Exponential of a standard normal random variable We know that $Z\sim N(0, 1)$. How do I prove that $e^Z$ has a mean of $e^{0.5}$? I have tried integrating $e^z$ times the pdf of $Z$ but I don't know why it isn't working out. 
Also what is the pdf of $\exp(Z)$?
 A: As @Glen_b mentions for self-study problems please show an attempt. Here is how you would get started:
$\begin{align*}
\text{E}\left[e^Z\right] &= \dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^z e^{-z^2/2}dz \\
&= \dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{-z^2/2 + z}dz \\
&= \dfrac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty e^{\tfrac{-1}{2}(z^2 - 2z)}dz \\
&=
\end{align*}$ 
Now try completeing the square in the exponential so you get an integral that looks like it is the PDF of a normal distribution with known mean and variance. The answer will be $e^{1/2}$
A: 
How do I prove that $e^Z$ has a mean of $e^{0.5}$? 

You can do it via integration, it's quite straightforward - you combine the $\exp$ terms into one exponent, complete the square and identify the density (which integrates to 1) leaving just a constant out the front.
An alternative is to do it via the MGF, which makes it trivial.

Also what is the pdf of $\exp(Z)$?

You just use straight up change of variable, keeping in mind the Jacobian. Or you can do it from first principles (Let $Y=\exp(Z)$ then $P(Y\leq y) = P(\exp(Z)\leq y) = ...$)
You get a particular case of the lognormal density.
A: Remember that the moment generating function of a normal random variable $Z\sim\textrm{N}(\mu,\sigma^2)$ is $\varphi_Z(t)=\mathrm{E}[e^{tZ}]=e^{\mu t + \sigma^2 t^2/2}$, and use that $\mathrm{E}[e^Z]=\varphi_Z(1)$.
Expanding a little bit
The following proposition, motivated by Huber's comment bellow, shows how this technique extends to the multivariate case.
Propositon. Let $(Z_1,\dots,Z_k)$ be a normal random vector with mean vector $m$ and covariance matrix $\Sigma=(\sigma_{ij})$. For the lognormal random vector $(U_1,\dots,U_k)=(e^{Z_1},\dots,e^{Z_k})$ we have $\mathrm{E}[U_i]=e^{m_i +\frac{1}{2}\sigma_{ii}}$ and $\mathrm{Cov}[U_i,U_j]=\mathrm{E}[U_i]\,\mathrm{E}[U_j](e^{\sigma_{ij}}-1)$, for $i,j=1,\dots,k$.
Proof. For $a=(a_1,\dots,a_k)^\top\in\mathbb{R}^k$, we have
$$
\mathrm{E}\!\left[\prod_{i=1}^k U_i^{a_i} \right] = \mathrm{E}\!\left[\prod_{i=1}^k e^{a_i Z_i} \right] = \mathrm{E}\!\left[e^{a^\top Z}\right] = e^{a^\top m +\frac{1}{2}a^\top\Sigma\,a} \, , \quad (*)
$$
in which we used the expression of the moment generating function of $(Z_1,\dots,Z_k)$. For $i=1,\dots,k$, choose $a\in\mathbb{R}^k$ such that its $i$-th component is equal to one and the other components are equal to zero. It follows from $(*)$ that $\mathrm{E}[U_i]=e^{m_i +\frac{1}{2}\sigma_{ii}}$. Now, for $i,j=1,\dots,k$, take $a\in\mathbb{R}^k$ such that its $i$-th and $j$-th components are equal to one and the others are equal to zero. It follows from $(*)$ that
$$
  \mathrm{E}[U_i\,U_j]=e^{m_i+m_j+\frac{1}{2}\sigma_{ii}+\sigma_{ij}+\frac{1}{2}\sigma_{jj}}=\mathrm{E}[U_i]\,\mathrm{E}[U_j]\,e^{\sigma_{ij}}
$$
and
$$
  \mathrm{Cov}[U_i,U_j]=\mathrm{E}[U_i\,U_j] - \mathrm{E}[U_i]\,\mathrm{E}[U_j] =\mathrm{E}[U_i]\,\mathrm{E}[U_j](e^{\sigma_{ij}}-1) \, .
$$
