2
$\begingroup$

the problem I am considering is as follows:

For every probability space $(\Omega , \mathcal F , \mathbb P)$; every separable $\mathbb R$-Banach space $(V, ||\cdot||_{V})$, and every centered Gaussian distributed random variable $X: \Omega \to V$, it holds that:

$limsup_{\epsilon \to 0^{+}} sup_{r \in \mathbb R} \Big[e^{\epsilon r^{2}} \mathbb P \Big(||X||_{V} \geq r\Big)\Big]$ is finite.

What I have tried is to employ Markov's Inequality.

Suppose, $r > 0$. (This positivity enables us to employ Markov's inequality.)

By Markov's inequality, we have:$ \quad e^{\epsilon r^{2}} \mathbb P \Big(||X||_{V} \geq r\Big) \leq \mathbb E(||X||_{V}) \frac {e^{\epsilon r^{2}}}{r}$.

Consequently, first taking the supremum on both side over $r \in \mathbb R$, and then letting $\epsilon \to 0^{+}$, we can conclude that the desired limit is finite since $\quad limsup_{\epsilon \to 0^{+}} sup_{r \in \mathbb R} \mathbb E(||X||_{V}) \frac {e^{\epsilon r^{2}}}{r}$ is $0$.

Is my argument okay ??

Please let me know if I am missing something!!

Thanks in advance!!!

P.S. (EDIT):- As ntt commented, if $r=0$, then the result is false.

$\endgroup$
  • 1
    $\begingroup$ If $r\leq 0$ then $e^{\epsilon r^2}P(\|X\|\geq r) = e^{\epsilon r^2}$, then its supremum should be infinity, right? $\endgroup$ – ntt Apr 21 '17 at 10:41
  • $\begingroup$ Ohh thanks... sorrry sorry !! I missed the modulus. Yuup, you are right. Thanks a ton. The question is the same, replacing $r$ by $|r|$, let me make the edit. $\endgroup$ – user92360 Apr 21 '17 at 10:48
  • 3
    $\begingroup$ Since $\lim_{|r|\to \infty}\frac{e^{\epsilon r^2}}{|r|} = +\infty$, you get $\sup_{r\in \mathbb R} \frac{e^{\epsilon r^2}}{|r|} = +\infty$. $\endgroup$ – ntt Apr 21 '17 at 11:10
  • $\begingroup$ @ntt , thanks !! $\endgroup$ – user92360 Apr 21 '17 at 11:14
  • 1
    $\begingroup$ Yes, the RHS is infinity, also you cannot conclude whether LHS is finite. It's only a remark that you cannot get the finiteness of the LHS via this way. $\endgroup$ – ntt Apr 21 '17 at 11:28
2
$\begingroup$

As mentioned in the comments, this is obviously false as written, but is true if we restrict the supremum to be over $r\ge0$. As the comments also noted, Fernique's theorem is the way to go. Fernique's theorem states that there exists $\alpha>0$ such that $\mathbb E[e^{\alpha\|X\|^2}]<\infty$. Then for $0<\epsilon<\alpha$ and $r>0$, we have $$e^{\epsilon r^2}\mathbb P(\|X\|\ge r)=e^{\epsilon r^2}\mathbb P(e^{\alpha\|X\|^2}\ge e^{\alpha r^2})\le e^{-(\alpha-\epsilon)\|r\|^2}\mathbb E[e^{\alpha\|X\|^2}]\le\mathbb E[e^{\alpha\|X\|^2}]$$ where the second to last step uses Markov's inequality. Taking suprema over $r\ge0$ and letting $\epsilon\to0$ completes the proof. Note that you could also fairly easily show that $$\lim_{\epsilon\to0}\limsup_{r\to\infty}e^{\epsilon r^2}\mathbb P(\|X\|\ge r)=0.$$

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.