2
$\begingroup$

The original problem states as below:

Suppose some random variable $X$ satisfies $\DeclareMathOperator*{\E}{\mathbb{E}} \E e^{\theta ^{2} X^{2}} \leq e^{c\theta^2}$ for some constant $c$ and $\forall \theta \in R$ show that $X$ is a bounded random variable ? i.e $\|X\|_{\infty} <\infty$

how to solve this one ?

Here is some of my attempt:

We want to bound $\DeclareMathOperator*{\E}{\mathbb{E}} (\E|X|^p)^{1/p}$

$$ \begin{align*} \DeclareMathOperator*{\E}{\mathbb{E}} \E|X|^p &= \int_0^\infty P(|X|^p>t)\mathrm{d}t\\ &= \int_0^\infty P(|X|>t^{1/p})\mathrm{d}t\\ &= \int_0^\infty P(e^{X^2}>e^{t^{2/p}})\mathrm{d}t\\ &\leq \int_0^\infty e^{-t^{2/p}} \E e^{X^2}\mathrm{d}t\\ &\leq e^{c}\int_0^\infty e^{-t^{2/p}} \mathrm{d}t \quad(\E e^{\theta ^{2} X^{2}} \leq e^{c\theta^2},take \ \ \theta=1)\\ &=e^c\Gamma(\frac{2/p+1}{2/p})\\ &\leq e^c (\frac{2+p}{2})^{\frac{2+p}{2}} \end{align*} $$

then we have: $$ \begin{align*} \DeclareMathOperator*{\E}{\mathbb{E}} (\E|X|^p)^{1/p} \leq e^{c/p}(\frac{2+p}{2})^{\frac{2+p}{2p}} \end{align*} $$ which do not converges as $ p \to \infty $ ,to here I think I fail to bound it

@Clement C. just check your hint , I think it is workable if I compute it right

$$ \begin{align*} \DeclareMathOperator*{\E}{\mathbb{E}} \underset{\theta}{inf} \frac{e^{c\theta^2}}{\theta^p}\Gamma(1+p/2)&= 2(ce)^{p/2} \frac{\Gamma(1+\frac{p}{2})}{(p/2)^{(p/2)}} \\ &= 2(ce)^{p/2}\frac{p}{2}\frac{\Gamma(\frac{p}{2})}{(p/2)^{(p/2)}} \\ &\leq (ce)^{p/2}P \end{align*} $$

and thus we have:

$$ \begin{align*} \DeclareMathOperator*{\E}{\mathbb{E}} (\E|X|^p)^{1/p} \leq (ce)^{1/2} p^{1/p} \leq (ce)^{1/2} e^{1/e} \end{align*} $$

which is bound by constant

$\endgroup$
  • $\begingroup$ Please read my answer also. If I didn't make t a mistake this is really trivial. $\endgroup$ – Kavi Rama Murthy Sep 11 '18 at 5:35
  • $\begingroup$ ShaoyuPei: And there is an even more trivial approach... Really, you accept answers much too quickly. $\endgroup$ – Did Sep 11 '18 at 10:31
  • $\begingroup$ Additionnally, I would be curious to know if you actually managed to complete the proof @ClementC. suggested... $\endgroup$ – Did Sep 11 '18 at 10:33
  • $\begingroup$ @Did Me too, to be honest. My solution (assuming is does go through without further assumptions on $c$) is not the most elegant nor shortest, merely one that mimicks the OP's attempt to fix it. $\endgroup$ – Clement C. Sep 11 '18 at 20:17
  • 1
    $\begingroup$ @ShaoyuPei Before accepting my answer, have you managed to make my suggestion go through? $\endgroup$ – Clement C. Sep 11 '18 at 20:25
1
$\begingroup$

Hint: You are not using the fact that the bound holds for every $\theta$. This is important.

To use it: Introduce "artificially" a $\theta$: for every $\theta>0$, \begin{align} \mathbb{E}|X|^p &= \int_0^\infty \mathbb{P}\{|X|^p>t\}\mathrm{d}t\\ &= \int_0^\infty \mathbb{P}\{|X|>t^{1/p}\}\mathrm{d}t\\ &=\int_0^\infty \mathbb{P}\{e^{\theta^2 X^2} > e^{\theta^2t^{2/p}}\} dt\\ &\leq \int_0^\infty e^{-\theta^2t^{2/p}} \mathbb{E}[e^{\theta^2 X^2}] dt \tag{Markov}\\ &\leq e^{c\theta^2}\int_0^\infty e^{-\theta^2t^{2/p}}dt \tag{assumption} \end{align} Now compute the integral as you did to get a final bound $$ \frac{e^{c\theta^2}}{\theta^p}\Gamma(1+p/2) \tag{$\dagger$} $$ which depends on $\theta$; then try to choose the best $\theta$ as a function of $p$ to optimize this bound.

Important: The above does not appear to go through. Namely, the minimum of $(\dagger)$ is achieved for $\theta^2 = p/(2c)$, and the final bound is asymptotically $\sqrt{p\pi} c^{p/2}$. Thus $(\mathbb{E}|X|^p)^{1/p}$ will only be bounded if $0\leq c<1$ (in which case the limit as $p\to\infty$ is $0$).

$\endgroup$
  • $\begingroup$ Please see if my answer is correct. $\endgroup$ – Kavi Rama Murthy Sep 11 '18 at 5:36
  • $\begingroup$ @KaviRamaMurthy Please see if mine is :) $\endgroup$ – Clement C. Sep 11 '18 at 20:28
  • $\begingroup$ @I think your solution is workable $\endgroup$ – ShaoyuPei Sep 15 '18 at 2:12
3
$\begingroup$

Low-tech approach: For every $\theta$ and every $a>c$, $$E(e^{\theta^2X^2})\geqslant E(e^{\theta^2X^2};X^2\geqslant a)\geqslant e^{\theta^2a}P(X^2\geqslant a)$$ hence $$P(X^2\geqslant a)\leqslant e^{-\theta^2a}e^{\theta^2c}$$ The RHS goes to $0$ when $\theta^2\to\infty$ hence $$P(X^2\geqslant a)=0$$ This holds for every $a>c$ hence $X^2\leqslant c$ almost surely, QED.

$\endgroup$
2
$\begingroup$

We have $Ee^{\theta^{2}(X^{2}-c)} \leq 1$. By Fatou's Lemma $E \lim \inf_{\theta \to \infty} e^{\theta^{2}(X^{2}-c)} \leq 1$. On the set $X^{2} >c$ the $\lim \inf $ is $\infty$. This implies that $X^{2} \leq c$ almost everywhere.

$\endgroup$

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.