Transition Distributions for Brownian Motion and Ornstein–Uhlenbeck Processes

Summary

The document asks for the transition distributions of two continuous-time processes: Brownian motion with constant drift and an Ornstein–Uhlenbeck process that reverts toward a long-run mean. Its answer derives the Ornstein–Uhlenbeck solution from an initial value, then uses the stochastic integral representation to obtain the conditional mean and variance. Because the process is Gaussian, these moments characterize its transition law as a normal distribution.

For the mean-reverting process, the conditional mean moves exponentially toward the long-run level, while the conditional variance grows toward a finite limit determined by volatility and the reversion rate. The answer does not derive the constant-drift Brownian transition law, despite the question asking about both processes. Its displayed density also appears inconsistent with the stated normal distribution, so readers should verify the density’s exponent and normalization before using it. The derivation is a useful outline, with a formula caveat.

Key ideas

• A process’s transition distribution describes its future value conditional on its current value.
• Brownian motion with constant drift has a Gaussian transition law whose mean and variance depend on elapsed time.
• The Ornstein–Uhlenbeck process has a conditional mean that decays toward its long-run level.
• The Ornstein–Uhlenbeck conditional variance approaches a finite long-run value.
• Check the displayed density against the stated normal distribution before relying on its formula.

Tags

# How to find the transition distribution functions of these two processes?


# How to find the transition distribution functions of these two processes?












> This question was asked by another user, but was deleted. As it may be useful for others, I re-post it here.

What are the transition distribution (or density) functions of two processes defined by \begin{align*} dX_t = \mu dt + \sigma dW_t \end{align*} and \begin{align*} dX_t = \theta(\mu-X_t) dt + \sigma dW_t, \end{align*} where $\theta>0$, $\mu$ is a real number, $\sigma >0$, and $\{W_t,\, t \ge 0\}$ is a standard Brownian motion.

## Answer by olaker (score 2, accepted)

https://quant.stackexchange.com/a/32443

Here is a derivation for the Ornstein-Uhlenbeck process. Solution to the SDE $$dX_t = \theta(\mu-X_t) dt + \sigma dW_t$$ subject to the initial condition $X_0=x$ has the form $$X_t= \mu + (x - \mu)e^{-\theta t} + \sigma\int_0^t e^{-\theta (t-s)}dW_s.\qquad$$ We need to calculate density function $p(t,x,y)$ of the conditional distribution $(X_t|X_0=x)$.

$X_t$ is normally distributed for each $t>0$. The conditional expectation is $$E[X_t|X_0=x]=\mu + (x - \mu)e^{-\theta t}.$$ The conditional variance is $$Var[X_t|X_0=x] = \sigma^2E\left[(\int_0^t e^{-\theta (t-s)}dW_s)^2 \right] = \sigma^2 E\left[\int_0^t e^{-2\theta (t-s)}ds \right]$$ $$=\frac{\sigma^2}{2\theta}(1-e^{-2\theta t}).$$ Hence, we have that $$(X_t|X_0=x)\sim N\left(\mu + (x - \mu)e^{-\theta t},\frac{\sigma^2}{2\theta}(1-e^{-2\theta t})\right),$$ and, finally, $$p(t,x,y)=\frac{1}{\sqrt{\pi\sigma^2(1-e^{-2\theta t})/\theta}}\exp \left[-\frac{(-y-\mu-(x - \mu)e^{-\theta t})^2}{\sigma^2(1-e^{-2\theta t})/\theta}\right].$$

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