Vtyjkyuk

The original problem states as below:

Suppose some random variable $X$ satisfies $DeclareMathOperator*EmathbbE E e^theta ^2 X^2 leq e^ctheta^2$ 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 $DeclareMathOperator*EmathbbE (E|X|^p)^1/p$

$$
beginalign*
DeclareMathOperator*EmathbbE E|X|^p &= int_0^infty P(|X|^p>t)mathrmdt\
&= int_0^infty P(|X|>t^1/p)mathrmdt\
&= int_0^infty P(e^X^2>e^t^2/p)mathrmdt\
&leq int_0^infty e^-t^2/p E e^X^2mathrmdt\
&leq e^cint_0^infty e^-t^2/p mathrmdt quad（E e^theta ^2 X^2 leq e^ctheta^2,take theta=1)\
&=e^cGamma(frac2/p+12/p)\
&leq e^c (frac2+p2)^frac2+p2
endalign*
$$

then we have:
$$
beginalign*
DeclareMathOperator*EmathbbE
(E|X|^p)^1/p leq e^c/p(frac2+p2)^frac2+p2p
endalign* $$
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

$$
beginalign*
DeclareMathOperator*EmathbbE
undersetthetainf frace^ctheta^2theta^pGamma(1+p/2)&= 2(ce)^p/2 fracGamma(1+fracp2)(p/2)^(p/2)
\ &= 2(ce)^p/2fracp2fracGamma(fracp2)(p/2)^(p/2)
\ &leq (ce)^p/2P
endalign* $$

and thus we have:

$$
beginalign*
DeclareMathOperator*EmathbbE
(E|X|^p)^1/p leq （ce)^1/2 p^1/p leq (ce)^1/2 e^1/e
endalign* $$

which is bound by constant

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 $theta>0$,
beginalign
mathbbE|X|^p &= int_0^infty mathbbPXmathrmdt\
&= int_0^infty mathbbPmathrmdt\
&=int_0^infty mathbbPe^theta^2 X^2 > e^theta^2t^2/p dt\
&leq int_0^infty e^-theta^2t^2/p mathbbE[e^theta^2 X^2] dt tagMarkov\
&leq e^ctheta^2int_0^infty e^-theta^2t^2/pdt tagassumption
endalign
Now compute the integral as you did to get a final bound
$$
frace^ctheta^2theta^pGamma(1+p/2) tag$dagger$
$$ 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 $sqrtppi c^p/2$. Thus $(mathbbE|X|^p)^1/p$ will only be bounded if $0leq c<1$ (in which case the limit as $ptoinfty$ is $0$).

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

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.