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Chapter 6: Problem 18

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a pdf \(f(x)\) that is symmetric about \(\mu\), so that \(\widetilde{X}\) is an unbiased estimator of \(\mu\). If \(n\) is large, it can be shown that \(V(\widetilde{X}) \approx 1 /\left(4 n[f(\mu)]^{2}\right)\). a. Compare \(V(\widetilde{X})\) to \(V(\bar{X})\) when the underlying distribution is normal. b. When the underlying pdf is Cauchy (see Example 6.7), \(V(\bar{X})=\infty\), so \(\bar{X}\) is a terrible estimator. What is \(V(\widetilde{X})\) in this case when \(n\) is large?

Short Answer

Expert verified
For normal distribution, \( V(\widetilde{X}) < V(\bar{X}) \). For Cauchy, \( V(\widetilde{X}) \approx \frac{\gamma^2}{4n} \), finite for large \( n \).

Step by step solution

01

Understand the Problem

We need to compare the variance of two estimators, \( \widetilde{X} \) and \( \bar{X} \), for the mean \( \mu \) of a symmetric distribution. Additionally, for the Cauchy distribution where the variance of \( \bar{X} \) is infinite, we need to calculate \( V(\widetilde{X}) \) when \( n \) is large.
02

Identify the Variance of the Sample Mean

For a normal distribution, the variance of the sample mean \( \bar{X} \) is given by \( V(\bar{X}) = \frac{\sigma^2}{n} \), where \( \sigma^2 \) is the population variance. This is a property of the sample mean for normally distributed data.
03

Use Given Approximation for \( V(\widetilde{X}) \)

The problem states that \( V(\widetilde{X}) \approx \frac{1}{4n[f(\mu)]^{2}} \). For normal distributions, the pdf at the mean is \( f(\mu) = \frac{1}{\sqrt{2\pi}\sigma} \). Substituting this into the formula gives \( V(\widetilde{X}) \approx \frac{1}{4n \left(\frac{1}{\sqrt{2\pi}\sigma}\right)^2} = \frac{\sigma^2}{2n\pi} \).
04

Compare \( V(\widetilde{X}) \) to \( V(\bar{X}) \) for Normal Distribution

For the normal distribution: \( V(\bar{X}) = \frac{\sigma^2}{n} \) and \( V(\widetilde{X}) \approx \frac{\sigma^2}{2n\pi} \). Clearly, \( V(\widetilde{X}) \) is smaller than \( V(\bar{X}) \) because \( 2\pi \approx 6.28 \), making \( \frac{1}{2\pi} \) a reducing factor.
05

Consider the Cauchy Distribution

For the Cauchy distribution, it's known that \( V(\bar{X}) = \infty \) due to the heavy tails and undefined variance. However, we can still use \( V(\widetilde{X}) \approx \frac{1}{4n[f(\mu)]^{2}} \). The pdf \( f(\mu) \) at the median for a Cauchy distribution is \( \frac{1}{\pi}\cdot\frac{1}{\gamma} \), where \( \gamma \) is the scale parameter. Substituting gives \( V(\widetilde{X}) \approx \frac{\gamma^2}{4n} \).
06

Conclude the Results

For the normal distribution, \( V(\widetilde{X}) \) is smaller than \( V(\bar{X}) \). For the Cauchy distribution with large \( n \), \( V(\widetilde{X}) \approx \frac{\gamma^2}{4n} \), which remains finite, making it a superior estimator to \( \bar{X} \) whose variance is infinite.

