What is the expectation, E(log(x-a)), when x is log normally distributed? Also x-a>0. I am looking for analytical solution or good numerical approximation.
The random variable [itex]X[/itex] is said to be log-normally distributed if [itex]\log X[/itex] is normally distributed (I know, it's a weird naming convention). In other words, [itex]X= e^Z[/itex], where [itex]Z\sim \mathcal N(\mu_Z,\sigma_Z^2)[/itex], a normal random variable. So then [itex]\mathbb E [\log X]= E[Z] = \mu_Z[/itex].