Gaussian Transition Laws For The Ornstein-Uhlenbeck Process

From the SDE to a Transition Law

Consider a process pulled toward a fixed level while receiving continuous random shocks. Its current state influences the conditional distribution of its future state: the mean moves toward the level, while the variance accumulates from the shocks over the time interval. The Ornstein-Uhlenbeck process makes both effects explicit, and its transition law is Gaussian.

Work on a filtered probability space $(\Omega,\mathcal{F},(\mathcal{F}_t)_{t\ge0},\mathbb{P})$ carrying a standard Brownian motion $B_t$ relative to the filtration. Fix constants $\kappa>0$, $\theta\in\mathbb{R}$, and $\sigma\in\mathbb{R}$, and let $X_0$ be $\mathcal{F}_0$-measurable. The Ornstein-Uhlenbeck SDE is

dX_t=\kappa(\theta-X_t)\,dt+\sigma\,dB_t.

Here $\kappa$ is the mean-reversion rate, $\theta$ is the level toward which the drift pulls the process, and $\sigma$ is the diffusion coefficient. The stochastic integral in the Ito solution is understood for progressively measurable integrands $H$ in $\mathcal{L}^2$, meaning $\mathbb{E}\int_0^T H_r^2\,dr<\infty$ for every finite $T$. In the formula below the integrand is deterministic and satisfies this condition. All stochastic integrals here are Ito integrals; no Stratonovich-to-Ito correction is being made.

Fix $0\le s<t$, and write $h=t-s$. The goal is to find the conditional law of $X_t$ given $\mathcal{F}_s$, and then express the result in terms of $X_s$. The key is to apply Ito's formula to an integrating-factor transformation. In the calculation, the quadratic-variation rules are $dB_t\,dB_t=dt$, $dB_t\,dt=0$, and $dt\,dt=0$.

Conditional Gaussian Transition Theorem

Theorem: For $\kappa>0$, $s<t$, and the SDE above, conditionally on $\mathcal{F}_s$, $X_t$ is normally distributed with mean $$ m_{s,t}=\theta+(X_s-\theta)e^{-\kappa(t-s)} $$ and variance $$ v_{s,t}=\frac{\sigma^2}{2\kappa}\left(1-e^{-2\kappa(t-s)}\right). $$ If $\sigma\ne0$, its conditional density at $y\in\mathbb{R}$ is $$ p_{s,t}(X_s,y)=\frac{1}{\sqrt{2\pi v_{s,t}}} \exp\left(-\frac{(y-m_{s,t})^2}{2v_{s,t}}\right). $$ The conditional distribution assertion holds almost surely; the formula defines a transition kernel for every starting state $x\in\mathbb{R}$ by replacing $X_s$ with $x$.

Proof. Set $F(u,x)=e^{\kappa u}(x-\theta)$. Its derivatives are $F_u(u,x)=\kappa e^{\kappa u}(x-\theta)$, $F_x(u,x)=e^{\kappa u}$, and $F_{xx}(u,x)=0$. Applying Ito's formula to $F(u,X_u)$, including the second-derivative correction explicitly, gives

dF(u,X_u)=F_u\,du+F_x\,dX_u+\tfrac12F_{xx}\sigma^2\,du =\kappa e^{\kappa u}(X_u-\theta)\,du +e^{\kappa u}\big(\kappa(\theta-X_u)\,du+\sigma\,dB_u\big) +\tfrac12(0)\sigma^2\,du =\sigma e^{\kappa u}\,dB_u.

Integrating from $s$ to $t$ and multiplying by $e^{-\kappa t}$ yields

X_t=\theta+e^{-\kappa h}(X_s-\theta) +\sigma\int_s^t e^{-\kappa(t-r)}\,dB_r.

For fixed $s,t$, the integrand $H_r=\sigma e^{-\kappa(t-r)}\mathbf{1}_{[s,t]}(r)$ is deterministic, progressively measurable, and square-integrable, since $\int_s^t H_r^2\,dr<\infty$. Brownian increments after $s$ are independent of $\mathcal{F}_s$, so the integral has conditional mean zero given $\mathcal{F}_s$. Its conditional variance, by the Ito isometry, is

\int_s^t \sigma^2 e^{-2\kappa(t-r)}\,dr =\frac{\sigma^2}{2\kappa}\left(1-e^{-2\kappa(t-s)}\right).

