Find the expectation and covariance of a stochastic process The problem is:
Let $W(t)$, $t ≥ 0$, be a standard Wiener process. Define a new stochastic process $Z(t)$ as $Z(t)=e^{W(t)-(1/2)\cdot t}$, $t≥ 0$. Show that $\mathbb{E}[Z(t)] = 1$ and use this result to compute the covariance function of $Z(t)$. 
I wonder how to compute and start with the expectation cause it is not any case with a formula to use. Any hint would be welcome. Thanks in advance!
 A: 1. Expectation
For showing that $E[Z(t)]=1$, you can apply some theorems regarding the log-normal distribution. A random variable $Y$ is log-normally distributed if $\log(Y)$ is normally distributed. This applies to your case because $\log[Z(t)] \sim N(- \frac{1}{2} t, t)$.  So, $Z(t)$ is log-normal. At first rewrite $W(t) - \frac{1}{2} t $ in the form of a standard normally distributed random variable:
$W(t) - \frac{1}{2} t  = \sqrt{t}X - \frac{1}{2} t $, with $X \sim N(0,1)$.
Then, you can apply the lemma that for a log-normally distributed random variable $Y = e^{\mu + X \sigma}$ the expectation is $E[Y] = e^{\mu + \frac{1}{2}\sigma^2} $.
In this case $\mu = - \frac{1}{2} t $ and $\sigma^2 = t$.
After replacing the parameters, one gets $E[Z(t)] = e^{- \frac{1}{2}t  + \frac{1}{2} t} = 1$.
2. Covariance
Now a starting point in thinking about the covariance is:
$Cov[Z(t),Z(s)] = E[Z(t)Z(s)] - E[Z(t)]E[Z(s)] =  E[Z(t)Z(s)] - 1.$
Further, use the property of the Wiener processes: $Cov[W(t),W(s)] = s$ for $s < t$. One finds also 
$\log[Z(t)Z(s)] = (W(t)+W(s)-\frac{t}{2} - \frac{s}{2})  \sim N(- \frac{t}{2} - \frac{s}{2}, t+3s)$. So, the product $Z(t)Z(s)$ is again a log-normally distributed random variable. Applying the theorem for expectation of log-normal random variables yields
$Cov[Z(t),Z(s)] =  E[Z(t)Z(s)] - 1= e^{- \frac{t}{2} - \frac{s}{2} + \frac{1}{2}(t + 3s) } -1 = e^s-1.$
Please note, it must hold $0 ≤ s < t$.
Appendix: Proof for normality of transformed process (requested in comment): 
Taking $\log$ of something cancels $e$: $\log (e^{W(t) - \frac{1}{2} t} ) = W(t) - \frac{1}{2} t $
Since we know $W(t) \sim N(0,t)$ and any linear transformation of normally distributed random variable is again normally distributed, we only need to find the expectation and variance of $(W(t) - \frac{1}{2} t)$ in order to characterize the distribution of it.
$E[W(t) - \frac{1}{2} t] = E[W(t)] - \frac{1}{2} t  = - \frac{1}{2} t$
$V[W(t) - \frac{1}{2} t] = V[W(t)]  = t$.
