Here is the full question .
We are interested in finding an estimator for [tex]Var (X_i )[/tex] and propose to use :
[tex]\hat {V} = \bar {X}_n (1- \bar {X} )_n[/tex]
Now; we are interested in the basis of [tex]\hat V[/tex]
Compute :
[tex]E \ \ [ \bar V] - Var (X_i) =[/tex]
Using this; find an unbiased estimator [tex][ \bar V][/tex] for [tex]p(1-p) \ if \ n \geq 2[/tex]
Write [tex]bar \ x{_n} \ for \ X_n[/tex]
Answer:
Step-by-step explanation:
[tex]\bar X_n = \dfrac{1}{n} {\sum ^n _ {i=1} } \\ \\ E(X_i) = - \dfrac{1}{n=1} \sum p \dfrac{1}{n}*np = \mathbf{p}[/tex]
[tex]V(\bar X_n) = V ( \dfrac{1}{n_{i=1} } \sum ^n \ X_i )} = \dfrac{1}{n^2} \sum ^n_{i=1} Var (X_i) \\ \\ = \dfrac{1}{n^2} \ \sum ^n _{i=1} p(1-p) \\ \\ = \dfrac{1}{n^2}*np(1-p) \\ \\ = \dfrac{p(1-p)}{n}[/tex]
[tex]E( \bar X^2 _ n) = Var (\bar X_n) + [E(\bar X_n)]^2 \\ \\ = \dfrac{p(1-p)}{n}+ p \\ \\ = p^2 + \dfrac{p(1-p)}{n} \\ \\ \\ \hat V = \bar X_n (1- \bar X_n ) = \bar X_n - \bar X_n ^2 \\ \\ E [ \hat V] = E [ \bar X_n - \bar X_n^2] \\ \\ = E[\bar X_n ] - E [\bar X^2_n] \\ \\ = p-(p^2 + \dfrac{p(1-p)}{n}) \\ \\ = p-p^2 -\dfrac{p(1-p)}{n}[/tex]
[tex]=p(1-p)[1-\dfrac{1}{n}] = p(1-p)\dfrac{n-1}{n}[/tex]
[tex]Bias \ (\bar V ) = E ( \hat V) - Var (X_i) \\ \\ = p(1-p) [1-\dfrac{1}{n}] - p(1-p) \\ \\ - \dfrac{p(1-p)}{n}[/tex]
Thus; we have:
[tex]E [\hat V] = p(1-p ) \dfrac{n-1}{n}[/tex]
[tex]E [\dfrac{n}{n-1} \ \ \bar V] = p(1 -p)[/tex]
[tex]E [\dfrac{n}{n-1} \ \ \bar X_n (1- \bar X_n )] = p (1-p)[/tex]
Therefore;
[tex]\hat V ' = \dfrac{n}{n-1} \bar X_n (1- \bar X_n)[/tex]
[tex]\mathbf{ \hat V ' = \dfrac{n \bar X_n (1- \bar X_n)} {n-1}}[/tex]
