I would like to determine the conditional expectation of $\mathbb{E}[\exp(z_{t+\Delta t})|z_t]$ and $\mathbb{E}[\exp(y_{t+\Delta t})|z_t]$. $x=(y,z)$ is a 2-dimensional Brownian motion which is normally distributed with parameters $(0,tI)$.

Are this followings true?

Because of the fact, that $x$ is a 2-dimensional Brownian motion, I know that $y, z$ are one dimensional Brownian motions and y and z are independent.

Therefore $\mathbb{E}[\exp(y_{t+\Delta t})|z_t]$=$\mathbb{E}[\exp(y_{t+\Delta t})]=\exp((t+\Delta t)^2/2)$.

$\mathbb{E}[\exp(z_{t+\Delta t})|z_t]=\mathbb{E}[\exp(z_{t+\Delta t}-z_t+z_t)|z_t]=\mathbb{E}[\exp(z_{t+\Delta t}-z_t)*\exp(z_t)|z_t]=\exp(z_t)*\mathbb{E}[\exp(z_{t+\Delta t}-z_t)]=\exp(z_t)*\exp((\Delta t)^2/2)$.

Thanks for you help

  • $\begingroup$ Yes and yes. $ $ $\endgroup$ – Did May 8 '17 at 19:47
  • $\begingroup$ @Did thanks a lot $\endgroup$ – Mathfreak May 8 '17 at 20:53
  • $\begingroup$ You can make your solution into an answer and accept it :) $\endgroup$ – user3658307 Aug 3 '17 at 6:31

the first and the second identity are true, because of the independency.


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