Prove that $\inf{[g(t): t \ge 0]} \le 1$ where $g(t)=\mathbb E(e^{tX})$.

Question

For $g(t)=\mathbb E(e^{tX})$ prove that $\inf{[g(t): t \ge 0]} \le 1$.

I was trying to expand $\mathbb E(e^{tx})$ like $$\mathbb E\left(1+tX+\frac{t^2 X^2}{2!}+\dots\right)= 1+t\mathbb E(X)+ \frac{t^2}{2}\mathbb E(X^2)+\dots$$ But cannot proceed after this. Please Help.

Answer 1

By definition of infremum, which is a lower bound of the given set, we have

$\inf\{g(t):t\ge 0\}\le g(0)=E(e^{0})=1$.

Answer 2

If you look at the expansion, you know $ t = 0$ is allowed. And $t = 0$ gives $1$. There might be another $t$ making the $E[e^{Xt}]$ smaller, and $\inf$ gives the $t$ making it as small as possible, thus you know for certain that it must be less or equal than $1$.