conditional expectation exponential function of Brownian motion

Question

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

Answer 1

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