# Moment generating function of standard normal

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?

## 1 Answer

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].

• Thank you! By the way...I was just wondering, when do we have issues regarding the negativity of $t$? Commented Jan 20, 2021 at 9:42
• 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 Commented Jan 20, 2021 at 9:49
• Thank you, that was really helpful! Commented Jan 20, 2021 at 12:58