Question #152807


Let A = 5 Consider the population A , A+2 ,A+4 , A+6 , A+8 Draw all posible sample of size 3 without replacement.If A is odd number verily the biasedness property of sample variance S^2



1
Expert's answer
2020-12-28T13:06:54-0500

Samples of size 3 without replacement:

{5,7,9}; {5,7,11}; {5,7,13}; {5,9,11};{5,9,13}; {5,11,13}; {7,9,11}; {7,9,13}; {7,11,13}; {9,11,13}

If A is odd number verily the biasedness property of sample variance S2.

Sn2=1n(n1)ij(XiXj)22=1n(n1)((n1)i=1nXi2ijXiXj)S^2_n=\frac{1}{n(n-1)}\sum_{i\neq j} \frac{(X_i-X_j)^2}{2} = \frac{1}{n(n-1)}((n-1)\sum_{i=1}^n X_i^2-\sum_{i\neq j} X_iX_j)

The expectation of sample variance:

E[Sn2]=i=1nE[Xi2]nijE[XiXj]n(n1)E[S^2_n]=\frac{\sum_{i=1}^n E[X_i^2]}{n}-\frac{\sum_{i\neq j} E[X_i*X_j]}{n(n-1)}

Since we have independent variables:

E[XiXj]=E[Xi]E[Xj]E[X_iX_j]=E[X_i]*E[X_j]

Let assume that:

E[Xi2]=μ2E[X_i^2]=\mu_2 and E[Xi]=μ1E[X_i]=\mu_1

Then:

i=1nE[Xi2]=nμ2\sum_{i=1}^n E[X_i^2]=n*\mu_2 and ijE[Xi]E[Xj]=n(n1)μ12\sum_{i\neq j} E[X_i]E[X_j]=n(n-1)\mu_1^2

Finally:

E[Sn2]=μ2μ12=Var[X]E[S_n^2]=\mu_2-\mu_1^2=Var[X]

So, S2 is unbiased.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS