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Math Help - Help with problem

  1. #1
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    Help with problem

    Hello,

    I have a doubt regarding the attached question. I've tried to figure out a way for the question which is to use the law of large numbers but there is a problem. LLN is mainly used for I.I.D but the question says only independent variables so i guess i cannot use the LLN...is there any other way to do this question??
    Please help.

    Thanks alot!
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  2. #2
    Moo
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    Hello,
    Quote Originally Posted by daknight1987 View Post
    Hello,

    I have a doubt regarding the attached question. I've tried to figure out a way for the question which is to use the law of large numbers but there is a problem. LLN is mainly used for I.I.D but the question says only independent variables so i guess i cannot use the LLN...is there any other way to do this question??
    Please help.

    Thanks alot!
    Haha, I know this problem ^^

    Use Chebyshev's inequality for \mathbb{P}\left(\left|\frac{\sum_{i=1}^n X_i}{n}-\mu\right|>\varepsilon\right) !


    You'll need to show that \mathbb{E}\left(\frac{\sum_{i=1}^n X_i}{n}\right)=\mu, which is very easy by using the linearity of the expectation.
    You'll also need to remember that \text{var}\left(\sum_{i=1}^n X_i\right)=\sum_{i=1}^n \text{var}(X_i), because the X_i's are independent.
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  3. #3
    MHF Contributor matheagle's Avatar
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    Don't have a cow, Moo, but the rv's don't have the same mean.

    \mu is the limit of the average of n means, i.e., \mu=\lim_{n\to\infty}{\sum_{k=1}^n \mu_k\over n} .
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  4. #4
    Moo
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    Yeah sorry, I didn't see that ><


    So then write :
    \frac{\sum_{i=1}^n X_i}{n}-\mu=\frac{\sum_{i=1}^n X_i}{n}-\frac{\sum_{i=1}^n \mu_i}{n}+\frac{\sum_{i=1}^n \mu_i}{n}-\mu

    Hence :
    \left|\frac{\sum_{i=1}^n X_i}{n}-\mu\right|\leq \left|\frac{\sum_{i=1}^n X_i}{n}-\frac{\sum_{i=1}^n \mu_i}{n}\right|+\left|\frac{\sum_{i=1}^n \mu_i}{n}-\mu\right|

    Now, use Chebyshev's inequality and the fact that \frac{\sum_{i=1}^n \mu_i}{n} converges (almost surely) to \mu
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  5. #5
    MHF Contributor matheagle's Avatar
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    since when do you apologize?
    is this a first?
    and the \mu_i's and \mu are just constants so they converge anyway you wish.
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  6. #6
    Moo
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    Quote Originally Posted by matheagle View Post
    since when do you apologize?
    is this a first?
    and the \mu_i's and \mu are just constants so they converge anyway you wish.
    They converge when n goes to infinity. That's what the text says.

    I said "almost surely", because it will help dealing with the epsilon's.



    As for apologizing : you don't know me enough to be able to tell !
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