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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 \(\mu^{2}\). [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 \(\mu^{2}\) ?

Short Answer

Expert verified
a. \(\bar{X}^2\) is not unbiased; b. \(k = \frac{1}{n}\).

Step by step solution

01

Define the Sample Mean

The sample mean \(\bar{X}\) is the average of \(n\) independent measurements: \(\bar{X} = \frac{1}{n} \sum_{i=1}^{n} X_i\). Given that each \(X_i\) is unbiased with mean \(\mu\), the expectation of the sample mean is \(E(\bar{X}) = \mu\).
02

Calculate the Expectation of the Square of the Sample Mean

Using the hint, \(E(\bar{X}^2) = V(\bar{X}) + [E(\bar{X})]^2\). The variance \(V(\bar{X})\) is given by \(\frac{\sigma^2}{n}\) because each \(X_i\) is independent with variance \(\sigma^2\). Therefore, \(E(\bar{X}^2) = \frac{\sigma^2}{n} + \mu^2\).
03

Determine If \(\bar{X}^2\) is an Unbiased Estimator

An estimator is unbiased for \(\mu^2\) if \(E(\bar{X}^2) = \mu^2\). From Step 2, \(E(\bar{X}^2) = \frac{\sigma^2}{n} + \mu^2\). The term \(\frac{\sigma^2}{n}\) indicates that \(\bar{X}^2\) is not an unbiased estimator for \(\mu^2\).
04

Use the Given Form to Create an Unbiased Estimator

We want \(E(\bar{X}^2 - kS^2) = \mu^2\). The sample variance \(S^2\) is \(\frac{1}{n-1}\sum_{i=1}^{n}(X_i - \bar{X})^2\) and \(E(S^2) = \sigma^2\).
05

Calculate the Expectation of the Adjusted Estimator

Substitute into the unbiased condition: \(E(\bar{X}^2 - kS^2) = E(\bar{X}^2) - kE(S^2) = \frac{\sigma^2}{n} + \mu^2 - k\sigma^2 = \mu^2\).
06

Solve for \(k\)

Set \( \frac{\sigma^2}{n} + \mu^2 - k\sigma^2 = \mu^2\) and solve for \(k\): \(k\sigma^2 = \frac{\sigma^2}{n}\). Thus, \(k = \frac{1}{n}\).

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Key Concepts

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

Sample Mean
The sample mean, represented as \( \bar{X} \), is a central concept in statistics. It is the average of a set of observations and offers a straightforward estimate of the population mean \( \mu \). By calculating the sample mean, we attempt to summarize the data with a single value that most closely represents the entire dataset.
When collecting \( n \) independent measurements \( X_{1}, X_{2}, \ldots, X_{n} \), the formula for the sample mean is:\[\bar{X} = \frac{1}{n} \sum_{i=1}^{n} X_i\]
This formula simply adds up all the observed values and divides by the number of observations.
  • The expectation of the sample mean, \( E(\bar{X}) \), is \( \mu \), indicating that \( \bar{X} \) is an unbiased estimator of the population mean.
  • An unbiased estimator means that the expected value of the estimator is equal to the true value of the parameter being estimated.
Variance
Variance is a key measure in statistics that tells us how spread out the values in a dataset are. It quantifies the average squared deviation of each number from the mean of the dataset.
For a sample with independent measurements \( X_{i} \), each having a variance \( \sigma^2 \), the variance provides insights into the consistency of the measurements.
  • High variance indicates that data points are spread out over a wider range of values.
  • Low variance indicates that data points tend to be close to the mean.
Variance is formally defined for a random variable \( Y \) as:\[V(Y) = E[(Y - E(Y))^2]\]
In the context of the sample mean \( \bar{X} \), the variance is reduced by a factor of \( n \), leading to:\[V(\bar{X}) = \frac{\sigma^2}{n}\]This adjustment reflects how the average of many observations tends to be more reliable than individual observations.
Expectation
Expectation, often referred to as expected value, is a fundamental concept in probability and statistics. It provides the long-term average or mean of a random variable's possible values, weighted according to their probabilities.
In simpler terms, the expectation of a random variable is the theoretical average value we would expect to happen if we could repeat a random experiment an infinite number of times.
For a random variable \( Y \), the expectation is denoted as \( E(Y) \). This concept is crucial, as it forms the basis for determining unbiased estimators. In assessing unbiasedness, we compare the expectation of an estimator with the true parameter value.
Expectations follow specific properties which simplify calculations:
  • Linearity: \( E(aY + bZ) = aE(Y) + bE(Z) \), where \( a \) and \( b \) are constants.
  • For a constant \( c \), \( E(c) = c \).
These properties allow us to simplify problems involving random variables and find meaningful interpretations.
Sample Variance
Sample Variance, denoted \( S^2 \), serves as an estimate of the variance \( \sigma^2 \) in a population. It measures the dispersion of a sample from its mean and provides insight into variability among sampled data points.
To calculate the sample variance, we use the formula:\[S^2 = \frac{1}{n-1} \sum_{i=1}^{n}(X_i - \bar{X})^2\]
This expression calculates the average squared differences from the mean, adjusted by \( n-1 \) to account for the degrees of freedom. This adjustment helps make \( S^2 \) an unbiased estimate of \( \sigma^2 \).
  • The sample variance provides a basis for confidence in sampling and insights into a population's variance.
  • The expectation of the sample variance, \( E(S^2) \), is exactly \( \sigma^2 \), reinforcing that it is an unbiased estimator.
These characteristics are crucial when estimating the accuracy and reliability of data collected from samples, allowing for informed decisions in both research and applied settings.

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Most popular questions from this chapter

Let \(X_{1}, \ldots, X_{n}\) be a random sample of component lifetimes from an exponential distribution with parameter \(\lambda\). Use the factorization theorem to show that \(\sum X_{i}\) is a sufficient statistic for \(\lambda\).

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