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!