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    Help understand probability in a simply random sample

    Quoted is an extract for Sample Survey Principles and Methods, Vic Barnett(2002) Pg 34 The concept of probability averaging only arises in relation to some prescribed probability sampling schemes. Thus, for simple random sampling we have the concept of the expected value of y_i, the ith...
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    'Alternate' proof that the expected value of the sample mean is the population mean

    It would be appreciated if someone could verify that this makes sense. By definition \bar{x} = \frac{\sum x_i}{n} So taking its expectation we get \bar{x} = \frac{1}{n} E[\sum x_i] Now, as we have a population of size N and a sample size of size n, we have {N\choose n} different samples and...
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    Help with proof that expected value of x_i is the populaton mean X bar

    I'm having a little trouble with the proof that the expected value of x_i is \bar{X} . What I have is E[x_i]=\sum_{j=1}^{N}X_j Pr(x_i=X_j) Then Pr(x_i=X_j) = 1/N This is the bit I can't understand, how does that probability evaluate to that value. I know the...