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Feb 20, 2018 at 19:22 comment added Dan Piponi It might be worth mentioning that many of the PDFs that arise in applications form exponential families. So computing the expected value of $\exp(tx)$ requires adding a simple $tx$ term into an already existing exponential and you find you have an integral that is no more difficult than proving the PDF itself integrates to 1. This makes the computing the MGF surprisingly tractable. Typically this is not the case when dealing with $(1-xt)^{-1}$. The former is also more likely to converge.
Aug 1, 2010 at 14:39 history answered Chris Godsil CC BY-SA 2.5