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Random Variables: Convergence in Probability?

  1. Jun 29, 2011 #1
    Definition: Let X1,X2,... be a sequence of random variables defined on a sample space S. We say that Xn converges to a random variable X in probability if for each ε>0, P(|Xn-X|≥ε)->0 as n->∞.
    ====================================

    Now I don't really understand the meaning of |Xn-X| used in the definition. Is |.| here the usual absolute value when we talk about real numbers? But Xn and X are functions, not real numbers.
    Also, when we talk about the probability of something, that something has to be subsets of the sample space Ω, but |Xn-X|≥ε does not look like the description of a subset of Ω to me.
    A random variable X is a function mapping the sample space Ω to the set of real numbers, i.e. X: Ω->R.
    The random variable X is the function itself, and X(ω) are the VALUES of the function which are real numbers. Only when we talk about X(ω) does it make sense to talk about the absolute value of real nubmers.

    So I assume the notation used in the definition above is really a shorthand for
    P({ω E Ω: |Xn(ω)-X(ω)|≥ε})?

    In other words, the notations P(|Xn-X|≥ε) and P({ω E Ω: |Xn(ω)-X(ω)|≥ε}) are interchangable? Am I right?


    I hope someone can clarify this! Thanks a lot!:smile:
     
  2. jcsd
  3. Jun 29, 2011 #2

    micromass

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    Hi kingwinner! :smile:

    It seems you answered your own question:

    You are 100% correct about this.
     
  4. Jun 30, 2011 #3
    I see, thanks!
    So I believe the |.| used in |X_n -X| above is just meant to be the usual absolute value for real numbers, and not some fancy metric, right?
     
  5. Jun 30, 2011 #4

    micromass

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    It's just the absolute value, don't worry :smile:
     
  6. Jun 30, 2011 #5
    By the way, does it matter whether we have ≥ε OR >ε in the definition? Why or why not?
     
  7. Jul 1, 2011 #6

    micromass

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    No, it doesnt. If [itex]P(|X_n-X|>\varepsilon)[/itex] converges to zero, then so does [itex]P(|X_n-X|\geq \varepsilon)[/itex].

    The reason is

    [tex]P(|X_n-X|>\varepsilon)\leq P(|X_n-X|\geq \varepsilon)\leq P(|X_n-X|>\varepsilon/2)[/tex]

    So since epsilon is arbitrary, all the sequences will converge together.
     
  8. Jul 2, 2011 #7
    Got it! Thanks!
     
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