https://file.aiquickdraw.com/mathbot/file/1725399711664kz51kr0h.jpg

The problem provided deals with the estimation of \(\mu = E(Y)\), where \(Y\) is a random variable.

### Problem Breakdown:
Given that \(n\) is even, the estimator for \(\mu\) is defined as:
\[
\hat{\mu}_n = \frac{1}{n} \left[ \frac{Y_1 + Y_2}{2} + \frac{Y_3 + Y_4}{2} + \ldots + \frac{Y_{n-1} + Y_n}{2} \right],
\]
where \(Y_i\) are independent and identically distributed (i.i.d.) with \(Y_i \sim IID \ Y\).

The tasks are:
1. Show that \(\hat{\mu}_n\) is unbiased.
2. Derive \(\text{Var}(\hat{\mu}_n)\).
3. Compare \(\hat{\mu}_n\) with the estimator \(\hat{\mu}_n^* = \frac{1}{n}(Y_1 + Y_2 + \ldots + Y_n)\) and determine which one is better, providing a reason.

### Solution:

#### Part (a) - Unbiasedness of \(\hat{\mu}_n\):

To show that \(\hat{\mu}_n\) is an unbiased estimator of \(\mu\), we need to calculate the expected value of \(\hat{\mu}_n\) and show that it equals \(\mu\).

\[
\hat{\mu}_n = \frac{1}{n} \left[ \frac{Y_1 + Y_2}{2} + \frac{Y_3 + Y_4}{2} + \ldots + \frac{Y_{n-1} + Y_n}{2} \right]
\]

This can be rewritten as:

\[
\hat{\mu}_n = \frac{1}{n} \left[ \frac{1}{2} \sum_{i=1}^{n} Y_i \right] = \frac{1}{2n} \sum_{i=1}^{n} Y_i
\]

Taking the expectation:

\[
E[\hat{\mu}_n] = \frac{1}{2n} \sum_{i=1}^{n} E[Y_i] = \frac{1}{2n} \cdot n \cdot E[Y] = \frac{n}{2n} \cdot \mu = \frac{1}{2} \cdot \mu = \mu
\]

Since the expected value \(E[\hat{\mu}_n] = \mu\), \(\hat{\mu}_n\) is an unbiased estimator of \(\mu\).

#### Part (b) - Variance of \(\hat{\mu}_n\):

Next, we calculate the variance of \(\hat{\mu}_n\):

\[
\text{Var}(\hat{\mu}_n) = \text{Var}\left(\frac{1}{2n} \sum_{i=1}^{n} Y_i\right)
\]

Since the \(Y_i\) are i.i.d., we have:

\[
\text{Var}(\hat{\mu}_n) = \frac{1}{(2n)^2} \sum_{i=1}^{n} \text{Var}(Y_i) = \frac{1}{4n^2} \cdot n \cdot \sigma^2 = \frac{\sigma^2}{4n}
\]

Where \(\sigma^2\) is the variance of \(Y\).

#### Part (c) - Comparison of \(\hat{\mu}_n\) and \(\hat{\mu}_n^*\):

Now, consider the variance of the estimator \(\hat{\mu}_n^*\):

\[
\text{Var}(\hat{\mu}_n^*) = \text{Var}\left(\frac{1}{n} \sum_{i=1}^{n} Y_i\right) = \frac{1}{n^2} \sum_{i=1}^{n} \text{Var}(Y_i) = \frac{1}{n^2} \cdot n \cdot \sigma^2 = \frac{\sigma^2}{n}
\]

Comparing the variances:
- \(\text{Var}(\hat{\mu}_n) = \frac{\sigma^2}{4n}\)
- \(\text{Var}(\hat{\mu}_n^*) = \frac{\sigma^2}{n}\)

Since \(\frac{\sigma^2}{4n} < \frac{\sigma^2}{n}\), the variance of \(\hat{\mu}_n\) is smaller, indicating that \(\hat{\mu}_n\) is a more efficient (i.e., better) estimator than \(\hat{\mu}_n^*\) based on the criterion of having a lower variance.

### Summary:
- **(a)** \(\hat{\mu}_n\) is unbiased.
- **(b)** \(\text{Var}(\hat{\mu}_n) = \frac{\sigma^2}{4n}\).
- **(c)** \(\hat{\mu}_n\) is better than \(\hat{\mu}_n^*\) because it has a smaller variance, making it more efficient.

Would you like further details on any of these parts?

### Relative Questions:
1. How does the efficiency of an estimator relate to its variance?
2. What conditions must hold for an estimator to be considered unbiased?
3. How does sample size affect the variance of an estimator?
4. Can an estimator with higher variance ever be considered better in certain contexts? If so, how?
5. What are some examples of other unbiased estimators in statistics?

### Tip:
When comparing estimators, always consider both bias and variance. An estimator with low variance but a significant bias may not always be preferable over a less biased, slightly higher variance alternative.

Consider the estimator of μ = E(Y) defined for even n as μ̂_n = (1/n) * [1/2 * (Y_1 + Y_2) + 1/2 * (Y_3 + Y_4) + ... + 1/2 * (Y_{n-1} + Y_n)] where Y_i ~ IID Y. a) Show that μ̂_n is unbiased. b) Derive var(μ̂_n). c) Which estimator of μ is better, μ̂_n or μ̂_n* = (1/n) * (Y_1 + Y_2 + ... + Y_n), Y_i ~ IID Y? Why?

Explore the unbiasedness and variance comparison between two estimators of μ = E(Y). Understand the key mathematical concepts and formulas involved in deriving the efficiency of estimators. Suitable for undergraduate statistics students.

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