$\operatorname{E}[e^{Xt}]$ is the moment generating function of $X$, which is either a discrete or continuous random variable. Is $\operatorname{E}[X]\in\mathbb{R}\Rightarrow \operatorname{E}[e^{Xt}] \in\mathbb{R}$ correct? If not, are there any circumstances under which it is correct?
$\operatorname{E}[X]\in\mathbb{R}\Rightarrow \operatorname{E}[e^{Xt}] \in\mathbb{R}$?
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probability-theory
random-variables
expectation
moment-generating-functions
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1When you say "$\in\mathbb{R}$", do you mean to emphasise that it's finite, or non-complex? – 2017-01-07
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0That it is finite. – 2017-01-07
2 Answers
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Nope: Let $X$ have pdf $x^{-3}\mathbb{I}[|x|>1]$. Then $X$ has mean $0$, but $\mathbb{E}[\exp(tX)]=\infty$ for all $t\in\mathbb{R}-0$.
In general, it's hard to guarantee the existence of $\mathbb{E}[\exp(tX)]$ - one can have finite moments of arbitrarily large order without necessarily having the MGF finite for any nonzero $t$. One can even have moments of all orders without the MGF existing for any nonzero $t$ - consider the lognormal distribution as described in Mark's answer elsewhere on this page.
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0You can, the LogNormal does. – 2017-01-07
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0@Mark thanks - saw your answer come up just as I finished editing. will make a note. +1 – 2017-01-07
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This may be a good resource on mgf's.
The short answer is no, and a specific counterexample is the log-normal distribution.
We say that $X$ is Log-Normal if $X = e^Y$ where $Y$ is normally distributed. It can be shown that: $$E[X^n] = e^{n\mu+\frac{1}{2}n^2\sigma^2}$$ Despite this, $E[e^{tX}]$ diverges for any positive $t$, so the mgf is undefined despite all moments being defined (and finite).