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.

  • 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

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 $$\lim_{\epsilon\to0}\limsup_{r\to\infty}e^{\epsilon r^2}\mathbb P(\|X\|\ge r)=0.$$


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