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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.

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2 Answers 2

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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.

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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.

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  • $\begingroup$ Can you please share why $[a,b] \subseteq [-C,C]$? Thanks! $\endgroup$
    – user321627
    Jan 9, 2017 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, 2017 at 5:59
  • $\begingroup$ No need for a PDF, which is fortunate since PDFs often do not exist. $\endgroup$
    – Did
    Jan 9, 2017 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, 2017 at 16:28

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