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Key Concepts

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

Symmetric Distribution
A symmetric distribution is one where the left and right sides of the distribution mirror each other around a central point, typically the mean or median.
This symmetry means that the probabilities equally distribute on both sides.
Commonly encountered symmetric distributions include the normal distribution, but it is important to understand that not all symmetric distributions are normal.
  • In symmetric distributions, the mean, median, and mode often coincide, creating a balanced shape.
  • Equal intervals on both sides of the center have the same probability of occurrence, making these distributions advantageous for predicting outcomes close to the mean.
  • Examples include the normal distribution, which is perfectly symmetric and bell-shaped, and the Cauchy distribution, which is symmetric but has a very different shape due to its heavy tails.
Understanding symmetric distributions plays a vital role in statistical estimation because it simplifies the assessment of estimators like the sample mean.
This symmetry is also crucial when discussing unbiased estimators, as they rely on predictable, stable distributions.
Normal Distribution
The normal distribution, also known as the Gaussian distribution, is a cornerstone of statistics. It's characterized by its bell-shaped curve, which is symmetric about the mean.
  • The normal distribution is defined by two parameters: the mean (渭) and the standard deviation (蟽).
  • The properties of the normal distribution allow the use of many statistical tools and techniques, particularly in inferential statistics.
  • One important feature is that the total area under the curve equals 1, representing a complete probability distribution for a continuous random variable.
  • About 68% of the data falls within one standard deviation of the mean, making this distribution highly predictable in terms of outcome ranges.
It's pivotal for processes such as quality control and finance because many natural phenomena tend to align with this distribution.
When estimating parameters like the sample mean and variance, the normal distribution provides a trustworthy framework due to its properties of symmetry and predictable spread around the mean.
Cauchy Distribution
The Cauchy distribution is a symmetric distribution characterized by its heavy tails and median-centered peak. It's crucial to understand that unlike the normal distribution, the variance of a Cauchy distribution is undefined.
  • This distribution is defined by the location parameter (usually the median) and a scale parameter \( \gamma \), which stretches or compresses the distribution.
  • It does not have a defined mean or standard deviation, which complicates the application of traditional statistical analysis methods.
  • Despite being symmetric, the presence of heavy tails means that extreme values have a higher probability of occurring than in a normal distribution.
  • In practice, the Cauchy distribution can model scenarios with large outliers or extreme values, often found in physics and finance.
The Cauchy distribution's peculiar properties make usual estimators like the sample mean ineffective.
As seen in the exercise, the sample mean variance is infinite, highlighting the need for alternative estimators such as the median to effectively estimate central tendencies.
Unbiased Estimator
An unbiased estimator is a statistical estimate that targets the actual parameter of a population without systematic errors. The key here is that its expected value equals the true parameter value it's estimating.
  • One of the primary advantages of an unbiased estimator is its long-term accuracy, as it provides the correct value on average, over numerous samples.
  • In the context of symmetric distributions, having an unbiased estimator ensures that the prediction aligns with the distribution's central value, which is crucial for accurate empirical analysis.
  • The sample mean for a normal distribution is an example of an unbiased estimator for the population mean.
However, unbiasedness alone does not guarantee efficiency; this is where the variance of the estimator becomes important.
For instance, while considering large samples, unbiased estimators like \( \widetilde{X} \) are preferred because they maintain a lower variance than biased estimators, offering more reliability.
Sample Mean Variance
Sample mean variance is a measure of how much the sample mean \( \bar{X} \) is expected to vary from the true population mean. It's a key concept in understanding the reliability of an estimator.
  • For a normal distribution with known variance \( \sigma^2 \), the variance of the sample mean is \( V(\bar{X}) = \frac{\sigma^2}{n} \), where \( n \) is the sample size.
  • This formula shows that as the number of samples increases, the variance of the sample mean decreases, leading to more precise estimations.
  • However, in some distributions, like the Cauchy distribution, the variance of the sample mean is not defined, making it a poor choice for estimating the mean.
In scenarios where variance can be optimized, choosing efficient estimators with lower variance can significantly improve the precision of statistical predictions.
Thus, when comparing \( \widetilde{X} \) to \( \bar{X} \) in the exercise, \( \widetilde{X} \) provides a much smaller variance, emphasizing its superiority as an estimator.

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Most popular questions from this chapter

Consider a random sample \(X_{1}, \ldots, X_{n}\) from the pdf $$ f(x ; \theta)=.5(1+\theta x) \quad-1 \leq x \leq 1 $$ where \(-1 \leq \theta \leq 1\) (this distribution arises in particle physics). Show that \(\hat{\theta}=3 \bar{X}\) is an unbiased estimator of \(\theta\).

Let \(X\) denote the proportion of allotted time that a randomly selected student spends working on a certain aptitude test. Suppose the pdf of \(X\) is $$ f(x ; \theta)=\left\\{\begin{array}{cc} (\theta+1) x^{\theta} & 0 \leq x \leq 1 \\ 0 & \text { otherwise } \end{array}\right. $$ where \(-1<\theta\). A random sample of ten students yields data \(x_{1}=.92, x_{2}=.79, x_{3}=.90, x_{4}=.65, x_{5}=.86, x_{6}=.47\) \(x_{7}=.73, x_{8}=.97, x_{9}=.94, x_{10}=.77\) a. Use the method of moments to obtain an estimator of \(\theta\), and then compute the estimate for this data. b. Obtain the maximum likelihood estimator of \(\theta\), and then compute the estimate for the given data.

Two different computer systems are monitored for a total of \(n\) weeks. Let \(X_{i}\) denote the number of breakdowns of the first system during the \(i\) th week, and suppose the \(X_{i}\) 's are independent and drawn from a Poisson distribution with parameter \(\mu_{1}\). Similarly, let \(Y_{i}\) denote the number of breakdowns of the second system during the \(i\) th week, and assume independence with each \(Y_{i}\) Poisson with parameter \(\mu_{2}\). Derive the mle's of \(\mu_{1}, \mu_{2}\), and \(\mu_{1}-\mu_{2}\).

a. Let \(X_{1}, \ldots, X_{n}\) be a random sample from a uniform distribution on \([0, \theta]\). Then the mle of \(\theta\) is \(\hat{\theta}=Y=\max \left(X_{i}\right)\). Use the fact that \(Y \leq y\) iff each \(X_{i} \leq y\) to derive the cdf of \(Y\). Then show that the pdf of \(Y=\max \left(X_{i}\right)\) is $$ f_{Y}(y)=\left\\{\begin{array}{cl} \frac{n y^{n-1}}{\theta^{n}} & 0 \leq y \leq \theta \\ 0 & \text { otherwise } \end{array}\right. $$ b. Use the result of part (a) to show that the mle is biased but that \((n+1) \max \left(X_{i}\right) / n\) is unbiased.

A diagnostic test for a certain disease is applied to \(n\) individuals known to not have the disease. Let \(X=\) the number among the \(n\) test results that are positive (indicating presence of the disease, so \(X\) is the number of false positives) and \(p=\) the probability that a disease-free individual's test result is positive (i.e., \(p\) is the true proportion of test results from disease-free individuals that are positive). Assume that only \(X\) is available rather than the actual sequence of test results. a. Derive the maximum likelihood estimator of \(p\). If \(n=20\) and \(x=3\), what is the estimate? b. Is the estimator of part (a) unbiased? c. If \(n=20\) and \(x=3\), what is the mle of the probability \((1-p)^{5}\) that none of the next five tests done on diseasefree individuals are positive?
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