4
$\begingroup$

Let $b_1,..b_n$ be real numbers and $\varepsilon_1,...,\varepsilon_n$ be independant Rademacher random variables. The Khintchine's inequality states that $$\mathrm{E}\left [ \left ( \sum_{i=1}^{n} b_i\varepsilon_i \right )^{2p}\right ]\leqslant \frac{\left ( 2p \right )!}{2^pp!}\left ( \sum_{i=1}^{n}b_i^2 \right )^p$$ for every integer $p \geqslant 1$.

I'm trying to prove that the constant $\frac{\left ( 2p \right )!}{2^pp!}$ is optimal, in the sense that it is impossible to obtain an inequality that holds for every Rademacher sum with a strictly smaller constant that does not depend on the dimension $n$.

Since $\frac{\left ( 2p \right )!}{2^pp!}$ is the $2p$-th moment of a standard normal variable, my idea was to approximate a well chosen Rademacher sum with a standard normal variable to obtain the optimality.

Let $b_1=...=b_n=1$. The central limit theorem ensures that $Z_n=\frac{1}{\sqrt{n}}\sum_{i=1}^{n}\varepsilon_i$ converges in distribution towards a random variable $X$ of distribution $\mathcal{N}(0,1)$.

If that implied that $$\lim_{n\rightarrow\infty}\mathrm{E}[Z_n^{2p}] = \mathrm{E}[X^{2p}]$$ then we would have $$\lim_{n\rightarrow\infty}\frac{1}{n^p}\mathrm{E}\left [ \left ( \sum_{i=1}^{n}\varepsilon_i \right )^{2p}\right ] = \frac{\left ( 2p \right )!}{2^pp!}$$ which proves the optimality.

So my question really is : is it true that $\lim_{n\rightarrow\infty}\mathrm{E}[Z_n^{2p}] = \mathrm{E}[X^{2p}]$ ? I don't think the dominated convergence theorem works here since $Z_n$ is not bounded.

The interpretation of convergence in distribution in terms of pointwise convergence of the characteristic functions yields $\forall t \in \mathbb{R}, \lim_{n\rightarrow\infty} \cos(\frac{t}{\sqrt{n}})^n=e^{-\frac{t^2}{2}}$. Could that be of any use ?

$\endgroup$
2
$\begingroup$

Indeed, the dominated convergence theorem does not work directly, since $\sup_n \lvert Z_n\rvert$ is not integrable.

However, we can conclude the wanted convergence if we can show that for each $p\geqslant 1$, $\left(Z_n^{2p}\right)$ is uniformly integrable, see here for the details.

If we show that $\sup_{n\geqslant 1}\mathbb E\left[Y_n^{2p}\right]$ is finite for each $p\geqslant 1$, we will get the uniform integrability. Indeed, if we want to show that $\left(Z_n^{2p_0}\right)$ is uniformly integrable for some $p_0$, we use the fact that $\left(Z_n^{2(p_0+1)}\right)$ is bounded in $\mathbb L^1$.

Now $\mathbb E\left[Y_n^{2p}\right]$ can be estimated by Khintichine inequality and can be bounded independently of $n$.

Alternatively, one can start from $$ \left\lvert\mathbb E\left[Y_n^{2p}\right]-\mathbb E\left[X^{2p}\right]\right\rvert =2p\left\lvert \int_0^\infty t^{2p-1}\left( \mathbb P\left(Y_n>t\right)-\mathbb P\left(X>t\right)\right)dt\right\rvert, $$ split the integral into two parts: on $(0,A)$, which can be bounded using the uniform convergence of the c.d.f. of $Y_n$ to that of $X$; for the integral on $(A,+\infty)$, one can use Hoeffding's inequality to show that its contribution vanishes as $A$ goes to infinity uniformly on $n$.

$\endgroup$
3
  • $\begingroup$ Got it - do you think there is a more elementary way of getting the convergence of moments in our specific case ? I don't believe I've seen this result before and it looks like it may not be that easy to prove at first glance $\endgroup$
    – backahast
    Nov 27 '20 at 14:36
  • $\begingroup$ @backahast I have added an "alternative" proof, at least without dealing with uniform integrability. $\endgroup$ Nov 28 '20 at 11:19
  • $\begingroup$ just a little detail, it should be $ \mathbb P\left(|Y_n|>t\right)-\mathbb P\left(|X|>t\right)$ instead of $ \mathbb P\left(Y_n>t\right)-\mathbb P\left(X>t\right)$, shouldn't it ? It works the same way anyway $\endgroup$
    – backahast
    Dec 9 '20 at 17:51

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.