91Ó°ÊÓ

Using a long rod that has length \(\mu\), you are going to lay out a square plot in which the length of each side is \(\mu\). Thus the area of the plot will be \(\mu^{2}\). However, you do not know the value of \(\mu\), so you decide to make \(n\) independent measurements \(X_{1}, X_{2}, \ldots X_{n}\) of the length. Assume that each \(X_{i}\) has mean \(\mu\) (unbiased measurements) and variance \(\sigma^{2}\). a. Show that \(\bar{X}^{2}\) is not an unbiased estimator for \(\mu^{2}\). \(\left[\right.\) Hint: For any rv \(Y, E\left(Y^{2}\right)=V(Y)+[E(Y)]^{2}\). Apply this with \(Y=\bar{X}\).] b. For what value of \(k\) is the estimator \(\bar{X}^{2}-k S^{2}\) unbiased for \(\mu^{2}\) ? [Hint: Compute \(E\left(\bar{X}^{2}-k S^{2}\right)\).]

Step by step solution

01

Understanding the Problem

We need to check if \(\bar{X}^2\) is an unbiased estimator for \(\mu^2\). Since we know \(\bar{X}\) is an unbiased estimator for \(\mu\), we start by confirming \(E(\bar{X}^2)\).

02

Calculate the Expectation of \(\bar{X}^2\)

\(\bar{X}^2\) is the square of the sample mean \(\bar{X} = \frac{1}{n} \sum_{i=1}^n X_i\). So, \(E(\bar{X}^2) = V(\bar{X}) + [E(\bar{X})]^2\), \(V(\bar{X}) = \frac{\sigma^2}{n}\), and \([E(\bar{X})]^2 = \mu^2\). Thus, \(E(\bar{X}^2) = \frac{\sigma^2}{n} + \mu^2\).

03

Conclusion for Part a

Since \(E(\bar{X}^2) = \frac{\sigma^2}{n} + \mu^2 eq \mu^2\), \(\bar{X}^2\) is not an unbiased estimator for \(\mu^2\).

04

Setting Up the Second Estimator

We want \(E(\bar{X}^2 - k S^2) = \mu^2\). Start by writing \(E(\bar{X}^2 - k S^2) = E(\bar{X}^2) - kE(S^2)\).

05

Calculate \(E(S^2)\)

The variance of the sample, \(S^2 = \frac{1}{n-1} \sum_{i=1}^n (X_i - \bar{X})^2\). \(E(S^2) = \sigma^2\).

06

Solve for k

From \(E(\bar{X}^2) - kE(S^2) = \mu^2\), substitute in the values to get \(\frac{\sigma^2}{n} + \mu^2 - k\sigma^2 = \mu^2\). Solving for \(k\), we have \(k = \frac{1}{n}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sample Variance

The concept of sample variance, denoted as \(S^2\), is crucial in statistics because it measures the variability in a sample data set. It quantifies how much each sample point differs from the sample mean. Calculating sample variance involves averaging the squared differences from the sample mean. For \(n\) observations, the formula is:
\[S^2 = \frac{1}{n-1} \sum_{i=1}^n (X_i - \bar{X})^2\]This formula uses \(n-1\) instead of \(n\) in the denominator, a method known as "Bessel's correction," ensuring that \(S^2\) is an unbiased estimate of the population variance. Without this correction, the calculation can underestimate the actual variance, particularly when dealing with small sample sizes.

• Sample variance is vital for understanding data dispersion in a dataset.
• It ensures a better representation of the variability of the population from which the sample is drawn.

Expectation

Expectation, often denoted as \(E\), is a key statistical concept that describes the average outcome if an experiment or process is repeated numerous times. When dealing with random variables, expectation helps predict what we can generally anticipate as a result. For a random variable \(Y\) with possible outcomes \(y_1, y_2, \ldots, y_k\) and respective probabilities \(p_1, p_2, \ldots, p_k\), the expectation is calculated as:
\[E(Y) = \sum_{i=1}^k y_i p_i\]In the context of the provided exercise, this concept helps evaluate the unbiased nature of estimators like \(\bar{X}^2\), where it becomes essential to understand whether the expectation equals the true parameter value.

• Expectation provides a foundation for evaluating estimator bias.
• It simplifies complex distributions into single values for easier interpretation.

Sample Mean

The sample mean, denoted as \(\bar{X}\), is one of the most straightforward and widely used statistical tools. It represents the central value of a sample from a larger population. Calculating the sample mean involves summing all observed values and dividing by the number of observations \(n\):
\[\bar{X} = \frac{1}{n} \sum_{i=1}^n X_i\]The sample mean serves as an unbiased estimator of the population mean \(\mu\), meaning that the expected value of the sample mean equals the population mean.

• The sample mean helps summarize data with a single metric indicating central tendency.
• It is foundational in numerous statistical analyses, providing insight into potential population characteristics based on samples.

Variance of Sample Mean

The variance of the sample mean \(V(\bar{X})\) measures the expected variability of sample means around the true population mean \(\mu\). It indicates how much we expect different sample means to differ from each other, if we draw multiple samples of size \(n\) from the same population.
The variance is calculated using the formula:
\[V(\bar{X}) = \frac{\sigma^2}{n}\]where \(\sigma^2\) is the population variance. This expression shows that even though each sample's observations may have variance \(\sigma^2\), the average (\(\bar{X}\)) is more stable, as its variance decreases with larger sample sizes.

• Larger sample sizes result in a smaller variance for the sample mean, making the mean a more reliable estimate of the population mean.
• Understanding the variance of the sample mean is essential for assessing the precision of the sample mean as an estimator.