2
$\begingroup$

I'm trying to understand this proof of Khintchine's inequality for the complex case ($a_n\in\mathbb{C}$). The author claims that the complex case follows from the real case by taking absolute values where appropriate, but I don't see where to place those absolut values, even after filling in the details of the proof:

Proof (for $a_n\in\mathbb{R}$): Let $\epsilon_1,\dots,\epsilon_n$ be some i.i.d. random variables with $\mathbb P(\epsilon_k=\pm1)=\frac12$ and let $a_1,\dots,a_n\in\mathbb{R}$ (w.l.o.g. $\sum_{k=1}^na_k^2=1$). For every $\lambda>0$ we have $$ \begin{align*} \mathbb E(\exp(\lambda\epsilon_ka_k)) &=\sum_{j=\pm1}\exp(j\lambda a_k)\cdot\overbrace{\mathbb P(\epsilon_k=j)}^{=1/2} =\cosh(\lambda a_k) =\sum_{\ell=0}^\infty\frac{(\lambda a_k)^{2\ell}}{(2\ell)!} \\ &\leq\sum_{\ell=0}^\infty\frac{(\lambda^2a_k^2)^\ell}{2^\ell\cdot\ell!} =\sum_{\ell=0}^\infty\frac{(\lambda^2a_k^2\,/\,2)^\ell}{\ell!} =\exp(\lambda^2a_k^2\,/\,2), \end{align*} $$ which gives us $$ \mathbb E\bigg(\exp\bigg(\lambda\sum_{k=1}^n\epsilon_ka_k\bigg)\bigg) =\prod_{k=1}^n\mathbb E(\exp(\lambda\epsilon_ka_k)) \leq\prod_{k=1}^n\exp(\lambda^2a_k^2\,/\,2) =\exp\bigg(\frac{\lambda^2}2\sum_{k=1}^na_k^2\bigg) =\exp(\lambda^2\,/\,2), $$ and thus $$ \begin{align*} \mathbb P\bigg(\bigg|\sum_{k=1}^n\epsilon_ka_k\bigg|\geq\lambda\bigg) &=2\mathbb P\bigg(\sum_{k=1}^n\epsilon_ka_k\geq\lambda\bigg) =2\mathbb P\bigg(\exp\bigg(\lambda\sum_{k=1}^n\epsilon_ka_k\bigg)\geq\exp(\lambda^2)\bigg) \\ &\overset{(*)}\leq\frac2{\exp(\lambda^2)}\mathbb E\bigg(\exp\bigg(\lambda\sum_{k=1}^n\epsilon_ka_k\bigg)\bigg) \leq\frac2{\exp(\lambda^2)}\exp(\lambda^2\,/\,2) =\exp(-\lambda^2\,/\,2), \end{align*} $$ where $(*)$ uses the fact that for every $\alpha>0$ and every real or complex (integrable) random variable $X$ we have $$ \mathbb P(|X|\geq\alpha) =\mathbb E\Big(1\cdot1_{\big\{\frac{|X|}\alpha\geq1\big\}}\Big) \leq\mathbb E\Big(\frac{|X|}\alpha\cdot1_{\big\{\frac{|X|}\alpha\geq1\big\}}\Big) \leq\mathbb E\Big(\frac{|X|}\alpha\Big) =\frac1\alpha\mathbb E|X|. $$ [... The rest of the proof is irrelevant for my question ...] $\blacksquare$


Now, my question is where do I have to put absolut values in the inequalities above to prove the complex case? In order for the first block to make sense in $\mathbb{C}$ we need to replace $(\lambda a_k)^{2\ell}$ with $|\lambda a_k|^{2\ell}$. We have $$ |\mathbb E(\exp(\lambda\epsilon_ka_k))| =|\cosh(\lambda a_k)| \leq\sum_{\ell=0}^\infty\frac{|\lambda a_k|^{2\ell}}{(2\ell)!} \leq\cdots\leq\exp(\lambda^2|a_k|^2\,/\,2), $$ which implies $$ \bigg|\mathbb E\bigg(\exp\bigg(\lambda\sum_{k=1}^n\epsilon_ka_k\bigg)\bigg)\bigg| =\prod_{k=1}^n|\mathbb E(\exp(\lambda\epsilon_ka_k))| \leq\cdots\leq\exp\bigg(\frac{\lambda^2}2\sum_{k=1}^n|a_k|^2\bigg) =\exp(\lambda^2\,/\,2). $$ But this last result leads me nowhere. I don't see how to translate the third block into complex numbers such that I can make use of the last result.

$\endgroup$
  • $\begingroup$ I don’t get the linked proof. It looks like it has its inequality the wrong way around in Jensen’s inequality, which would be $\mathbb{E}\exp(X) \geq \exp(\mathbb{E}X)$, not $\leq$. $\endgroup$ – WimC Jun 10 at 11:17
  • $\begingroup$ Well, maybe you understand my more-fleshed-out part of the proof, which doesn't require the knowledge of the article (I just linked to it as a source for the claim that the complex case should be as easy as the real case) $\endgroup$ – Cubi73 Jun 10 at 11:29
  • $\begingroup$ I see it now, thanks. $\endgroup$ – WimC Jun 10 at 13:29

Your Answer

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

Browse other questions tagged or ask your own question.