Check Cochran,Sampling Techniques 3rd ed. Best wishes, David Booth

If the population is finite iof size N then the population is of the form

{ Y1 , Y2 , Y3 , ....................... , YN }

Now, if { y1 , y2 , ............... , yn } is a random sample (with out replacement) of size n drawn from this finite population then the sample mean square given by { Square of (y1 - m) + Square of (y2 - m) + ............... ............... + Square of (yn - m) } / (n - 1) where m is the sample mean,

tends to the population mean square

{ Square of (Y1 - M) + Square of (Y2 - M) + ..........................

.............. + Square of (YN - M) } / (N - 1)

but does not tend to the population variance

{ Square of (Y1 - M) + Square of (Y2 - M) + ..........................

.............. + Square of (YN - M) } / N

Therefore, how can the sample mean square be consistent estimator of the population variance?

The answer of this question is not available in the book

"Sampling Techniques" 3rd ed.

by Cochran.

If

{ y1 , y2 , ............... , yn }

is a random sample (with out replacement) of size n drawn from a finite population

{ Y1 , Y2 , Y3 , ....................... , YN }

of size N then the sample mean square given by

{ Square of (y1 - m) + Square of (y2 - m) + ...............

................ + Square of (yn - m) } / (n - 1)

where m is the sample mean,

is no doubt an unbiased estimator of the population variance since it has been proved that

E( sample mean square) = population variance .

But, my point is whether this sample mean square is a consistent estimator of the population variance.

For the sample mean square to be consistent estimator of the population variance, as per the logic behind the consistency, it should be equal to the population variance given by

{ Square of (Y1 - M) + Square of (Y2 - M) + ......................

.............. + Square of (YN - M) } / N when the sample contains all the elements of the population.

But in this case, when all the elements of the population are considered to belong to the sample then the sample mean square becomes equal to the population mean square given by

{ Square of (Y1 - M) + Square of (Y2 - M) + ...................... .............. + Square of (YN - M) } / (N - 1) but does not become equal to the population variance. Therefore, the sample mean square can not be consistent estimator of the population variance in this case.

This is the reason behind my question.

@ Theo K. Dijkstra,,

The sample variance given by

{ Square of (y1 - m) + Square of (y2 - m) + ...............

................ + Square of (yn - m) } / n

where m is the sample mean,

is no doubt a consistent estimator of the population variance whether the population is finite or infinite.

But, the sample mean square given by

{ Square of (y1 - m) + Square of (y2 - m) + ...............

................ + Square of (yn - m) } / (n - 1)

is, as I have found, not a consistent estimator of the population variance if the population is finite.

Of course, if the population is infinite then the sample mean square given above will be a consistent estimator of the population variance.

I have mathematically shown in my earlier answer that if

{ y1 , y2 , ............... , yn }

is a random sample drawn from a finite population then it is not correct to accept the sample mean square as a consistent estimator of the population variance.

However, it can be mathematically shown that in this case it is correct to accept the sample variance as a consistent estimator of the population variance.

Now, I invite opinions from the thinkers/researchers on what happens about the sample mean square and sample variance in this case if the parent population is an infinite one.