Variance Estimator in Simple Random Sampling Without Replacement

Let $(y_1,...,y_n)$ be a simple random sample without replacement from population $(x_1,...,x_N).$ Then the population mean and variance are, respectively, $$begin{align}mu:&={1over N}sum_{i=1}^Nx_i\ sigma^2:&={1over N}sum_{i=1}^N(x_i-mu)^2.end{align}$$ Following is a sketch of how to show that $$begin{align}Eleft({N-1over N}{1over n-1}sum_{i=1}^n(y_i-bar{y})^2right)=sigma^2.end{align}$$

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Aside: Some authors differ on the definition of "population variance", taking it to be the quantity $$S^2:={Nover N-1}sigma^2= {1over N-1}sum_{i=1}^N(x_i-mu)^2,$$ presumably to allow the above unbiasedness result to be written as follows:

$$begin{align}Eleft({1over n-1}sum_{i=1}^n(y_i-bar{y})^2right)=S^2.end{align}$$

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By the OP's identity (as originally posted, which is proved here),

$$begin{align}Eleft(frac{1}{n} sum_{i =1}^{n} (y_{i} - bar{y})^2right) &= frac{1}{2n^2} sum_{i =1}^{n}sum_underset{Large{jneq i}}{j=1}^{n} E(y_i - y_j)^2\ &={1over 2n^2} n(n-1)E(y_1-y_2)^2\ &={1over 2n^2} n(n-1)Eleft((y_1-mu)-(y_2-mu)right)^2\ &={1over 2n^2} n(n-1)Eleft((y_1-mu)^2+(y_2-mu)^2-2(y_1-mu)(y_2-mu)right)\ &={1over 2n^2} n(n-1),2(sigma^2-text{cov}(y_1,y_2))\ &={1over 2n^2} n(n-1),2(sigma^2-(-{sigma^2over N-1}))\[2ex] &={n-1over n}{Nover N-1}sigma^2. quadquadquadquadquadquadquadquadtext{QED}end{align}$$ In the above, the covariance term is obtained as follows, because each of the $N(N-1)$ possible outcomes for $(y_1-mu)(y_2-mu)$ is equally likely: $$begin{align}text{cov}(y_1,y_2) &=Eleft((y_1-mu)(y_2-mu)right)\ &=frac{1}{N(N-1)} sum_{i =1}^{N}sum_underset{Large{jneq i}}{j=1}^{N} (x_i-mu)(x_j-mu)\ &=frac{1}{N(N-1)} (-Nsigma^2)\ &=-{sigma^2over N-1} end{align}$$ where we have used $$sum_{i =1}^{N}sum_underset{Large{jneq i}}{j=1}^{N} (x_i-mu)(x_j-mu)=-Nsigma^2$$ which is a consequence of the following identity: $$begin{align}0^2=left(sum_{i=1}^N(x_i-mu)right)^2 &=sum_{i=1}^N(x_i-mu)^2 + sum_{i =1}^{N}sum_underset{Large{jneq i}}{j=1}^{N} (x_i-mu)(x_j-mu)tag{*}\ &=Nsigma^2 + sum_{i =1}^{N}sum_underset{Large{jneq i}}{j=1}^{N} (x_i-mu)(x_j-mu).end{align}$$

Note that (*) is just a special case (with $z_i=x_i-mu$, so $sum z_i=0$) of the general identity $$left(sum_{i=1}^N z_iright)^2 =sum_{i=1}^Nz_i^2 + sum_{i =1}^{N}sum_underset{Large{jneq i}}{j=1}^{N}z_iz_j. $$

Sources:

http://dept.stat.lsa.umich.edu/~moulib/sampling.pdf https://issuu.com/patrickho77/docs/mth_432a_-_introduction_to_sampling