/*! 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 28 Which of the following are chara... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 28

Which of the following are characteristic functions: (a) \(\phi(t)=|-| t \mid\) if \(|t| \leq 1, \phi(t)=0\) otherwise, (b) \(\phi(t)=\left(1+t^{4}\right)^{-1}\) (c) \(\phi(t)=\exp \left(-t^{4}\right)\), (d) \(\phi(t)=\cos t\) (c) \(\left.\phi(t)=2(1-\cos t) / t^{2}\right\\}\)

Short Answer

Expert verified
Only function (c) is a characteristic function.

Step by step solution

01

Introduction to Characteristic Functions

A characteristic function \( \phi(t) \) of a random variable \( X \) is defined as \( \phi(t) = E[e^{itX}] \), where \( t \) is a real number.
02

Analyze Function (a)

The function \( \phi(t) = |-| t \mid \) does not satisfy the properties of a characteristic function because it is not continuous at \( t = 0 \) and is not non-negative definite.
03

Analyze Function (b)

For the function \( \phi(t) = (1 + t^4)^{-1} \), the function is not necessarily a characteristic function since its Laplace transform does not necessarily guarantee positivity and unit value at \( t = 0 \).
04

Analyze Function (c)

The function \( \phi(t) = e^{-t^4} \) is continuous and has \( \phi(0) = 1 \). Its form \( e^{-at^4} \) is a typical example of a characteristic function due to its non-negative definiteness.
05

Analyze Function (d)

The function \( \phi(t) = \cos t \) is not a characteristic function. Even though it is continuous and \( \cos(0) = 1 \), it does not satisfy non-negative definiteness across all \( t \).
06

Analyze Function (e)

The function \( \phi(t) = \frac{2(1 - \cos t)}{t^2} \) does not meet the characteristics of a characteristic function. It is not continuous at \( t = 0 \) and does not exhibit non-negative definiteness.
07

Conclusion

Characteristic functions must be non-negative definite, continuous at \( t = 0 \), and have \( \phi(0) = 1 \). Based on the analysis, function (c) is the only characteristic function.

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 91影视!

Key Concepts

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

Random Variable
A random variable is a fundamental concept in probability theory. It's essentially a numerical representation of the outcome of a random event. Think of it as a variable whose possible values are outcomes of a random phenomenon.

To break it down further, here are some key points about random variables:
  • Types: They can be discrete or continuous. Discrete random variables have distinct values like 0, 1, 2, etc., often counted. Continuous ones can take any value within a range, resembling measurements like height or temperature.
  • Function of outcomes: Random variables are typically denoted by letters such as X or Y, and they can be functions depending on the context and the outcomes of a probabilistic scenario.
  • Probability distribution: Every random variable has an associated probability distribution that provides probabilities of its possible values or ranges of values.
Understanding random variables is crucial because they help quantify and model real-world uncertainties. In calculating characteristic functions, random variables serve as the input, and the function captures their probabilistic distribution.
Laplace Transform
The Laplace transform is a powerful mathematical tool primarily used for transforming a function of a real variable (usually time) into a function of a complex variable (frequency). It's especially useful in solving differential equations and analyzing linear time-invariant systems.

