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

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

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

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

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

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

    Now after this in the book it shows this step

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

    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 $\displaystyle \bar X={ \sum_{i=1}^n X_i \over n}$

    The sample mean is the average of n, not an infinite number of observations.
    and it's THEOREM
    $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n} \mu$ = n $\displaystyle . \frac{1}{n} \mu$
    you didnt understand my question . I want to knw from where did the 'n' come from
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  4. #4
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    Quote Originally Posted by wolfyparadise View Post
    $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n} \mu$ = n $\displaystyle . \frac{1}{n} \mu$
    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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    $\displaystyle \sum_{n=1}^3k=k+k+k=3k$

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

    $\displaystyle \sum_{n=1}^{100}\mu=100\mu$
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  6. #6
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by wolfyparadise View Post
    $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n} \mu$ = n $\displaystyle . \frac{1}{n} \mu$
    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

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

    and

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

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