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

  • 4
    $\begingroup$ A trivial lower bound is $2^{-n}e^n$, so this is false. $\endgroup$ Jan 23, 2016 at 4:13
  • $\begingroup$ @ChristianRemling Thanks for your comment! I've modified the question a little bit. What if $C < \log 2$? In this case, the trivial lower bound $2^{-n} \exp(Cn)$ goes to zero. Thanks! $\endgroup$
    – Steve
    Jan 23, 2016 at 4:42

2 Answers 2


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.

  • 1
    $\begingroup$ Both links are broke. $\endgroup$
    – Hans
    Jan 7, 2018 at 22:06

EDITED: As pointed out by Anthony and John, my 2am solution was anything but. In summary, the conjecture is TRUE for $C$ smaller than approximately 0.6880137 and false for larger $C$.

The exact value is $$ S(C,n) = \sum_{k=0}^n 2^{-n}\binom{n}{k} \exp\bigl(Cn^{-3}(n-2k)^4\bigr)$$ since the product of the first two terms is the probability that $\sum_i X_i=n-2k$.

Consider the term $k=(\tfrac12+\beta)n$, where $-\frac12\le\beta\le\frac12$. Stirling's approximation for constant $\beta\ne\pm\frac12$ is $$2^{-n}\binom nk \approx \bigl(2\pi(\tfrac12-\beta)(\tfrac12+\beta)n\bigr)^{-1/2} \bigl(2 (\tfrac12-\beta)^{\tfrac12-\beta} (\tfrac12+\beta)^{\tfrac12+\beta}\bigr)^{-n}$$

Therefore the value of the term is close to $$\bigl(2\pi(\tfrac12-\beta)(\tfrac12+\beta)n\bigr)^{-1/2} \exp\bigl(f(\beta)n\bigr),$$ where $$ f(\beta) = -(\tfrac12-\beta)\ln(\tfrac12-\beta) - (\tfrac12+\beta)\ln(\tfrac12+\beta) - \ln 2 + 16C\beta^4. $$ (Note that there is an order $1/n$ term from Stirling's formula missing. Since it doesn't depend on $C$ and the value of the sum is 1 for $C=0$, the effect of the missing term is easily corrected at the end.)

The function $f(\beta)$ has a local maximum of value 0 at $\beta=0$ for every $C$. There is a constant $C_0\approx 0.6880137$ (note: less than $\ln 2$) so that for $C\lt C_0$ we have $f(\beta)\lt 0$ for all $\beta\ne 0$. For $C\gt C_0$ there are places where $f(\beta)\gt 0$ and the sum will be exponential.

Therefore, when $C\lt C_0$ the sum is dominated by terms near $\beta=0$. Around there, $$f(\beta) = -2\beta^2 + (16C-\tfrac43)\beta^4 + O(\beta^6),$$ so the term is close to $$(\pi n/2)^{-1/2} e^{-2\beta^2 n}\bigl(1+(16C-\tfrac43)\beta^4n\bigr) $$ with terms for $\beta=O(n^{-1/2})$ mattering. Approximating the sum by an integral, I get $$ S(C,N) = 1 + 3C/n + O(n^{-2}),$$ for $C\lt C_0$.

ADDED: The exact value of $C_0$ is $$\frac{1}{64\beta_0^3}\ln\Bigl(\frac{1+2\beta_0}{1-2\beta_0}\Bigr),$$ where $\beta_0$ is the zero near 0.4953 of $$\bigl(-\tfrac12+\tfrac34\beta\bigr)\ln(1-2\beta) + \bigl(-\tfrac12-\tfrac34\beta\bigr)\ln(1+2\beta).$$ To 50 digits: $C_0 = 0.68880137394879099153980106986892429419403651844842$.

  • $\begingroup$ It seems like Stirling's formula would be better to use than a normal approximation. $\endgroup$ Jan 23, 2016 at 16:27
  • 2
    $\begingroup$ But only small $\alpha$ are available to you, right? $\endgroup$ Jan 23, 2016 at 18:07
  • $\begingroup$ @Anthony: now I am confused. was I right before or am I right now? $\endgroup$
    – John Jiang
    Jan 23, 2016 at 23:03
  • $\begingroup$ This proof seems to make the same mistake as I did with my previous deleted one: first alpha can't be larger than 1 as Anthony pointed out. Then for small alpha actually the square dominates the fourth power, so the lower bound is less than 0. Am I right? $\endgroup$
    – John Jiang
    Jan 23, 2016 at 23:55
  • $\begingroup$ @John : Yes, that problem is there. Now that I have woken up, sort of, I'll see if I can fix it. $\endgroup$ Jan 24, 2016 at 0:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.