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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:

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

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

Fred

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1 Answer 1

No. If e.g. $n$ is even and $a_i = \frac 1 {\sqrt{n}}$, then $v $ has an atom at $0$ hence the LHS does not converge to $0$ as $ t \rightarrow \infty$ while the RHS does.

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Thanks for replying. Could you please let me know what you mean by 'atom at 0'? thanks –  Fred Mar 1 '13 at 2:33
    
and also why doesn't the LHS converge to 0?. I am a bit confused: On one hand, $e^{-t v^{2}}$ goes to zero almost surely and it is bounded above by 1, so I can apply DCT to get the convergence to zero in mean. On the other hand, I ran a numerical test, and you are right that particular vector of $a_{i}$'s does not go to zero. Could you please explain what I am missing? Thanks a lot –  Fred Mar 1 '13 at 3:38
    
Upon further reading about atoms, I think I understand what you meant: I was so carried away with the details of the computations that I forgot my measure on the right hand side is different than the left hand side. So $e^{-t v^{2}}$ does not go to zero almost everywhere with respect to its discrete measure, as like you said, point 0 has none-zero measure!!!...I think I get it now...but is there any hope for this inequality to be correct for, say, $ t \in [0,1]$? –  Fred Mar 1 '13 at 4:06

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