Does bounded almost surely imply bounded in Lp?

[wayneckm]

Hello all,

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

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

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

Thanks.

Wayne

 

[SW VandeCarr]

Hello all,

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

Pr(|X|\leq M)=1.

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.

K\leq M

 

[g_edgar]

Hello all,

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

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

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