If $E e^{\theta ^{2} X^{2}} \leq e^{c\theta^2}$ for every $\theta$, then $X$ is almost surely bounded

Question

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

Answer 1

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 $0$).

Answer 2

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.

Answer 3

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.