Fork me on GitHub
Math for the people, by the people.

User login

Inequality for square of the subgaussian distributions

Primary tabs

Inequality for square of the subgaussian distributions

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 aaa be unit vector in nsuperscriptn\mathcal{R}^{{n}} and wisubscriptwiw_{{i}}, i=1,2,,ni12normal-…ni=1,2,...,n, be nnn i.i.d *Rademacher* rv’s. Also let v=inaiwivsuperscriptsubscriptinsubscriptaisubscriptwiv=\sum_{{i}}^{{n}}a_{{i}}w_{{i}}. I know that 0<t<12,?(etv2)?(etz2)formulae-sequencefor-all0t12?superscriptetsuperscriptv2?superscriptetsuperscriptz2\forall 0<t<\frac{1}{2},\;{\mathbb{E}}(e^{{tv^{{2}}}})\leq{\mathbb{E}}(e^{{tz^% {{2}}}}), where zzz is standard normal r.v. and independent of wisubscriptwiw_{{i}}’s.

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

t>0,?(e-tv2)?(e-tz2)formulae-sequencefor-allt0?superscriptetsuperscriptv2?superscriptetsuperscriptz2\forall t>0,\;{\mathbb{E}}(e^{{-tv^{{2}}}})\leq{\mathbb{E}}(e^{{-tz^{{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:

?v(e-tv2)=?z?v(ei2tvz)=?zi=1n?wi(ei2taiwiz)=?zi=1ncos(2taiz)subscript?vsuperscriptetsuperscriptv2subscript?zsubscript?vsuperscriptei2tvzsubscript?zsuperscriptsubscriptproducti1nsubscript?subscriptwisuperscriptei2tsubscriptaisubscriptwizsubscript?zsuperscriptsubscriptproducti1ncos2tsubscriptaiz{\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) (1)

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


I already wrote an answer to your post about a month ago. For some reason it got deleted. So, here I go again ….

Actually I forgot what I wrote at that time. I remember I did the calculations to prove it. You can calculate the moment generating functions on both sides, which means you basically calculate the characterstic function and leave out the complex i. Then the claim follows immediately modulo calculations ;)

Hope that helps, and I hope it stays this time ;)


Subscribe to Comments for "Inequality for square of the subgaussian distributions"