Hi all,

For my research I am trying to bound some exponential moments of subgaussian r.v.'s. And I am stuck with proving one of such inequalities. More specifically:

Let $a$ be unit vector in $\mathcal{R}^{n}$ and $w_{i}$, $i =1,2,...,n$, be $n$ i.i.d Rademacher rv's. Also let $v = \sum_{i}^{n} a_{i}w_{i}$. I know that $\forall 0 < t < \frac{1}{2} , \; {\mathbb E} (e^{tv^{2}}) \leq {\mathbb E}(e^{t z^{2}})$, where $z$ is standard normal r.v. and independent of $w_{i}$'s.

Now my question is: would this inequality also works if we change the sign on $t$? i.e.:

$$ \forall t > 0 , \; {\mathbb E} (e^{-tv^{2}}) \leq {\mathbb E}(e^{-t z^{2}}) $$

I have run many numerical experiments and it seems to be correct, but I am yet to prove it.

What I have done so far is as follows:

\begin{equation} {\mathbb E_v}(e^{-tv^{2}}) = {\mathbb E_{z}}{\mathbb E_{v}}(e^{i\sqrt{2t}vz}) = {\mathbb E_{z}} \prod_{i=1}^{n} {\mathbb E_{w_{i}}}(e^{i\sqrt{2t}a_{i} w_{i} z}) = {\mathbb E_{z}} \prod_{i=1}^{n} (cos(\sqrt{2t}a_{i}z)) \end{equation}

but I am stuck here (not even sure if what I have done is going to get me anywhere at all). This must be something that someone out there should know about, I am hoping.

Any help, suggestion or pointers would be greatly appreciate it.

Cheers and thanks for reading


  • $\begingroup$ For what it's worth, we have $\mathbb E_z\mathbb E_v e^{i\sqrt{2t}vz} = \mathbb E_z \mathbb E_{(w_j)} e^{i\sqrt{2t}\sum_ja_jw_jz} = \prod_j \mathbb E_{w_j}\mathbb E_z e^{i\sqrt{2t}a_jw_jz} = \prod_j\mathbb E_{w_j}e^{-ta_j^2w_j^2} = e^{-t\sum_j a_j^2} = e^{-t}$ $\endgroup$
    – ansobol
    Feb 28 '13 at 5:32
  • $\begingroup$ Thanks for replying but I think what you suggested is not quite correct (and please correct me if I am wrong): taking ${\mathbb E_{z}}$ inside the $\prod_{i=1}^{n} {\mathbb E_{w_{i}}} \cdots$ would require the terms inside the product to be independent, which is clearly not the case here (all the terms have $z$ in them. $\endgroup$
    – Fred
    Feb 28 '13 at 8:04

This would be a fascinating inequality. Unfortunately, it is false in general. Indeed, let $n$ and the $a_i$'s be such that $p:=P(v=0)>0$; e.g., one can take $n=2$ and $a_1=a_2=1/\sqrt2$. Then $Ee^{-tv^{2}}\ge p>0$ for all $t>0$, whereas $Ee^{-tz^{2}}\to0$ as $t\to\infty$ (by the dominated convergence or direct calculation of $Ee^{-tz^{2}}$).


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