Does bounded almost surely imply bounded in Lp?

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The discussion centers on the concept of "bounded almost surely" in relation to random variables and their implications for boundedness in Lp spaces. It is established that if a random variable X(ω) is bounded almost surely, it implies that Pr(|X| ≤ M) = 1, meaning it is bounded except on a set of zero measure. Furthermore, if X is bounded almost surely, it is also bounded in Lp for any p ≥ 1, provided that the constant K is finite. The distinction between constants K and K(ω) is clarified, emphasizing that K(ω) is not useful for a single random variable.

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wayneckm
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Hello all,

I am a bit confused by the concept of "bounded almost surely".

If a random variable [tex]X(\omega)[/tex] is bounded a.s., so this means (i) [tex]X \leq K[/tex] for some constant [tex]K[/tex] ? or some [tex]K(\omega)[/tex]?

Also, if it is bounded almost surely, does that mean it is also bounded in [tex]L^{p}[/tex]? Apparently if case (i) is true, then it should be also bounded in [tex]L^{p}[/tex]?

Thanks.

Wayne
 
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wayneckm said:
Hello all,

I am a bit confused by the concept of "bounded almost surely".

[tex]Pr(|X|\leq M)=1[/tex].

Almost surely=almost everywhere which excludes sets of zero measure.

If L means sets in Lebesgue measure then sets of zero measure would be excluded, so I believe it would be bounded in L if it's bounded in M.

[tex]K\leq M[/tex]
 
Last edited:
wayneckm said:
Hello all,

I am a bit confused by the concept of "bounded almost surely".

If a random variable [tex]X(\omega)[/tex] is bounded a.s., so this means (i) [tex]X \leq K[/tex] for some constant [tex]K[/tex] ? or some [tex]K(\omega)[/tex]?

The [tex]K(\omega)[/tex] version would be worthless for a single random variable [tex]X[/itex], just take [tex]K(\omega) = |X(\omega)|[/tex]. Now "bounded almost surely" where you talk about a <i>sequence</i> of random variables is another question.[/tex]
 

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