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Thread: 'Alternate' proof that the expected value of the sample mean is the population mean

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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 of those, {N-1\choose n-1} contain each of the values X_1, X_2,...,X_N.

    Then clearly,

    \sum{x_i}={N-1\choose n-1}\sum{X_j}

    So, the expectation of \sum{x_i} will be given by,

    E[\sum{x_i}]=\frac{{N-1\choose n-1}\sum{X_j}}{{N\choose n}}

    \hspace{17mm}=\frac{n}{N} \sum{X_j}

    Therefore

    E[\bar{x}]=\frac{1}{n}(\frac{n}{N} \sum{X_j})

    \hspace{10mm}=\bar{X}
    Last edited by sid9221; 04-06-2014 at 10:56 AM.

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    Re: 'Alternate' proof that the expected value of the sample mean is the population me

    You aren't being very clear about what the summations are over and what terms are involved. I also don't understand what you're trying to say when you claim that
    Quote Originally Posted by sid9221 View Post
    {N-1\choose k-1} contain each of the values X_1, X_2,...,X_N.
    What? You have from X1 up to Xn but the count implies you're looking at the number of combinations of size k-1 from a set of size N-1.

    Overall I think you need to be more explicit in what you're referring to which might clear up some of the ambiguity that I see in the proof.
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    Re: 'Alternate' proof that the expected value of the sample mean is the population me

    Quote Originally Posted by Dason View Post
    You aren't being very clear about what the summations are over and what terms are involved. I also don't understand what you're trying to say when you claim that


    What? You have from X1 up to Xn but the count implies you're looking at the number of combinations of size k-1 from a set of size N-1.

    Overall I think you need to be more explicit in what you're referring to which might clear up some of the ambiguity that I see in the proof.
    Apologies, the k's were all typos, then were meant to be n's. Does it make sense now.

    I'm working on the 'Latex' error bit. Its meant to be [[N-1 Choose n-1 ]*sum X_j]/[N Choose n]

    The summations for x_i go from 1..n and for X_j go from 1..N. Trying to update the latex versions, doesn't seem to be working..

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    Re: 'Alternate' proof that the expected value of the sample mean is the population me


    I guess I understand your question a little bit more now.

    In your question, you have a population with N distinct elements:

    \{x_1, x_2, \ldots, x_N\}

    Now sampling n \in \{1, 2, \ldots, N\} elements out of N without replacement, forming a set of random sample

    \{X_1, X_2, \ldots, X_n\}

    with the assumption that each element is equally likely to be sampled.

    Now, the population mean is defined by

    \mu = \frac {1} {N} \sum_{i=1}^N x_i

    and the sample mean is defined by

    \bar{X} = \frac {1} {n} \sum_{i=1}^n X_i

    OP may want to calculate the expected value of the sample mean without invoking the linearity of expectation, but just rely on the first principle with combinatorics argument.

    As mentioned by OP, the number of ways to form such a random sample is M = \binom {N} {n} and by the equally likely assumption, each way has a equal probability of \frac {1} {M}

    So the expected value may be expressed like this:

    E[\bar{X}] = \frac {1} {M} \sum_{j=1}^M \frac {1} {n} \sum_{i=1}^n x_i^{(j)} = \frac {1} {Mn} \sum_{i=1}^N a_ix_i

    where a_i are the number of times that the elements x_i appear in the above summation.

    The number of ways of forming a random set of n elements, with the condition that a particular elements x_i must be included in the set, is

    \binom {N - 1} {n - 1}

    and thus we have

    a_i = \binom {N - 1} {n - 1}, i = 1, 2, \ldots, N

    Finally, putting these things together,

    E[\bar{X}] 
= \frac {\displaystyle \binom {N - 1} {n - 1}} {\displaystyle  \binom {N} {n} \times n}\sum_{i=1}^N x_i = \frac {1} {N} \sum_{i=1}^N x_i = \mu

    and this should be what OP trying to argue for.

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