0
$\begingroup$

I read somewhere that it is trivial to see that if a sequence of random variables are uniformly bounded, i.e., $|X_j| \leq C$ for all $j = 1,2,\ldots$, then $E(X_j^4)$ exists. I am not sure where this result is from, could someone guide me? I can prove it but am not sure if this is the result of some convergence theorem. thanks.

$\endgroup$
5
$\begingroup$

In General, if $\left|X\right|$ is bounded by a constant $C>0$, i.e $\left|X\right| \leq C$ a.s, then $$ \Bbb E [|X^n|] =\Bbb E [|X|^n] \leq \Bbb E[C^n] =C^n <\infty, $$ for all $n \in \Bbb N$. So all moments exist.

$\endgroup$
2
$\begingroup$

You shouldn't need any convergence theorems. If $|X| \le C$, then its pdf $f_X(x)$ it is necessarily zero outside of $[-C,C]$, and $\int_{-C}^C f_X(x) \; dx =1$; i.e. $P( -C \le X \le C)=1$. $$ |E[X^4]| = \left\vert \int_{-\infty}^\infty x^4 f_X(x) \; dx \right\vert = \left\vert \int_{-C}^C x^4 f_X(x) \; dx\right\vert \le \int_{-C}^C |x|^4 f_X(x) \; dx \le C^4 \int_{-C}^C f_X(x) \; dx = C^4. $$ and this is finite.

$\endgroup$
  • $\begingroup$ Can you please share why $[a,b] \subseteq [-C,C]$? Thanks! $\endgroup$ – user321627 Jan 9 '17 at 5:44
  • $\begingroup$ Looking back the $[a,b]$ part isn't needed and I'm just overcomplicating things. Let me know if the update makes sense. $\endgroup$ – user365239 Jan 9 '17 at 5:59
  • $\begingroup$ No need for a PDF, which is fortunate since PDFs often do not exist. $\endgroup$ – Did Jan 9 '17 at 9:16
  • $\begingroup$ Thanks @Did I didn't think about that. user321627 you should accept Cettt answer - it is an improvement upon mine. $\endgroup$ – user365239 Jan 9 '17 at 16:28

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.