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I have a simple question about moment generating functions(MGFs).

Does the interval on which a MGF is defined corresponds to the support of the random variable?

For instance, considering a standard logistic density function

$f(x)= \frac{e^{x}}{(1+e^x)^2}$

where $x \in(-\infty,+\infty)$. Does the interval on which the MGF of $x$ is defined corresponds to $\mathbb{R}$?

Maximilian
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    In the example you give, the mgf is $\phi(t)=E[\exp(tX)]$ which is obtained by integrating $\exp(tx)$ against $f(x).$ What do you get for the integral when $t\ge 1$? – whuber Jan 24 '23 at 21:28
  • To be honest, I'm not sure about the answer – Maximilian Jan 24 '23 at 21:49
  • Use the fact that for large enough $x$ and any $t\ge 1,$ $\exp(x)f(x) \ge 1/2.$ $x\ge 1$ works. (If that's not obvious, graph this function.) Thus the mgf must exceed $\int_1^\infty (1/2),\mathrm dx,$ which obviously is infinite. Consequently the mgf cannot be defined for $t\ge 1.$ How does this compare to the support of $f$? – whuber Jan 24 '23 at 21:53
  • I see the function takes values $\ge$ 1/2 for large $x$ and $t \ge 1$. Why are you saying the mgf must exceed the integral you just wrote? – Maximilian Jan 24 '23 at 22:20
  • Because the mgf is the area under a higher curve. – whuber Jan 25 '23 at 00:14
  • I see. What’s the connection with the support of $x$? Could you also provide another example in an answer so that I can vote it – Maximilian Jan 25 '23 at 07:41
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    There is no connection with the support except in some simplistic ways, such as bounded support trivially guarantees the mgf is defined everywhere. The deeper connection is that the domain of the mgf tells us about how the probability behaves "at infinity." The mgf is nothing other than the Laplace transform, about which you can find a great deal of literature dating back several hundred years if you are curious about its properties. – whuber Jan 25 '23 at 12:06

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