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

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

\(\hspace{10mm}=\bar{X}\)

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

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

\(\hspace{10mm}=\bar{X}\)

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