Does a prediction interval have to contain the mean? I am having a huge problem with a conceptual problem that I came up with.
Say a company has a distribution that is highly skewed. Something similar to an exponential or lognormal only more extreme.  Now pretend the distribution is so skewed that the mean of the distribution is higher than the 99% Percentile of the distribution.  (Aka 1-2 extreme higher values caused the mean to be extremely high compared to the rest of distribution).
By definition, if this distribution was used to forecast a future value (aka a random sample from the distribution) would it be true that mean would not be in the 95% Prediction interval?
In my brain, a 95% predition interval is a range that 95% of all future values will fall between. For any distribution this should exactly equal the .025 Percentile on the lower bound, and the .975 percentile on the upper bound... If the mean is higher than the .975 Percentile, then the mean would not be within the '95% prediction interval'.
Am I thinking of this incorrectly?  It seems strange to report a forecast as 


*

*Mean Forecasted Value:  6,000,0000

*95% Prediction Interval:   [400,5000].

 A: No, a prediction interval need not contain the mean.  I think some of your confusion might be mixing prediction intervals and confidence intervals.  While the goal of a prediction interval is to contain with some certainty future values of the random variable, the goal of a confidence interval is to contain the true mean of distribution.
As you mentioned in highly skewed distributions these ideas seem to be at odds with each other.  The important thing is to recognize the value in each of the statistics provided.
The predictive value of the mean is:
1) Cumulative: As more samples come in, their average will tend toward the true mean.  So if the cumulative value is of interest (for instance, if you're gambling and dealing with winnings or losses you're interested in cumulative effects) then the mean is very useful.
2) Minimizes Squared Residuals: While squared residuals are a somewhat arbitrary quantity of interest it is worthwhile to know what your prediction is minimizing.
If however your goal is to minimize the absolute error in your predictions, the mean forecasted value of 6,000,000 is not what I would go with.
A: Consider the distribution of possible returns in the St Petersburg paradox: 
Prob(1)=1/2
Prob(2)=1/4
Prob(4)=1/8
...
Prob(2^n)=1/2^(n+1)
The mean diverges and is outside of any reasonable prediction interval.  (The median is 1 in this case, but I don't know what I'd use for my point forecast.  Maybe Stephan Kolassa, see above, has a suggestion.)
There's another complication:  Let's say you want a 95% prediction interval for some distribution (other than the one I just mentioned).  Do you go from the 2.5%tile to the 97.5%tile or the 0 to the 95th or the 5th to the 100th or....?  The answer probably depends on why you are asking the question.  
