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 $0\leq C \leq \log 2$?



Asymptotics of functional of i.i.d. Rademacher random variables
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](http://www.ifp.illinois.edu/~srikant/ECE567/Fall09/cramer-many-sources.pdf). 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](http://statweb.stanford.edu/~adembo/stat-310a/lnotes.pdf). 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.


