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Two Probability Questions

  1. Oct 7, 2009 #1
    The problem statement, all variables and given/known data
    Let {Xi} be a sequence of uniform random variables on [0,1]. What is P(sup {Xi} = 1) and P(Xi = Xj), i ≠ j?

    The attempt at a solution
    If sup{Xi} = 1, then either Xi is 1 for some i or there is an increasing sequence of the Xi's that converges to 1. The probability of the former is 0. I don't know how to calculate the probability of the latter.

    For the second probability, if things were finite I would condition on the value of Xj and compute the probability that way. But in this case, I have no idea what to do.
     
  2. jcsd
  3. Oct 7, 2009 #2

    jbunniii

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    Consider [itex]P(\sup\{x_i\} \neq 1)[/itex]. If the supremum is not 1, it must be less than one, say [itex]\sup\{x_i\} = \alpha < 1[/itex]. If this is the case, then none of the [itex]x_i[/itex] can be greater than [itex]\alpha[/itex]. What is the probability of that?

    For the second question, note that [itex]x_i = x_j[/itex] if and only if [itex]x_i - x_j = 0[/itex].
     
  4. Oct 7, 2009 #3
    Good tip. For a particular i, [itex]P(X_i > \alpha) = 1 - P(X_i \le \alpha) = 1 - \alpha[/itex], so the probability that [itex]X_i < \alpha[/itex] for all i must be [itex]\prod_i (1 - \alpha) = 0[/itex]. Hmm...this can't be right.

    Right. I don't know why this didn't occur to me. Thanks.
     
  5. Oct 7, 2009 #4

    jbunniii

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    That's what I get. This shows that [itex]P(\sup\{x_i\} \leq \alpha) = 0[/itex] for all [itex]\alpha < 1[/itex]. So [itex]P(\sup\{x_i\} = 1)[/itex] must be 1.

    It makes sense that this must be the case. If I choose infinitely many uniform numbers in [0,1], why should ALL of these numbers avoid some interval [itex][\alpha,1][/itex] with nonzero probability? What's special about the rightmost edge of the interval? Nothing at all.
     
    Last edited: Oct 7, 2009
  6. Oct 7, 2009 #5
    You're right. My intuition about such things is, as you can tell, not very good. Thanks.
     
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