Consider a sequence $(X_k^{(n)})_{k\leqslant n}$ of nonnegative measurable functions from a probability space such that $\mathbb{E}X_k^{(n)}=p_n$ and $\|X_k^{(n)}\|_\infty\leqslant 1$. Suppose that $np_n\to 0$ as $n\to\infty$ and additionally assume that $X^{(n)}$ is $\beta$-mixing with summable mixing coefficients for every $n\geqslant 1$.
I am interesting in bounding the Laplace transform of $S_n=\sum_{i=1}^nX_i^{(n)}$, that is \begin{equation} \mathscr{L}_\lambda(S_n)=\mathbb{E}\mathrm{e}^{\lambda S_n} \end{equation} for some $\lambda>0$. To be precise, under what conditions, both on the mixing structure as on the coefficients $p_n$, is the Laplace transform above uniformly bounded in $n\geqslant1$?
Indeed, by a simple application of the generalised Hölder's inequalities, one can obtain \begin{equation} \mathscr{L}_\lambda(S_n)\leqslant\Big(\prod_{i=1}^n\mathrm{e}^{\lambda nX_i}\Big)^{1/n}\leqslant1+p_n\mathrm{e}^{\lambda n} \end{equation} which shows that $\mathscr{L}_\lambda(X)<\infty$ granted $p_n$ converges exponentially fast, which is a strong assumption to make. Additionally, since a sum of independent Bernoulli random variables can be approximated by Poisson variables, another approach would be to investigate whether this yields similar approximations under mixing, at least an approximation with a bounded Laplace transform.
Now using a rather complicated sequence of arguments, including the Fuk-Nagaev inequality, Marcinkiewicz–Zygmund type inequalities and Viennet–Delyon’s inequality, I was able to show that $\mathscr{L}_\lambda(S_n)<\infty$ uniformly in $n$ if the underlying sequence is geometrically mixing. This was essentially needed in order to ensure that $\lambda^p\|S_n\|_p^p/p!$ is a summable sequence, otherwise the Taylor expansion would diverge.
However, as $\mathbb{E}S_n\to 0$ and as $\|S_n\|_\infty<\infty$ for every finite $n\geqslant 1$, then I would hypothesise that perhaps a finite bound may also be obtainable when we do not assume that $X$ is (sub)geometrically absolutely regular, however I am unable to (dis)prove this so far.
