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Does bounded almost surely imply bounded in Lp?

  1. Jun 21, 2010 #1
    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]?


  2. jcsd
  3. Jun 21, 2010 #2
    [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: Jun 22, 2010
  4. Jun 23, 2010 #3
    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 sequence of random variables is another question.
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