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As far as I know, given a random variable $X$, we define its moment generating function as $$M_X(t) = \mathbb{E} \left[ e^{ tX} \right] \ \ , \ t \in \mathbb{R}$$ I read that MGF for a general random variable may not be defined for negative values of $t$. What about if $X \sim N(0,1)$? Can we be sure its MGF is well defined for negative t?

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For $X \sim N(0,1)$ we have $Ee^{tX}=e^{t^{2}/2}$ for all real numbers $t$. One way to prove this is to use the characteristic function and use a basic result from Complex Analysis [The Identity Theorem].

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  • $\begingroup$ Thank you! By the way...I was just wondering, when do we have issues regarding the negativity of $t$? $\endgroup$ Commented Jan 20, 2021 at 9:42
  • $\begingroup$ If $f(x) =\frac 1 {x^{2}}$ for $x <-1$ and $0$ for $x \geq -1$ then $f$ is a density function. The MGF exists in this case if and only if $t\geq 0$. @YodaAndFriends $\endgroup$ Commented Jan 20, 2021 at 9:49
  • $\begingroup$ Thank you, that was really helpful! $\endgroup$ Commented Jan 20, 2021 at 12:58

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