It remains to justify that this integral is Gaussian, not merely to calculate its first two moments. Approximate its deterministic integrand in $L^2([s,t])$ by step functions. The integral of each step function is a finite linear combination of Brownian increments on disjoint intervals, hence is normal and independent of $\mathcal{F}_s$. The Ito isometry gives convergence of these integrals in $L^2(\mathbb{P})$, and therefore in probability. Their characteristic functions converge to the characteristic function of the limiting integral; their variances converge to the displayed variance, so the limit is centered normal with that variance and remains independent of $\mathcal{F}_s$. Combining this fact with the explicit solution proves the conditional Gaussian law. If $\sigma\ne0$ and $t>s$, the variance is positive, giving the stated density. $\square$

The transition density is a density with respect to Lebesgue measure. If $\sigma=0$, the conditional law instead puts all its mass at $\theta+(X_s-\theta)e^{-\kappa h}$, so a Lebesgue density of the displayed form does not exist. At $t=s$, the conditional law is likewise a point mass at $X_s$. These cases are described by the same mean and variance formulas, with zero variance, but require a point-mass interpretation.

Worked Transition Calculations

Worked Example: A Conditional Density at a Specified Time

Take $\kappa=2$, $\theta=1$, $\sigma=2$, $s=0$, $t=\ln 2$, and suppose $X_0=0$. Find the conditional mean, variance, and density of $X_t$.

First, $e^{-\kappa(t-s)}=e^{-2\ln2}=1/4$. Thus the conditional mean is $1+(0-1)(1/4)=3/4$. Also, $e^{-2\kappa(t-s)}=e^{-4\ln2}=1/16$, so the conditional variance is

\frac{2^2}{2\cdot2}\left(1-\frac1{16}\right) =1\cdot\frac{15}{16} =\frac{15}{16}.

Therefore the standard deviation is $\sqrt{15}/4$, and the density is

p_{0,\ln2}(0,y) =\frac{4}{\sqrt{30\pi}} \exp\left(-\frac{(y-\tfrac34)^2}{2(15/16)}\right) =\frac{4}{\sqrt{30\pi}} \exp\left(-\frac{8(y-\tfrac34)^2}{15}\right).

Check: At the mean $y=3/4$, the exponent is zero, so the density is $4/\sqrt{30\pi}$, which is exactly $1/(\sqrt{2\pi}\sqrt{15/16})$. The mean is between the initial value $0$ and the mean-reversion level $1$, as expected over a positive time interval.

Worked Example: A One-Sided Transition Probability

Let $\kappa=1$, $\theta=0$, $\sigma=\sqrt{2}$, $X_0=0$, and $t=\ln2$. Calculate $\mathbb{P}(X_t>\sqrt{3/4}\mid\mathcal{F}_0)$.

The mean is $0+(0-0)e^{-\ln2}=0$. The variance is

\frac{(\sqrt2)^2}{2\cdot1}\left(1-e^{-2\ln2}\right) =1\left(1-\frac14\right)=\frac34.

Hence $X_t$ conditionally has standard deviation $\sqrt{3/4}$. Standardizing the strict tail event gives

\mathbb{P}\left(X_t>\sqrt{3/4}\mid\mathcal{F}_0\right) =\mathbb{P}(Z>1) =1-\Phi(1),

where $Z$ is standard normal and $\Phi$ is its cumulative distribution function. Numerically, $1-\Phi(1)\approx0.1587$. Strict versus non-strict inequality gives the same probability here because the conditional law has a density and assigns zero probability to a single point.

Check: the threshold is exactly one conditional standard deviation above the mean, since $(\sqrt{3/4}-0)/\sqrt{3/4}=1$. The resulting upper-tail probability is therefore the standard normal upper tail at $1$, not at a rescaled threshold.

Worked Example: Probability of an Interval

Take $\kappa=1$, $\theta=2$, $\sigma=2$, $X_0=0$, and $t=\ln2$. Find the probability that $X_t$ lies between $1-\sqrt{3/2}$ and $1+\sqrt{3/2}$, with both endpoints included.

Since $e^{-\ln2}=1/2$, the mean is $2+(0-2)/2=1$. The variance is

\frac{2^2}{2\cdot1}\left(1-e^{-2\ln2}\right) =2\left(1-\frac14\right) =\frac32.

The standard deviation is $\sqrt{3/2}$. Subtracting the mean and dividing by this standard deviation turns the lower and upper endpoints into $-1$ and $1$, respectively. Thus

\mathbb{P}\left(1-\sqrt{\frac32}\le X_t\le1+\sqrt{\frac32}\mid\mathcal{F}_0\right) =\Phi(1)-\Phi(-1) =2\Phi(1)-1 \approx0.6827.

The endpoints do not affect the value because a nondegenerate Gaussian distribution assigns probability zero to each endpoint.

Check: the interval is centered at the computed mean $1$, and its half-width is exactly the computed standard deviation $\sqrt{3/2}$. Its standardized endpoints are therefore precisely $-1$ and $1$.

