Why does $E\exp(\theta X_i)<\infty$ imply that $EX_i^+<\infty$

Question

Why does $E\exp(\theta X_i)<\infty$ imply that $EX_i^+<\infty$ for $\theta>0$?

Is it because of: $\exp(\theta X_i)\ge|1\pm\theta X_i|$, on $\theta X_i>-1$

So then $E\exp(\theta X_i)\ge E|1+\theta X_i|,\quad E\exp(\theta X_i)\ge E|1-\theta X_i|$ and both imply $E|\theta X_i|<\infty$ on $\theta X_i>-1$ do you agree ?

Answer 1

Let $x$ be a real number.

1. If $x < 0$, then $\theta \cdot x^+ = 0 \leq e^{\theta x}$

2. If $x \geq 0$, then using $e^y \geq 1+y$ for all $y$, we get $\theta x^+ \leq 1 + \theta x^+ \leq e^{\theta x^+} = e^{\theta x}$.

We use this to derive $$ \Bbb{E}(\theta \cdot X_i^+ ) \leq \Bbb{E} e^{\theta X_i} < \infty, $$ which easily implies $\Bbb{E}(X_i^+) < \infty$.