{2\kappa s}\,ds<\infty$, and for $s<t$, the increment $\int_s^t e^{\kappa u}\,dB_u$ has conditional expectation zero given $\mathcal{F}_s$. Hence $\mathbb{E}[N_t\mid\mathcal{F}_s]=N_s$.

For each fixed $t$, $M_t$ is also a well-defined square-integrable Itô integral: its deterministic integrand on $[0,t]$ is progressively measurable, and $\int_0^t e^{-2\kappa(t-s)}\,ds<\infty$. But as a process in $t$, $M_t=e^{-\kappa t}N_t$ is generally not a martingale. For $0\le s<t$, the martingale property of $N$ gives $$ \mathbb{E}[M_t\mid\mathcal{F}_s] =e^{-\kappa t}N_s =e^{-\kappa(t-s)}M_s. $$ If $\kappa>0$ and $s>0$, then $\mathbb{E}[M_s^2]=\int_0^s e^{-2\kappa(s-u)}\,du>0$, so $M_s$ is not zero almost surely. Since $e^{-\kappa(t-s)}\ne1$, the conditional expectation is not $M_s$ almost surely. Thus $M$ is not a martingale in this case. This distinction matters: the integrating-factor integral $N$ is a martingale, while the convolution appearing in $X_t$ generally is not.

Common Pitfall: Forgetting the Exponential Kernel

A tempting but incorrect answer is $X_t=\theta+(x_0-\theta)+\sigma B_t$, on the grounds that the noise is additive. That expression omits the mean-reverting drift and the time-dependent weighting of past noise. For example, with $\kappa=1$, $\theta=0$, $x_0=1$, and $\sigma=1$, the incorrect expression is $1+B_t$. Its differential is $dB_t$, but the required drift is $-X_t\,dt$, which is not zero in general. The correct formula is $X_t=e^{-t}+\int_0^t e^{-(t-s)}\,dB_s$; its drift term is present, as the integrating-factor verification shows.

A second error is to call $\int_0^t e^{-\kappa(t-s)}\,dB_s$ a martingale because it is an Itô integral. For a fixed terminal time $t$, it is indeed an Itô integral with a deterministic $\mathcal{L}^2$ integrand. But when $t$ varies, the integrand itself changes with $t$; the conditional expectation calculation above shows why the resulting process is generally not a martingale. Keep the fixed-time integral and the process-level martingale claim separate.

The explicit representation is useful because it turns the SDE into a deterministic decay term plus a weighted accumulation of Brownian increments. The weight $e^{-\kappa(t-s)}$ discounts noise from earlier times when $\kappa>0$. This formula is also a direct route to further properties of the process, but deriving its transition laws and covariance structure is a separate calculation.

Check Your Understanding

Use the integrating-factor derivation and verify each proposed conclusion directly.

For $\kappa=3$, $\theta=2$, $\sigma=0$, and $x_0=8$, write the explicit solution and check its initial value.
For general $\kappa$, identify the transformed process that is a square-integrable martingale and state the integrability condition for its integrand on $[0,T]$.
Compute $\mathbb{E}[M_t\mid\mathcal{F}_s]$ for $M_t=\int_0^t e^{-\kappa(t-u)}\,dB_u$, where $0\le s<t$, and explain when the result rules out the martingale property.
Set $\kappa=0$ in the explicit formula and simplify it for arbitrary $\theta,\sigma,x_0$.
Apply the integrating factor to show why a proposed solution with no mean-reverting drift cannot satisfy the equation for general $\kappa>0$.