Let's explore some basics of the Laplace transform:
  • Definition: The Laplace transform of a function \( f(t) \), where \( t \) 鈮� 0, is given by the integral \( \, L\{f(t)\} = \int_{0}^{\infty} e^{-st} f(t) \, dt \), provided the integral converges.
  • Applications: It simplifies solving differential equations by converting them into algebraic equations that are easier to manipulate.
  • Relation to characteristic functions: Characteristic functions are a specific type of transform similar to the Laplace transform, but they focus on probability distributions instead of general functions of time.
Through the Laplace transform, we can gain insights into the stability and dynamics of a system, making it a critical concept in engineering and physics.
Non-negative Definiteness
Non-negative definiteness is a concept related to functions or matrices that this function or matrix should not produce negative values in a specific context.For characteristic functions, non-negative definiteness ensures that the sum of the function's weighted terms is non-negative.In simpler terms:
  • Complex weights: If a function is non-negatively definite, then for any set of complex weights \( a_1, a_2, \ldots, a_n \), the weighted sum \( \Sigma_{i,j} a_i \overline{a_j} \phi(t_i - t_j) \) is non-negative.
  • Importance: This property is essential for characteristic functions because it maintains the probability's non-negative nature, ensuring that probabilities do not contradict basic probability rules.
  • Evaluation: To assess this in potential characteristic functions, test for positivity in the resulting sums across all t-values.
Non-negative definiteness serves as a foundational requirement for functions modeling probability distributions, enhancing their validity and applicability.
Continuity at Zero
Continuity at zero is a critical property for characteristic functions. It guarantees that the function does not have any sudden jumps or breaks, specifically at the point where \( t = 0 \).

The significance of continuity at zero includes:
  • Definition: A function \( \phi(t) \) is continuous at zero if for all \( \epsilon > 0 \), there exists a \( \delta > 0 \) such that whenever \( |t| < \delta \), it follows that \( |\phi(t) - \phi(0)| < \epsilon \).
  • Implication for characteristic functions: At \( t = 0 \), a characteristic function \( \phi(t) \) must be equal to 1, that is \( \phi(0) = 1 \). Continuity at zero ensures smoothness around this critical point.
  • Role in probability theory: This property helps ensure the consistency and reliability of the probability distribution being modeled by the characteristic function.
Overall, continuity at zero ensures the fundamental and expected property of characteristic functions, connecting back to their original definitions and purposes.

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 coin is tossed repeatedly, and heads turns up with probability \(P\) on each toss. Let \(h_{n}\) be the probability of an even number of heads in the first \(n\) tosses, with the convention that 0 is an even number. Find a difference equation for the \(h_{n}\) and deduce that they have generating function \(\frac{1}{2}\left\\{(1+2 p s-s)^{-1}+(1-s)^{-1}\right\\}\)

If \(X\) is an integer-valued random variable with characteristic function \(\phi\), show that $$ P(X=k)=\frac{1}{2 \pi} \int_{-\pi}^{\pi} e^{-i t k} \phi(t) d t $$ What is the corresponding result for a random variable whose distribution is arithmetic with span \(\lambda\) (that is, there is probability one that \(X\) is a multiple of \(\lambda\), and \(\lambda\) is the largest positive number with this? property)?

Prove the local central limit theorem for sums of random variables taking integer values. You may assume for simplicity that the summands have span 1 , in that \(\operatorname{ged}\\{|x|: P(X=x)>0\\}=1\).

Let \(X\) have density function \(f\) and characteristic function \(\phi\), and suppose that \(\int_{-\infty}^{\infty}|\phi(t)| d t<\infty\). Deduce that $$ f(x)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} e^{-i t x} \phi(t) d t $$

The cumulant generating function \(K_{X}(\theta)\) of the random variable \(X\) is defined by \(K_{X}(\theta)=\) \(\log \mathrm{E}\left(e^{\theta X}\right)\), the logarithm of the moment generating function of \(X\). If the latter is finite in a neighbourtsood of the origin, then \(K_{X}\) has a convergent Taylor expansion: $$ K_{X}(\theta)=\sum_{n=1}^{\infty} \frac{1}{n !} k_{n}(X) \theta^{n} $$ and \(k_{n}(X)\) is called the \(n\)th camulant (or semi-invariant) of \(X\). (a) Express \(k_{1}(X), k_{2}(X)\), and \(k_{3}(X)\) in terms of the moments of \(X\). (b) If \(X\) and \(Y\) are independent random variables, show that \(k_{n}(X+Y)=k_{n}(X)+k_{n}(Y)\).
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Pure Maths

Read Explanation
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.