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.