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Math Help - Central Limit Theorem

  1. #1
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    Central Limit Theorem

    Let X_1, X_2, ... be a sequence of independent random variables with E(X_i) = \mu_i and  Var(X_i) = (\sigma_i)^2. Show that if  n^{-1} \sum_{i=1}^{n} \mu_i \rightarrow \mu and n^{-2}\sum_{i=1}^{n}{\sigma_i}^2\rightarrow 0, then  \bar{X} \rightarrow \mu in probability.
    Last edited by h2osprey; February 6th 2010 at 11:08 PM.
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  2. #2
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by h2osprey View Post
    Let X_1, X_2, ... be a sequence of independent random variables with E(X_i) = \mu_i and  Var(X_i) = (\sigma_i)^2. Show that if  n^{-1} \sum_{i=1}^{n} \mu_i \rightarrow \mu, then  \bar{X} \rightarrow \mu in probability.
    you want P\left(\left|{\sum_{i=1}^n X_i\over n}-\mu\right|>\epsilon\right)\to 0 for all \epsilon>0

    Try adding and subtracting {\sum_{i=1}^n \mu_i\over n}

     P\left(\left|{\sum_{i=1}^n X_i\over n}-\mu\right|>\epsilon\right)

    \le  P\left(\left| {\sum_{i=1}^n X_i\over n} -{\sum_{i=1}^n \mu_i\over n}\right|>\epsilon/2\right) +P\left(\left| {\sum_{i=1}^n \mu_i\over n} -\mu\right|>\epsilon/2\right)
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    Quote Originally Posted by matheagle View Post
    you want P\left(\left|{\sum_{i=1}^n X_i\over n}-\mu\right|>\epsilon\right)\to 0 for all \epsilon>0

    Try adding and subtracting {\sum_{i=1}^n \mu_i\over n}

     P\left(\left|{\sum_{i=1}^n X_i\over n}-\mu\right|>\epsilon\right)

    \le  P\left(\left| {\sum_{i=1}^n X_i\over n} -{\sum_{i=1}^n \mu_i\over n}\right|>\epsilon/2\right) +P\left(\left| {\sum_{i=1}^n \mu_i\over n} -\mu\right|>\epsilon/2\right)
    Ah I suppose after that you just apply Chebyshev's inequality and the proof is done.

    Thing is, I'm not quite sure how the inequality you mentioned holds? I can see intuitively how it works, but is there a theorem or proof or something for that?

    EDIT: Nvm, figured it out. Thanks for everything!

    Also, I edited the question, left out the fact that the variances converge (I suppose this is for applying Chebyshev's inequality later.)
    Last edited by h2osprey; February 6th 2010 at 11:39 PM.
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  4. #4
    MHF Contributor matheagle's Avatar
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    I KNEW you left something out!
    Second term goes to zero since they are constants and the different goes to zero.
    And you square the absolute value and use markov/chebyshev's in the first.
    I didn't think this was complete.
    Either you needed that the sum of the variances divided by the square of n to zero
    OR another way was that all the variances were equal OR the sup of the variances was bounded....
    But something was off.
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    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by h2osprey View Post
    Thing is, I'm not quite sure how the inequality you mentioned holds? I can see intuitively how it works, but is there a theorem or proof or something for that?
    IF P(|A+B|>2c) then P(\{|A|>c\} \cup \{|B|>c\})

    So P(\{|A|>c\} \cup \{|B|>c\}) \le P(|A|>c)+P(|B|>c)
    Last edited by matheagle; February 7th 2010 at 08:03 AM.
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  6. #6
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    Yup that's what I got thanks a lot for the help!

    Sorry for the missing variances thing.. it was part of the previous question so I left it out by accident
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  7. #7
    MHF Contributor matheagle's Avatar
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    I couldn't finish your problem since it didn't seem correct, without some more info.
    That's why I stopped after using the triangle inequality.
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