$x_i$ - value of the ith observation in the sample

$X_j$ - value of the jth observation in the population

$\bar{x}$ is the sample mean

$\bar{X}$ is the population mean

This follows directly from an earlier question Help with proof that the expected value of $x_i$ is $\bar{X}$

I was hoping I could clarify that this alternative proof is consistent. Particularly when the expectation of $E[x_i]$ is defined as $\frac{{N-1\choose n-1}\sum{X_j}}{N\choose n}$

By definition $\bar{x} = \frac{\sum x_i}{n}$

So taking its expectation we get


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,

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

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


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




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