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Math Help - Stuck at the therem

  1. #1
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    Stuck at the therem

    Hi guys i am trying to solve the following theorem

    Theorem : If X_1,X_2,...X_n constitute the random sample from an infinite population with the mean \mu and variance \sigma^2, then

    E[\bar X] = \mu and var(\bar X) = \frac{\sigma^2}{n}

    Solution: Let Y = \bar X
    Y = \sum a_i \ X_i

    \therefore E[\bar X] = E \left( \sum a_i \ X_i \right)

    Let a_i = \frac{1}{n} and as E[\bar X] = \mu therefore substituting these terms in the above equation we get

    E[\bar X] = \sum_{n=1}^{\infty} \frac{1}{n} \mu

    Now after this in the book it shows this step

    \sum_{n=1}^{\infty} \frac{1}{n} \mu = n. \frac{1}{n} \mu...............How did he get 'n' instead of \sum_{n=1}^{\infty}
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  2. #2
    MHF Contributor matheagle's Avatar
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    Because \bar X={ \sum_{i=1}^n X_i \over n}

    The sample mean is the average of n, not an infinite number of observations.
    and it's THEOREM
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    Quote Originally Posted by matheagle View Post
    Because \bar X={ \sum_{i=1}^n X_i \over n}

    The sample mean is the average of n, not an infinite number of observations.
    and it's THEOREM
    \sum_{n=1}^{\infty} \frac{1}{n} \mu = n . \frac{1}{n} \mu
    you didnt understand my question . I want to knw from where did the 'n' come from
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  4. #4
    Moo
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    Quote Originally Posted by wolfyparadise View Post
    \sum_{n=1}^{\infty} \frac{1}{n} \mu = n . \frac{1}{n} \mu
    you didnt understand my question . I want to knw from where did the 'n' come from
    Because you're summing n times a 'constant' (which doesn't depend on k)
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  5. #5
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    \sum_{n=1}^3k=k+k+k=3k

    \sum_{n=1}^{100}k=100k

    \sum_{n=1}^{100}\mu=100\mu
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  6. #6
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by wolfyparadise View Post
    \sum_{n=1}^{\infty} \frac{1}{n} \mu = n . \frac{1}{n} \mu
    you didnt understand my question . I want to knw from where did the 'n' come from
    I certainly DID understand your question.
    YOU don't have the correct definition of the sample mean.

    The sample mean is not an infinite sum

    \bar X={X_1+\cdots + X_n \over n}

    and

    E(\bar X)={E(X_1)+\cdots + (X_n) \over n}

    ={\mu+\cdots + \mu \over n}={n\mu\over n}=\mu
    Last edited by matheagle; January 2nd 2010 at 04:29 PM.
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  7. #7
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    thanks
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