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Math Help - Proving the R.V. condition

  1. #1
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    Proving the R.V. condition

    Hello all, I have a problem that I'm not too clear about.

    It is as follows:

    Let X be a random variable. Show that for every x E R,
    B(x) = { w E Omega: X(w) < x} E F

    *Where E = element of; Omega = collection of objects, the sample space; w = a member of Omega, elementary or outcome; and F = a field

    My solution begins with setting,
    B_n(x) = {x E Omega: X(w) <= x - 1/n}

    Then I am not too sure how to go about proving it...

    I was thinking along the lines of showing equivalence between B(x) and B_n(x), since B_n(x) is already defined to be in the field...

    Please help!
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  2. #2
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    Joined
    Oct 2009
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    posted solution

    Actually, here is the solution that was written on the board, but I wasn't able to follow it at some parts.

    Let x E R. Let n >= 1 be an integer.

    Set B_n(x) = {xE Omega: X(w) <= x - 1/n} E F

    Then B_n(x) is a subset of B_n for all n >=1.
    So, the Union where n=1 to infinity B_n(x) is a subset of B(x). Show B(x) = Union from n=1 to infinity of B_n(x).

    Claim: B(x) is s subset of Union from n=1 to infinity of B_n(x). Else there exists w E B(x) s.t. w E/ Union from n=1 to infinity of B_n(x)

    So for every n >=1, [w E Union from n=1 to infinity of B_n(x)]^C or X(w) > X - 1/n.

    (***) Letting n --> infinity, we see

    X(w) >= x

    Consequently, if w E B(x)
    X(w) < x and X(w) >= x

    This is impossible and thereforefore B(x) = Union from n=1 to infinity of B_n(x) E F as B_n(x) E F for each n.

    (***) This is where I get completely lost.

    Also, ^C represents the complement
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