# Does bounded almost surely imply bounded in Lp?

1. Jun 21, 2010

### 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

2. Jun 21, 2010

### SW VandeCarr

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

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.

$$K\leq M$$

Last edited: Jun 22, 2010
3. Jun 23, 2010

### g_edgar

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