/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 55 Show directly from the pdf that ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Show directly from the pdf that the mean of a \(t_{1}\) (Cauchy) random variable does not exist.

Short Answer

Expert verified
The mean does not exist due to the divergence of the integral.

Step by step solution

01

Understand the Probability Density Function

The probability density function (pdf) of a Cauchy distribution is given by \( f(x) = \frac{1}{\pi (1 + x^2)} \). This is a symmetric distribution around zero with heavy tails.
02

Calculate the Mean Value

The mean of a probability distribution is given by the integral \( \mu = \int_{-\infty}^{\infty} x f(x) \, dx \). For the Cauchy distribution, this becomes \( \mu = \int_{-\infty}^{\infty} \frac{x}{\pi (1 + x^2)} \, dx \).
03

Analyze the Convergence of the Integral

Consider the integral \( \int_{-\infty}^{\infty} \frac{x}{\pi (1 + x^2)} \, dx \). We can split this integral into two parts: \( \lim_{A \to \infty} \int_{-A}^{A} \frac{x}{\pi (1 + x^2)} \, dx \).
04

Evaluate the Symmetry of the Integrand

The function \( \frac{x}{1 + x^2} \) is odd, meaning \( f(x) = -f(-x) \). Integrating an odd function over symmetric limits like \((-A, A)\) results in zero value, provided the limits approach infinity symmetrically.
05

Divergence of Principal Value

Although the integral of an odd function over symmetric limits results in zero, the principal value doesn't imply convergence of the mean. In fact, due to heavy tails, any attempt to evaluate this suggests divergence or improper behavior.
06

Conclusion on the Existence of the Mean

Since the integrand does not converge properly due to the infinite extent of the distribution's tails, the mean of the Cauchy distribution cannot exist in a traditional sense.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Probability Density Function
The Cauchy distribution's probability density function (pdf) is a fundamental aspect that defines its unique characteristics. Mathematically, the pdf is expressed as: \[ f(x) = \frac{1}{\pi (1 + x^2)} \]
This mathematical expression shows us how probabilities are distributed across different values of the random variable.
For the Cauchy distribution, the pdf has heavy tails, meaning it assigns more probability to extreme values compared to distributions like the Normal distribution.
Also, this distribution is not concentrated around any particular value.
Understanding the functional form of the pdf helps frame our perception of events' likelihoods in the context outlined by the Cauchy distribution.
It's crucial to remember this distribution does not have a well-defined variance or mean, unlike many other distributions.
Symmetric Distribution
A distribution is called symmetric when its shape on the left side of a central point is a mirror image of its shape on the right side.
The Cauchy distribution is symmetric about zero. This symmetry implies certain properties, such as:
  • Equal probabilities for negative and positive deviations from the center.
  • The center acts as a perfect balance point.
This symmetrical nature of the Cauchy distribution is important when analyzing it because the symmetry affects how we calculate and interpret integrals over the distribution.
For instance, an integral of an odd function over symmetrical limits will lead to specific outcomes like a zero result, significant when assessing the average behavior or mean.
Divergence of Integral
The divergence of an integral can indicate that a mean, or other expected value, may not exist.
When evaluating the mean (\[ \mu = \int_{-\infty}^{\infty} x f(x) \, dx \]) of the Cauchy distribution, the integral does not converge.
Instead, it splits across infinite limits, creating a situation where the tails of the distribution heavily influence the integral's behavior.
Often, attempts to compute such integrals using the limits to infinity—specifically with heavy-tailed distributions like the Cauchy—reveal improper convergence.
This non-convergence, or divergence, indicates that no finite mean value can be established, reflecting the infinite spread and influence of the distribution's tails.
Mean Value Calculation
To calculate the mean value of a probability distribution, an integral of the form \( \mu = \int_{-\infty}^{\infty} x f(x) \, dx \) must converge.
For the Cauchy distribution, substituting its pdf into this formula yields \( \mu = \int_{-\infty}^{\infty} \frac{x}{\pi (1 + x^2)} \, dx \).
Despite the integral of the symmetric and odd function over symmetric limits giving a principle value of zero, this does not ensure true convergence.
The principal value being zero reflects symmetry but does not yield an existent mean.
This affirms the mathematical observation that the Cauchy distribution, due to its form and tail behavior, lacks a defined, traditional mean, aligning with its characterization as a distribution with undefined moments.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A company maintains three offices in a region, each staffed by two employees. Information concerning yearly salaries (1000's of dollars) is as follows: $$ \begin{array}{lcccccc} \text { Office } & 1 & 1 & 2 & 2 & 3 & 3 \\ \text { Employee } & 1 & 2 & 3 & 4 & 5 & 6 \\ \text { Salary } & 29.7 & 33.6 & 30.2 & 33.6 & 25.8 & 29.7 \end{array} $$ a. Suppose two of these employees are randomly selected from among the six (without replacement). Determine the sampling distribution of the sample mean salary \(\bar{X}\). b. Suppose one of the three offices is randomly selected. Let \(X_{1}\) and \(X_{2}\) denote the salaries of the two employees. Determine the sampling distribution of \(X\). c. How does \(E(X)\) from parts (a) and (b) compare to the population mean salary \(\mu\) ?

