2

For a Brownian motion $(X_t)$ where $0≤t≤1$ define

$(Y_t)=e^{-t}X_{e^{2t}}$ for $t\in R$

What is the distribution of $Y_t$ for a given $t \in R$?

It should be distributed like an Ornstein-Uhlenbeck process but I have no clue why and how to show it. Any answers or comments are more than welcome!

Thanks in advance

Jynne94.
  • 21
  • 1
  • Well, normal distribution has the following scaling property: if $X$ is a random variable with $X \sim \mathcal{N}(0, \sigma)$, then $aX \sim \mathcal{N}(0, a^2 \sigma)$. Now, what can we conclude from this, knowing that $X_{e^{2t}} \sim \mathcal{N}(0, e^{2t})$? – Sangchul Lee Jan 25 '17 at 17:08
  • Do you understand why $Y_t$ is an Ornstein-Uhlenbeck process? – Conrado Costa Aug 01 '17 at 10:10

0 Answers0