Lemma 2.7.2 in probability theory and examples by Rick Durrett Part of the proof is in the following.


Here are my confusions:

*

*I would like to know why it's necessary to prove (i) $\kappa$ is continuous at 0.

*$\theta_- = \inf\{\theta : E(e^{\theta X}) < \infty\}$. From the definition, the RHS of (*) should not be $L^1$. Therefore, I think dominated convergence theorem doesn't work here.

 A: I think you need continuity to ensure that the first derivative $\kappa'(\theta)$ exists.
I don't really get your second question: which definition are you referring to? Anyway the function $1 + e^{\theta_0 x}$ is $L^1$ in fact, at least for $\theta > 0$ we have:
$$
\mathbb{E}[1 + e^{\theta_0 X}] = 1 + \mathbb{E}[e^{\theta_0 X}] < 1 + \mathbb{E}[e^{\theta_{\text{-}} X}] < \infty
$$
A: *

*Durrett is aiming to show (i)-(iii) to prove the Lemma. Assume we have (i)-(iii), let's try to conclude the statement of the Lemma:

By (ii), we can write for $0<\epsilon \leq\theta<\theta_+$ (as we only know differentiability inside the interval $(0, \theta_0)$) that
$$
\int_\epsilon^\theta \kappa'(x) \textrm{d}x = \kappa(\theta)-\kappa(\epsilon).
$$
By (i), we can have an improper Riemann integral defined as:
$$
\int_0^\theta \kappa'(x) \textrm{d}x = \kappa(\theta)-\lim_{\epsilon\to0}\kappa(\epsilon) = \kappa(\theta)- \kappa(0) = \kappa(\theta)
$$ and by (iii) you can see that
$$
a\theta - \kappa(\theta) = \int_0^\theta a-  \kappa'(x) \textrm{d}x \geq  \theta (a-\mu)/2
$$
whenever $\theta$ is so small such that $a-  \kappa'(x)>(a-\mu)/2$ for all $x\in(0, \theta)$. These steps show how to conclude the Lemma based on the properties (i), (ii) and (iii).


*The assumption $0<\theta<\theta_0<\theta_−$ in the proof is a typo, it should be replaced by $0<\theta<\theta_0<\theta_+$. In fact, since $\theta_- \leq 0$, the original assumption would be an empty premise, hence anything derived from it is not useful to the proof. If you replace it by $0<\theta<\theta_0<\theta_+$ though, it all makes sense.

You can see that $\theta_- \leq 0$ based on the proof that you outlined in About moment generating function.
