How can the formula for the expectation of a log-normal random variable be dimensionally sound? If $X\sim\mathcal{LN}({\mu,\sigma^2})$, then $\mathrm{E}[X]=e^{\mu+\sigma^2/2}$.  My question is: what right do we have to add a mean and variance together?  If $X$ has physical dimensions, then the expression $\mu + \sigma^2/2$ is incoherent.  So what gives? 
 A: Tongue in cheek: this sum is allowed only in free countries where this is actually considered as a basic human right.
Seriously: you are not adding a mean and a variance since $\mu$ is not the mean and $\sigma^2$ is not the variance of a lognormal variate. The mean is, as you said $\mathrm{E}[X]=e^{\mu+\sigma^2/2}$, and the variance is $\mathrm{Var}[X]=(e^{\sigma^2}-1)e^{2\mu+\sigma^2}$. Then $\mathrm{E}[X]$ and $\sqrt{\mathrm{Var}[X]}$ have the same units. You can read more on this in the following link:
http://en.wikipedia.org/wiki/Variance#Units_of_measurement
A: I believe that $\mu$ and $\sigma$ don't have the same units in this case, but $\mu$ and $\sigma^2$ do.
Take for example the stochastic differential equation of a geometric brownian motion used to model stock prices in the Black–Scholes model:
$ dS_t = \mu S_t\,dt + \sigma S_t\,dW_t $
The stock price $S_t$ follows a log-normally distributed random variable.
Call $D_i$ the unit used to measure the variable $i$. Dimensional homogeneity implies that $D_\mu D_t = 1$, so if your measuring the time in days, the unit of $\mu$ is $1/day$.
Dimensional homogeneity also implies that $D_\sigma D_{W_t} = 1$. $W_t$ is a Wiener process with variance $t$, so $D_{W_t}^2=D_t$. Therefore, $D_\sigma=1/\sqrt{D_t}$. If you are measuring the time in days, the unit of $\sigma$ is $1/\sqrt{day}$.
Note that this interpretation is consistent with the formula used to convert daily volatility $\sigma_{daily}$ to annualized volatility $\sigma_{annual}$, assuming that a year has 252 trading days: $\sigma_{annual} = \sigma_{daily}\sqrt{252}$.
