Expected value nd variance of mean estimator

[safina]

Homework Statement

A sample of size n is drawn from a population having N units by simple random sampling without replacement. A sub-sample of size n_{1} units is drawn from the n units by simple random sampling without replacement. Let \bar{y_{1}} denote the mean based on n_{1} units and \bar{y_{2}} based on (n-n_{1}) units.
Consider the estimator \hat{\overline{Y}} = w\bar{y_{1}} + (1-w)\bar{y_{2}}.
Show that E[\hat{\overline{Y}}] =\overline{Y} and obtain its variance.

Homework Equations

The Attempt at a Solution

E[\hat{\overline{Y}}] = E[w\bar{y_{1}} + (1-w)\bar{y_{2}}]
= w E[\bar{y_{1}}] + (1-w) E[\bar{y_{2}}]
= w\overline{Y}_{1} + (1-w)\overline{Y}_{2}
Why I did not arrive at the rigth answer which is \overline{Y}?

 

[vela]

What exactly do you mean by \bar{Y}, \bar{Y}_1, and \bar{Y}_2?

 

[safina]

What exactly do you mean by \bar{Y}, \bar{Y}_1, and \bar{Y}_2?

They are not stated in the problem, but I think \bar{Y} is the overall mean, \bar{Y}_1 is the mean of n1 units, and \bar{Y}_2?[/QUOTE] is the mean of the remaining (n-n1) units

 

[vela]

Then what's the difference between \bar{y}_1 and \bar{Y}_1?

 

[safina]

Then what's the difference between \bar{y}_1 and \bar{Y}_1?

I am sorry, I mean \bar{y}_1 is the mean of the n1 units and \bar{y}_2 is the mean of the remaining (n-n1) units.

 

[vela]

OK, let's try this instead. You have

\bar{y}_1 = \frac{1}{n_1}\sum_{i=1}^{n_1} y_i

So evaluate its expected value:

E[\bar{y}_1] = E\left[\frac{1}{n_1}\sum_{i=1}^{n_1} y_i\right] = \cdots\,?

What do you get?

 

[safina]

Ok, I figured them out.

E\left[\widehat{\bar{Y}}\right]=E\left[w\bar{y_{1}}+(1-w)\bar{y_{2}}\right]
=wE\left[\bar{y_{1}}\right]+(1-w)E\left[\bar{y_{2}}\right]
=\left(\frac{n_{1}}{n}\right) E\left[\frac{1}{n_{1}}\sum^{n_{1}}_{1}y_{i}\right] +\left(\frac{n-n_{1}}{n}\right)E\left[\frac{1}{n-n_{1}}\sum^{n}_{n_{1}+1}y_{i}\right]
=\frac{1}{n}\left[E\left\{\sum^{n_{1}}_{1}y_{i}+\sum^{n}_{n_{1}+1}y_{i}\right\}\right]
=\frac{1}{n}\sum^{n}_{1}E\left[y_{i}\right]
=\frac{1}{n}\sum^{n}_{1}\overline{Y}
=\overline{Y}

Thank you so much Vela!