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    A random sample (X1, X2, X3...Xn) is taken from a population of unknown mean \mu and variance \sigma^2

    I have to find the bias of \mu* where \mu*= 	\sum Xi / (n-1)

    The bias is = E\mu*- \mu* , right?
    So I have:  1/(n-1) E( \sum Xi)- \sum Xi / (n-1)

    But I'm not really sure what to do after that? I have to find the variance and mse as well after but need the bias first...

    Any help appreciated, my stats lecturer is terrible and I've never done this topic before

    cheers!
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    (Original post by narc)
    The bias is = E\mu*- \mu* , right?
    No, it's E(mu*) - mu.

    So I have:  1/(n-1) E( \sum Xi)- \sum Xi / (n-1)
    Well, the hard part is working out E( \sum X_i) right? So why don't you think about what the sum of X_i is?

    \sum X_i = X_1 + X_2 + ... X_n surely?

    So you need to work out E(\mu*) = \frac{1}{n-1}E(\sum X_i) = \frac{1}{n-1}E(X_1 + X_2 + ... + X_n)
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    (Original post by Swayum)
    No, it's E(mu*) - mu.



    Well, the hard part is working out E( \sum X_i) right? So why don't you think about what the sum of X_i is?

    \sum X_i = X_1 + X_2 + ... X_n surely?

    So you need to work out E(\mu*) = \frac{1}{n-1}E(\sum X_i) = \frac{1}{n-1}E(X_1 + X_2 + ... + X_n)
    Hmm, do I assume that  E X_i = \mu because the sample is random?
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    I have no clue to be honest.
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    Ah. Well...thank you anyway
 
 
 
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