Composition of Transition Kernels

The transition formulas are consistent across successive time intervals. If the process starts at $x$, let $P_h(x,\cdot)$ denote its law after elapsed time $h$. For $\sigma\ne0$, this is the Gaussian kernel with mean $\theta+(x-\theta)e^{-\kappa h}$ and variance $\frac{\sigma^2}{2\kappa}(1-e^{-2\kappa h})$. The next result records how two such transitions combine.

Proposition: For $h,k\ge0$, the transition kernel satisfies $$ P_hP_k(x,\cdot)=P_{h+k}(x,\cdot). $$ This holds for every $x\in\mathbb{R}$, including the point-mass cases $h=0$, $k=0$, or $\sigma=0$.

Proof. First suppose $h,k>0$ and $\sigma\ne0$. Write $q_h=e^{-\kappa h}$ and $q_k=e^{-\kappa k}$. Under the first transition the intermediate state $Y$ has mean $\theta+(x-\theta)q_h$ and variance $v_h=\frac{\sigma^2}{2\kappa}(1-q_h^2)$. Given $Y$, the second transition has mean $\theta+(Y-\theta)q_k$ and variance $v_k=\frac{\sigma^2}{2\kappa}(1-q_k^2)$. The second-stage innovation is independent of the first-stage state, by independent Brownian increments. Consequently the final mean is

\theta+\big(\theta+(x-\theta)q_h-\theta\big)q_k =\theta+(x-\theta)q_hq_k =\theta+(x-\theta)e^{-\kappa(h+k)}.

The final variance is the variance of $q_kY$ plus the second-stage variance:

q_k^2v_h+v_k =\frac{\sigma^2}{2\kappa}\left(q_k^2(1-q_h^2)+(1-q_k^2)\right) =\frac{\sigma^2}{2\kappa}(1-q_h^2q_k^2) =\frac{\sigma^2}{2\kappa}\left(1-e^{-2\kappa(h+k)}\right).

A sum of independent Gaussian variables is Gaussian, so these matching mean and variance identify $P_{h+k}(x,\cdot)$. If $h=0$ or $k=0$, the corresponding transition leaves the state unchanged, so the identity follows directly. If $\sigma=0$, both transitions are deterministic contractions toward $\theta$; composing them gives $\theta+(x-\theta)e^{-\kappa h}e^{-\kappa k}=\theta+(x-\theta)e^{-\kappa(h+k)}$. This proves all cases. $\square$

Common Mistake: Forgetting the Squared Decay

A frequent error is to calculate the conditional variance using the decay factor $e^{-\kappa(t-r)}$ only once, as though variances accumulated linearly in the integrand. The Ito isometry squares the integrand. The relevant integral is $\sigma^2\int_s^t e^{-2\kappa(t-r)}\,dr$, not $\sigma^2\int_s^t e^{-\kappa(t-r)}\,dr$.

For instance, with $\kappa=1$, $\sigma=1$, and $t-s=\ln2$, the incorrect unsquared calculation would give $\int_0^{\ln2}e^{-u}\,du=1/2$. The correct variance is

\int_0^{\ln2}e^{-2u}\,du =\frac{1-e^{-2\ln2}}{2} =\frac{1-\tfrac14}{2} =\frac38.

The distinction matters because each shock at time $r$ is multiplied by $e^{-\kappa(t-r)}$ in the future state, and its contribution to variance is the square of that multiplier. The conditional mean uses the single decay factor $e^{-\kappa(t-s)}$; the conditional variance uses the squared factor inside its time integral.

The transition law is useful whenever a conditional probability or expectation over a future state is needed: it replaces a path-dependent SDE calculation with an ordinary Gaussian calculation. In particular, one can evaluate threshold probabilities by standardizing with $m_{s,t}$ and $\sqrt{v_{s,t}}$, provided $t>s$ and $\sigma\ne0$. The deterministic cases must instead be tested directly against the point mass.

Check Your Understanding

Use the transition formulas and the conditions under which a Gaussian density exists.

For $\kappa=2$, $\theta=0$, $\sigma=1$, $X_s=3$, and $t-s=\ln2$, compute the conditional mean and variance of $X_t$.
For a nondegenerate transition, express $\mathbb{P}(X_t\le a\mid\mathcal{F}_s)$ using $\Phi$, $m_{s,t}$, and $v_{s,t}$.
Explain why the transition law is a point mass when $\sigma=0$, even though its mean formula remains valid.
Verify the composition of conditional means for elapsed times $h$ and $k$ by multiplying their decay factors.
For $\kappa=1$, $\sigma=2$, and $t-s=\ln2$, compute the conditional variance and compare it with the result obtained by incorrectly omitting the square on the exponential integrand.