# Derivation of an alternative representation of the Ornstein-Uhlenbeck process

The Ornstein-Uhlenbeck process can be defined as:

$X_t = e^{-\lambda t} \left( X_0 + \int_0^t e^{\lambda s} dB_s \right)$

where $\lambda > 0$ and $\{ B_t \}_{t \geq 0}$ is the standard Brownian motion. The process has an alternative represenation as follows:

$X_t = \frac{1}{\sqrt{2\lambda}} e^{\lambda t} \hat{B}_{e^{-2 \lambda t}}$

The formula can be found in "Aspects of Brownian Motion" by R. Mansuy and M. Yor. It is also mentioned in Wikipedia in a slightly different form, where one should let $\mu = 0$, $\sigma = 1$, and $\theta = \lambda$.

The question is how to derive this representation. I know that it should be done by means of the It$\hat{\text{o}}$ formula, but cannot get it till the end.

Thank you.

Regards, Ivan

UPDATE: It turned out that I had missed one condition that $X_0 \sim \mathcal{N}(0, \frac{1}{2 \lambda})$. In this case, it becomes quite straight-forward to check the desired representation (the mean and variance as mentioned in the comments by @mike and @Sasha). Thanks for the answers.

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when dealing with gaussian process mean and covariance tell the whole story – mike May 2 '12 at 11:02

Assuming $X_0$ is a fixed constant, $X_t$ is a Gaussian process, meaning that $(X_{t_1}, X_{t_2}, \ldots, X_{t_n})$ is a Gaussian random vector for arbitrary $n \in \mathbb{N}$ and arbitrary distinct times $t_k$.
Distribution of such a vector is determined by vector mean and the covariance matrix. These are obtained from Ito isometry and the property that expectation of Ito integral is zero: $$\mathbb{E}(X_t) = X_0 \mathrm{e}^{-\lambda t} \qquad \mathbb{Cov}(X_t, X_s) = \mathrm{e}^{-\lambda(t+s)} \int_0^{\min(t,s)} \mathrm{e}^{2\lambda u} \mathrm{d} u = \mathrm{e}^{-\lambda(t+s)} \frac{\exp(2 \lambda \min(t,s))-1}{2 \lambda}$$
Now observe that $Y_t = \mathrm{e}^{-\lambda t} \left( X_0 + \frac{1}{\sqrt{2 \lambda}} \hat{B}_{\exp(2 \lambda t)-1}\right)$ is another Gaussian process, whose mean function and covariance function are exactly the same, hence $X_t$ and $Y_t$ are equal in distribution.
Sasha, thanks for the detailed answer. I just don't see how the expectation and covariance of $X_t = \frac{1}{\sqrt{2\lambda}} e^{\lambda t} \hat{B}_{e^{-2 \lambda t}}$ become what you wrote. Actually, in the book there are two equal representations: $X_t = \frac{1}{\sqrt{2 \lambda}} e^{-\lambda t} \tilde{B}_{e^{2 \lambda t}} = \frac{1}{\sqrt{2 \lambda}} e^{\lambda t} \hat{B}_{e^{-2 \lambda t}}$. I'm interested in the later one. Probably, the part $e^{-\lambda t} X_0$ is missing in those representations, but what about the covariance? Thank you. – Ivan May 3 '12 at 8:50