2
$\begingroup$

Going through a stats course and having some difficulties with point estimation. It makes sense in basic examples but when introducing "without replacement" I'm getting stuck.

A box has 4 numbered balls, 1,2,3 and 4. A sample of 2 is drawn without replacement. What is the expected sample mean?

For 1 ball its $\frac{1}{4} \sum X_i$ which is $2.5$. Simple.

Once one ball is extracted, the probabilities change for the others, but as I don't know which ball number was removed, I can't use the same formula.

Would really appreciate some advice and pointers. Thanks.

$\endgroup$
1
$\begingroup$

There are $\binom{4}{2} = 6$ ways to pick 2 balls from 4. Each pair thus has a $\frac{1}{6}$ probability of being picked. The possible averages are $\frac{1+2}{2}, \frac{1+3}{2},\frac{1+4}{2}, \frac{2+3}{2},\frac{2+4}{2},\frac{3+4}{2}$. Denoting the first ball by $X_1$ and the second by $X_2$, by the definition of expectation, \begin{align*} E[\frac{X_1+X_2}{2}] = \frac{1}{6} (3/2 + 4/2 + 5/2 + 5/2 + 6/2 + 7/2) \end{align*}

$\endgroup$
  • $\begingroup$ awesome, i was thinking about it incorrectly, i should have been working out teh probability of each combination! that has helped a lot! thanks! for large numbers of combinations this would be quite cumbersome, is there an area i should resarch for a more elegant way to do this, say if my set of numbered balls was from 1 to 10000000 ? $\endgroup$ – wilson_smyth Jun 4 '17 at 9:08
1
$\begingroup$

The expected sample mean will always be the actual mean (here, $2.5$). This is because the expected value of the first ball is $2.5$ and the expected value of the second ball (if you don't know what the first ball was) is also $2.5$. It is always true that $E[X_1+X_2]=E[X_1]+E[X_2]$, even when $X_1$ and $X_2$ are not independent, and so $E[\frac12(X_1+X_2)]=\frac12(E[X_1]+E[X_2])$. This works for any sample size.

$\endgroup$
0
$\begingroup$

Following Especially Lime's answer, one can show that the mean of a sample without replacement from a finite population is indeed an unbiased estimate of the population mean:

If $x_1, x_2, \dots x_N$ is the population and $X_1, X_2, \dots X_n$ is a sample without replacement where $n \leq N$, then $$ E\left[ \frac{1}{n} \sum^n_{i=1} X_i \right] = \mu $$ where $\mu =\frac{1}{N} \sum^N_{i=1}x_i$. Proof:

Let $1_{\lbrace x_i \in S \rbrace}$ be a random indicator variable indicating whether $x_i$ is in the sample $S$ of size $n$. We have that $$ \sum_{i=1}^{n} X_i = \sum_{i = 1}^N x_i 1_{\lbrace x_i \in S \rbrace} $$ Since $n$ and $N$ are fixed and every sample is equally likely, we know that $$ E[1_{\lbrace x_i \in S \rbrace}] = \frac{n}{N}. $$ Therefore, $$ E \left[\frac{1}{n} \sum_{i=1}^{n} X_i \right] = E \left[\frac{1}{n} \sum_{i = 1}^N x_i 1_{\lbrace x_i \in S \rbrace} \right] = \frac{1}{n} \sum_{i = 1}^N x_i E[1_{\lbrace x_i \in S \rbrace}] = \mu $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.