Chapter 7, Survey Sampling

Solution 51


To prove that bias of is of the order of , we need to show contains one term equals to and remaining terms contains or higher powers in denominator.

Given:

Proof:

Last step is reduced by removing summation and division by because every terms occurs times in the sum.

Thus all the terms contains or higher powers in the denominator.

$$\tag*{$\blacksquare$} $$