TITLE: Asymptotics of functional of i.i.d. Rademacher random variables
QUESTION [11 upvotes]: Let $X_1,\ldots, X_n$ be i.i.d. Rademacher random variables. That is, $\operatorname{Pr}(X_i = 1) = \operatorname{Pr}(X_i = -1) = 1/2$. I was wondering if the following argument is true:
$$
\mathbb{E} \exp\biggl( C\cdot \left(\sum_{i=1}^n X_i\right)^4\big/n^3 \biggr) = 1 + O(1/n),
$$
where $C \geq 0$ is a constant.
I did some numerical simulations and the results validated this argument. I would appreciate it if anyone can give a proof.
If my conjecture is correct, this problem reveals an interesting phenomenon of probability theory. In fact, if we replace $\sum_{i=1}^n X_i /\sqrt{n}$ by a standard norm random variable, the expectation is $\infty$. This means that we cannot simply plug in Gaussian approximation here. 
P.S.: As pointed out by Christian Remling, this problem has trivial lower bound $2^{-n} \exp(C n)$. So the expectation will explode when $C > \log 2$. What if $0\leq C \leq \log 2$?

REPLY [4 votes]: I believe the conjecture is true for sufficiently small $C$. Previously I was trying to disprove it using Large deviation theory. But I missed a sign at the last step. But the same argument can be turned around.
In compact form, one gets
$$ \lim \frac{1}{n} \log P(\sum_i X_i > \alpha n) = -I(\alpha),$$
where $I(\alpha) = \sup_\theta (\alpha \theta - \log \mathbb{E} e^{\theta X_1})$. One can compute 
$$I(\alpha) = (\frac{1}{2} - \alpha)\log (\frac{1}{2} - \alpha) + (\frac{1}{2} + \alpha)\log(\frac{1}{2} + \alpha) + \log 2.$$
More precisely, the lower bound (Cramer's inequality) states that for any $\epsilon > 0$, as long as $n$ is sufficient large, we get $P(\sum_i X_i \ge \alpha n) \ge e^{-n (I(\alpha) + \epsilon)}$; see this lecture notes. The upper bound is without the $\epsilon$ and can be proved easily using exponential Tchebyshev. We only need the upper bound here.
Now write the original expectation as $\int_0^1 1 + P(|\sum X_i | > \alpha n) \frac{d}{d\alpha} e^{C \alpha^4 n} d\alpha$, by the so-called integration by parts formula in probability; see page 66 lemma 1.4.30 here. The latter is bounded by
$$ \int_0^1 1 d\alpha + \int_0^1  4 \alpha^3 C n e^{(C \alpha^4 - I(\alpha))n} d\alpha.$$
After evaluating the third moment of a Gaussian of variance $n^{-1}$ near $\alpha = 0$, one sees that the integral near $0$ goes to zero as $C n^{-2}$, and the rest of the integral is clearly very small.
Note this argument doesn't work for Gaussian because the upper bound of integration is $\infty$ in that case. This is a good check of my often reckless arithmetic.