Solution: Ornstein-Uhlenbeck via Itô

Exercise: Ornstein-Uhlenbeck via Itô: solve $dX_t = -\kappa X_t\,dt + \sigma\,dW_t$

Part 1

With $V(t, X) = e^{\kappa t} X$ : $V_t = \kappa e^{\kappa t}X$ , $V_X = e^{\kappa t}$ , $V_{XX} = 0$ . The coefficients of the SDE are $a = -\kappa X_t$ and $b = \sigma$ . Itô's lemma:

d(e^{\kappa t}X_t) = \left(\kappa e^{\kappa t}X_t + (-\kappa X_t)\cdot e^{\kappa t} + \tfrac{1}{2}\sigma^2\cdot 0\right)dt + \sigma\cdot e^{\kappa t}\,dW_t

The drift is $\kappa e^{\kappa t}X_t - \kappa e^{\kappa t}X_t = 0$ . So:

d(e^{\kappa t}X_t) = \sigma e^{\kappa t}\,dW_t

Part 2

Integrating both sides from $0$ to $t$ :

e^{\kappa t}X_t - X_0 = \sigma\int_0^t e^{\kappa s}\,dW_s

Multiplying through by $e^{-\kappa t}$ :

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

The stochastic-integral integrand is $\sigma e^{-\kappa(t-s)}$ — an exponentially-decaying weight of past Brownian innovations.

Part 3

The integral $\int_0^t e^{-\kappa(t-s)}\,dW_s$ is a Gaussian (it is a Wiener integral of a deterministic function), so $X_t$ is Gaussian with mean $x_0 e^{-\kappa t}$ . Itô isometry:

\operatorname{Var}(X_t) = \sigma^2\int_0^t e^{-2\kappa(t-s)}\,ds = \sigma^2\cdot\frac{1 - e^{-2\kappa t}}{2\kappa}

t \to \infty

e^{-2\kappa t} \to 0

and

\operatorname{Var}(X_t) \to \sigma^2/(2\kappa)

. The process has a stationary distribution

\mathcal{N}(0,\,\sigma^2/(2\kappa))

: mean-reversion (

-\kappa X\,dt

) prevents the variance from growing without bound.

Part 4

For the ODE

\dot x = -\kappa x + f(t)

, the integrating factor is

e^{\kappa t}

; multiplying gives

(e^{\kappa t}x)' = e^{\kappa t}f(t)

, which integrates directly. The OU SDE is exactly this, with

f(t)\,dt

replaced by

\sigma\,dW_t

. Itô's lemma reproduces the integrating-factor calculation without generating a correction term precisely because

V = e^{\kappa t}X

is linear in

X

, so

V_{XX} = 0

. Linear-in-state transformations of an Itô SDE are Itô-neutral — the stochastic chain rule agrees with the ordinary one for them.

Alternative approach

A second route is to note that

X_t

is a Gaussian process (it is a deterministic-coefficient linear SDE driven by Brownian motion). Because it is Gaussian, it is characterised by its mean function

m(t) = \mathbb{E}[X_t]

and covariance function

c(s, t) = \operatorname{Cov}(X_s, X_t)

. Taking expectations in the SDE gives

m' = -\kappa m

, so

m(t) = x_0 e^{-\kappa t}

. A similar argument on

d(X^2)

— which does pick up an Itô correction

\sigma^2\,dt

— gives the variance ODE directly. Both routes confirm the same answer.

Takeaways

The integrating-factor trick carries over from ODE to SDE whenever the transformation is linear in state. Linear means no Itô correction.
Mean-reversion ( $\kappa > 0$ ) produces a stationary variance. Without it (e.g. Brownian motion itself, $\kappa = 0$ ) the variance grows linearly in $t$ and there is no stationary distribution.
The OU process is Gaussian. This makes calibration and simulation easy — all you need are the two closed-form moments from parts 2 and 3.
Vasicek short-rate model = OU process with a non-zero mean level. Adding a constant drift $\kappa\theta$ shifts the stationary mean to $\theta$ ; the variance is unchanged. Every OU-based short-rate model inherits these properties.