# Optimality of Khintchine's inequality, convergence in distribution and convergence of moments

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 ?

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 $$n$$.

• 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 Nov 27 '20 at 14:36
• @backahast I have added an "alternative" proof, at least without dealing with uniform integrability. Nov 28 '20 at 11:19
• 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 Dec 9 '20 at 17:51