Question

We are interested in finding an estimator for Var (Xi), and propose to use V=-n (1-Xn). 0/2 puntos (calificable) Now, we are interested in the bias of V. Compute: E [V]-Var (Xi)-[n Using this, find an unbiased estimator V for p (1 - p) if n22. rite barX_n for Л n . 72 1--X 7t

Answer 1

Here is the full question .

We are interested in finding an estimator for
`Var (X_i )` and propose to use :

`\hat {V} = \bar {X}_n (1- \bar {X} )_n`

Now; we are interested in the basis of
`\hat V`

Compute :

`E \ \ [ \bar V] - Var (X_i) =`

Using this; find an unbiased estimator
`[ \bar V]` for
`p(1-p) \ if \ n \geq 2`

Write
`bar \ x{_n} \ for \ X_n`

Answer:

Explanation:

`\bar X_n = (1)/(n) {\sum ^n _ {i=1} } \\ \\ E(X_i) = - (1)/(n=1) \sum p (1)/(n)*np = \mathbf{p}`

`V(\bar X_n) = V ( (1)/(n_(i=1) ) \sum ^n \ X_i )} = (1)/(n^2) \sum ^n_(i=1) Var (X_i) \\ \\ = (1)/(n^2) \ \sum ^n _(i=1) p(1-p) \\ \\ = (1)/(n^2)*np(1-p) \\ \\ = (p(1-p))/(n)`

`E( \bar X^2 _ n) = Var (\bar X_n) + [E(\bar X_n)]^2 \\ \\ = (p(1-p))/(n)+ p \\ \\ = p^2 + (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 + (p(1-p))/(n)) \\ \\ = p-p^2 -(p(1-p))/(n)`

`=p(1-p)[1-(1)/(n)] = p(1-p)(n-1)/(n)`

`Bias \ (\bar V ) = E ( \hat V) - Var (X_i) \\ \\ = p(1-p) [1-(1)/(n)] - p(1-p) \\ \\ - (p(1-p))/(n)`

Thus; we have:

`E [\hat V] = p(1-p ) (n-1)/(n)`

`E [(n)/(n-1) \ \ \bar V] = p(1 -p)`

`E [(n)/(n-1) \ \ \bar X_n (1- \bar X_n )] = p (1-p)`

Therefore;

`\hat V ' = (n)/(n-1) \bar X_n (1- \bar X_n)`

`\mathbf{ \hat V ' = \frac{n \bar X_n (1- \bar X_n)} {n-1}}`