/*! 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 15 Let \(Y\) have a distribution fu... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Let \(Y\) have a distribution function given by $$F(y)=\left\\{\begin{array}{ll} 0, & y<0 \\ 1-e^{-y^{2}}, & y \geq 0 \end{array}\right.$$ Find a transformation \(G(U)\) such that, if \(U\) has a uniform distribution on the interval \((0,1), G(U)\) has the same distribution as \(Y\)

Short Answer

Expert verified
The transformation is \(G(U) = \sqrt{-\ln(1-U)}\).

Step by step solution

01

Understand the Problem

We are given a cumulative distribution function (CDF) for the random variable \(Y\) and need to find a transformation \(G(U)\) of a uniform variable \(U\) on \((0,1)\) that results in \(Y\). This means finding an inverse transformation of the CDF.
02

Define the Transformation

Since \(U\) is uniformly distributed on \((0,1)\), the transformation \(G(U)\) should produce the same distribution as \(Y\). The general approach is to use the inverse of the CDF: set \(G(u) = F^{-1}(u)\).
03

Express \(U\) as a Function of \(Y\)

For \(U = F(Y)\), we have \(U = 1 - e^{-Y^2}\). Solving this for \(Y\), we rearrange to find \(Y\) in terms of \(U\).
04

Solve for \(Y\) in terms of \(U\)

Rearrange \(U = 1 - e^{-Y^2}\) to get \(e^{-Y^2} = 1 - U\). Taking the natural logarithm of both sides, we find \(-Y^2 = \ln(1-U)\). Hence, \(Y = \sqrt{-\ln(1-U)}\), noting \(Y\geq 0\). This gives us \(G(U) = \sqrt{-\ln(1-U)}\).
05

Verify Transformation

Check that if \(U\) follows a uniform distribution on \((0, 1)\), then \(G(U)\) follows the distribution of \(Y\) by substituting \(G(U)\) back into \(F(y)\) and ensuring it satisfies \(U = F(G(U))\).

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.

Cumulative Distribution Function (CDF)
The cumulative distribution function (CDF) is a fundamental concept in probability. It gives the probability that a random variable will take a value less than or equal to a certain number. Mathematically, for a continuous random variable, it is represented as:\[ F(y) = P(Y \leq y) \]The CDF is useful because it captures all the information about the probability distribution of a random variable. In the context of our exercise, the variable \(Y\) has a CDF given by:\[ F(y) = \begin{cases} 0, & \text{if } y < 0 \ 1 - e^{-y^2}, & \text{if } y \geq 0 \end{cases} \]This means that for values of \(Y\) less than 0, the probability is 0, and for values equal to or greater than 0, the probability evolves according to the exponential function \(1 - e^{-y^2}\). This CDF allows us to understand the spread and behavior of the random variable \(Y\).
Uniform Distribution
A uniform distribution is a type of probability distribution where every outcome in a specified range has an equal chance of occurring.In the problem, the uniform distribution is specified on the interval \(0, 1\). This means any number in this interval is equally likely to be picked, making it perfect for generating random numbers.- **Characteristics:** - **Constant Probability Density:** For a uniform distribution on \(0, 1\), the probability density function (PDF) is constant at \(1\). - **Distribution on Interval:** The variable \(U\) is uniformly distributed and this ensures a flat distribution over its range.Uniform distribution serves as a base for many random transformations, including those needed for simulating other distributions; this technique is evident in approaches like the inverse transform sampling. By using a uniform distribution, we can derive other more complex distributions using transformations.
Inverse Transformation
Inverse transformation is a mathematical technique used to obtain samples from a specific probability distribution from a uniform distribution.Given a cumulative distribution function (CDF) \(F(y)\), the inverse CDF (or quantile function), denoted as \(F^{-1}\), maps uniform samples to the desired distribution. - **Process Overview:** - Compute \(U = F(Y)\) - Solve for \(Y = F^{-1}(U)\)In the exercise, our aim was to find a function \(G(U)\) such that if \(U\) comes from a uniform distribution on \(0, 1\), then \(G(U)\) has the same distribution as \(Y\).This is achieved by expressing \(Y\) in terms of \(U\) as follows:\[ U = 1 - e^{-Y^2} \]Re-arranging gives:\[ Y = \sqrt{-\ln(1-U)} \]This equation shows \(G(U)\) ensures that \(G(U)\) matches the distribution of \(Y\), demonstrating the power and application of inverse transformations.
Continuous Random Variables
Continuous random variables are those that can take any real value within a certain range. They are modeled using continuous probability distributions.- **Key Features**: - **Infinite Possible Values:** Unlike discrete random variables, continuous random variables can take an infinite number of possible values. - **Probability Density Function (PDF):** The PDF describes the likelihood of the random variable falling within a particular range of values, rather than taking on any one value.In the context of this problem, the random variable \(Y\) is continuous. Its probability distribution is described by the CDF we discussed earlier. Understanding the nature of continuous random variables is crucial for working with transformations, especially for simulations and statistical modeling using distributions like the normal distribution and exponential distribution.Continuous variables are versatile, and using them with transformation techniques allows for a broad array of applications in probability and statistics.

