Suppose $\{\xi_{i}\}$ be i.i.d r.v.'s, and $\text{Var}(\xi_{1})>0$. Let $\phi(\theta) = E\exp(\theta\xi_1) < \infty$ for $\theta \in (-\delta,\delta)$, $\delta >0 $, and let $\psi(\theta) := (\log\phi(\theta) )$ and $S_n = \sum^n_{i=1}\xi_{i}$
$\textbf{(a)}$ Prove that $\psi$ is strictly convex;
$\textbf{(b)}$ Show that $E\sqrt{X^\theta_n} \rightarrow 0$ and $X^\theta_n \rightarrow 0$ a.s. with $X^\theta_n := \exp(\theta S_n - n\psi(\theta))$.
This is the original question from the "Probability: Theory and Examples by Rick Durrett" Chapter5 Exercise 7.4. https://services.math.duke.edu/~rtd/PTE/PTE4_1.pdf
My attempt:
$\textbf{(a)}$ To show $\psi$ is strictly convex that means we should show $\psi^{''} (\theta) > 0$. Since $$\psi^{'}(\theta) = \frac{\phi^{'}(\theta)}{\phi(\theta)}$$, then $$\Big(\frac{\phi^{'}(\theta)}{\phi(\theta)}\Big)^{'} = \frac{\phi^{''}(\theta)}{\phi(\theta)} - \Big(\frac{\phi^{'}(\theta)}{\phi(\theta)}\Big)^{2}$$
From this step, I can not show why $$\frac{\phi^{''}(\theta)}{\phi(\theta)} - \Big(\frac{\phi^{'}(\theta)}{\phi(\theta)}\Big)^{2} > 0$$
$\textbf{Can I get some helpes from how to show the last inequality holds?}$
$\textbf{(b)}$ $$ \sqrt{X^\theta_n} = \exp(\theta/2 S_n - n/2\psi(\theta))$$ $$\Rightarrow \sqrt{X^\theta_n} = \sqrt{X^{\theta/2}_n}\exp(n\psi(\theta/2)- \psi(\theta)/2)$$ By $\psi(\theta)$ is strictly convex from $\textbf{(a)}$ and $\psi(0)=0$, so $\psi(\theta/2) - \psi(\theta)/2 < 0$, so we have $$E\sqrt{X^\theta_n} = E \sqrt{X^{\theta/2}_n}\exp(n\psi(\theta/2)- \psi(\theta)/2)$$
$\textbf{I know $X^{\theta/2}_n$ is a martingale, but how can I conclude that $E\sqrt{X^\theta_n} \rightarrow 0$ from the above equality}$ $\textbf{and how to use $E\sqrt{X^\theta_n} \rightarrow 0$ to get $X^\theta_n \rightarrow 0$ a.s.?}$
Thank U!