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Almost equal random variables

  1. Aug 4, 2009 #1
    I've been trying to solve the following question: Let X be a random variable s.t. Pr[|X|<+\infty]=1. Then for every epsilon>0 there exists a bounded random variable Y such that P[X\neq Y]<epsilon.

    The ideia here would be to find a set of epsilon measure so Y would be different than X in that set. However, it is not clear even that such a measurable set exists...

    Any help?
  2. jcsd
  3. Aug 4, 2009 #2
    To show there is some interval (-a, a) such that [tex]\int_{-a}^a f(x) dx > 1-\epsilon[/tex], apply the definition of an improper integral to [tex]\int_{-\infty}^\infty f(x) dx = 1[/tex]. Have Y=X over (-a,a).
  4. Aug 5, 2009 #3
    For a large positive number [itex]\lambda[/itex], define [itex]Y = X[/itex] on the set [itex]\{|X| \le \lambda\}[/itex] and [itex]Y=0[/itex] on the remaining part of the sample space. Now all you have to do is choose [itex]\lambda[/itex] large enough to get the conclusion you want.
  5. Aug 5, 2009 #4
    It would remain to show that such $\lambda$ exists. We can definte the sets A_n=\{\omega\in \Omega: |X(\omega)|<n\}. Since A_n is measurable, as X is a random variable, and A_n \uparrow \{\omega: |X|<+\infty}, we have that \lim P(A_n) = P(\{\omega: |X|<+\infty}. Therefore, for all \epsilon, there is $m$ large enough so that P(A_n)>1-\epsilon. Then we can define Y=X in A_m and Y=0 otherwise.

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