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

Customers arrive at a department store checkout counter according to a Poisson distribution with a mean of 7 per hour. In a given two-hour period, what is the probability that 20 or more customers will arrive at the counter?

Let \(Y\) be a random variable with probability density function given by $$f(y)=\left\\{\begin{array}{ll} 2(1-y), & 0 \leq y \leq 1 \\ 0, & \text { elsewhere } \end{array}\right.$$ a. Find the density function of \(U_{1}=2 Y-1\) b. Find the density function of \(U_{2}=1-2 Y\) c. Find the density function of \(U_{3}=Y^{2}\) d. Find \(E\left(U_{1}\right), E\left(U_{2}\right),\) and \(E\left(U_{3}\right)\) by using the derived density functions for these random variables. e. Find \(E\left(U_{1}\right), E\left(U_{2}\right),\) and \(E\left(U_{3}\right)\) by the methods of Chapter 4.

Suppose that two electronic components in the guidance system for a missile operate independently and that each has a length of life governed by the exponential distribution with mean 1 (with measurements in hundreds of hours). Find the a. probability density function for the average length of life of the two components. b. mean and variance of this average, using the answer in part (a). Check your answer by computing the mean and variance, using Theorem 5.12 .

The length of time that a machine operates without failure is denoted by \(Y_{1}\) and the length of time to repair a failure is denoted by \(Y_{2} .\) After a repair is made, the machine is assumed to operate like a new machine. \(Y_{1}\) and \(Y_{2}\) are independent and each has the density function $$f(y)=\left\\{\begin{array}{ll} e^{-y}, & y>0 \\ 0, & \text { elsewhere } \end{array}\right.$$ Find the probability density function for \(U=Y_{1} /\left(Y_{1}+Y_{2}\right),\) the proportion of time that the machine is in operation during any one operation-repair cycle.

A member of the Pareto family of distributions (often used in economics to model income distributions) has a distribution function given by $$F(y)=\left\\{\begin{array}{ll} 0, & y<\beta \\ 1-\left(\frac{\beta}{y}\right)^{\alpha}, & y \geq \beta \end{array}\right.$$ where \(\alpha, \beta>0\) a. Find the density function. b. For fixed values of \(\beta\) and \(\alpha\), find a transformation \(G(U)\) so that \(G(U)\) has a distribution function of \(F\) when \(U\) has a uniform distribution on the interval (0,1) c. Given that a random sample of size 5 from a uniform distribution on the interval (0,1) yielded the values. 0058, .2048,7692,2475 and \(.6078,\) use the transformation derived in part \((\mathrm{b})\) to give values associated with a random variable with a Pareto distribution with \(\alpha=2, \beta=3\)

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.