Suppose your waiting time for a bus in the morning is uniformly distributed on \([0,8]\), whereas waiting time in the evening is uniformly distributed on \([0,10]\) independent of morning waiting time. a. If you take the bus each morning and evening for a week, what is your total expected waiting time? [Hint: Define rv's \(X_{1}, \ldots, X_{10}\) and use a rule of expected value.] b. What is the variance of your total waiting time? c. What are the expected value and variance of the difference between morning and evening waiting times on a given day? d. What are the expected value and variance of the difference between total moming waiting time and total evening waiting time for a particular week?

A rock specimen from a particular area is randomly selected and weighed two different times. Let \(W\) denote the actual weight and \(X_{1}\) and \(X_{2}\) the two measured weights. Then \(X_{1}=W+E_{1}\) and \(X_{2}=W+E_{2}\), where \(E_{1}\) and \(E_{2}\) are the two measurement errors. Suppose that the \(E_{i}\) 's are independent of each other and of \(W\) and that \(V\left(E_{1}\right)=V\left(E_{2}\right)=\sigma_{E}^{2} .\) a. Express \(\rho\), the correlation coefficient between the two measured weights \(X_{1}\) and \(X_{2}\), in terms of \(\sigma_{W}^{2}\), the variance of actual weight, and \(\sigma_{X}^{2}\), the variance of measured weight. b. Compute \(\rho\) when \(\sigma_{W}=1 \mathrm{~kg}\) and \(\sigma_{E}=.01 \mathrm{~kg}\).

A box contains ten sealed envelopes numbered 1 , \(\ldots, 10\). The first five contain no money, the next three each contain \(\$ 5\), and there is a \(\$ 10\) bill in each of the last two. A sample of size 3 is selected with replacement (so we have a random sample), and you get the largest amount in any of the envelopes selected. If \(X_{1}, X_{2}\), and \(X_{3}\) denote the amounts in the selected envelopes, the statistic of interest is \(M=\) the maximum of \(X_{1}, X_{2}\), and \(X_{3}\). a. Obtain the probability distribution of this statistic. b. Describe how you would carry out a simulation experiment to compare the distributions of \(M\) for various sample sizes. How would you guess the distribution would change as \(n\) increases?

A student has a class that is supposed to end at 9:00 a.m. and another that is supposed to begin at 9:10 a.m. Suppose the actual ending time of the 9 a.m. class is a normally distributed rv \(X_{1}\) with mean 9:02 and standard deviation 1.5 min and that the starting time of the next class is also a normally distributed rv \(X_{2}\) with mean 9:10 and standard deviation \(1 \mathrm{~min}\). Suppose also that the time necessary to get from one classroom to the other is a normally distributed rv \(X_{3}\) with mean \(6 \mathrm{~min}\) and standard deviation \(1 \mathrm{~min}\). What is the probability that the student makes it to the second class before the lecture starts? (Assume independence of \(X_{1}, X_{2}\), and \(X_{3}\), which is reasonable if the student pays no attention to the finishing time of the first class.)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.