/*! 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 4 Let \(X\) be a chi-square random... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 3: Problem 4

Let \(X\) be a chi-square random variable with \(n(\geq 1)\) degrees of freedom with the pdf: $$ f(x)= \begin{cases}\frac{x^{n / 2-1}}{2^{n / 2} \Gamma(n / 2)} e^{-x / 2}, & x>0, \\ 0, & \text { elsewhere. }\end{cases} $$ Find the pdf of the random variable \(Y=\sqrt{X / n}\).

Short Answer

Expert verified
The PDF of the random variable Y is \( f_Y(y) = \frac{n^{n/2} y^{n-1}}{2^{n/2-1} \Gamma(n/2)} e^{-ny^2/2}, \) for \( y > 0. \)

Step by step solution

01

Understand the given PDF

The given PDF for the chi-square random variable X with n degrees of freedom is:\[f(x) = \begin{cases}\frac{x^{n / 2-1}}{2^{n / 2} \Gamma(n / 2)} e^{-x / 2}, & x>0, \ 0, & \text { elsewhere. }\end{cases}\]This PDF applies for values of x greater than 0 and is zero elsewhere.
02

Define the transformation

We are given the transformation \(Y = \sqrt{X / n}\). To find the PDF of Y, first rewrite it as \(X = n Y^2\) and note the relationship between X and Y.
03

Calculate the Jacobian

The Jacobian of the transformation \(X = nY^2\) is computed as follows:\[\frac{dX}{dY} = \frac{d(nY^2)}{dY} = 2nY.\]
04

Substitute and adjust the PDF

Substitute \(X = nY^2\) into the original PDF of X:\[f(x) = \frac{(nY^2)^{n / 2-1}}{2^{n / 2} \Gamma(n / 2)} e^{-nY^2 / 2}, \text{for } x > 0.\]Adjust the notation for the transformation and include the Jacobian:\[f_Y(y) = f_X(nY^2) \times |2nY| = \frac{(nY^2)^{n / 2-1}}{2^{n / 2} \Gamma(n / 2)} e^{-nY^2 / 2} \times 2nY.\]
05

Simplify the expression

Simplify the expression step-by-step:\[f_Y(y) = 2nY \cdot \frac{(nY^2)^{n / 2-1}}{2^{n / 2} \Gamma(n / 2)} e^{-nY^2 / 2}\]\[= \frac{2nY \cdot (nY^2)^{n / 2-1}}{2^{n / 2} \Gamma(n / 2)} e^{-nY^2 / 2}\]\[= \frac{2nY \cdot n^{n / 2-1}Y^{n-2}}{2^{n / 2} \Gamma(n / 2)} e^{-nY^2 / 2}\]Combine constants and exponents:\[= \frac{2n^{n / 2} Y^{n-1}}{2^{n / 2} \Gamma(n / 2)} e^{-nY^2 / 2}.\]\[\text{Since } 2n^{n/2} = 2 (2^{n/2})n^{n/2},\] the final simplified form is: \[f_Y(y) = 2 \cdot f_X(ny^2) y\]
06

Accounting for domain of Y

Since Y = \sqrt{X/n} and X > 0, it follows that Y > 0. The final PDF of Y is valid for Y > 0.\[f_Y(y) = \frac{n^{n/2} y^{n-1}}{2^{n/2-1} \Gamma(n/2)} e^{-ny^2/2}, \text{for } y > 0.\]

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.

Probability Density Function
A Probability Density Function (PDF) describes the likelihood of a continuous random variable taking on a particular value. It is a function that must remain non-negative for all possible values and must integrate to 1 over its domain, ensuring proper normalization.
In this exercise, the given PDF is for a chi-square distribution with n degrees of freedom:
\[ f(x) = \begin{cases} \frac{x^{n / 2-1}}{2^{n / 2} \Gamma(n / 2)} e^{-x / 2}, & x>0, \ 0, & \text { elsewhere. } \end{cases}\]This PDF is only positive when x > 0, illustrating that the chi-square distribution only takes positive values. For a given x, the probability density is determined by the function shown, combining the power term, exponential decay, and normalization constants. This form is crucial in many statistical tests and procedures.
Jacobian Transformation
The Jacobian Transformation is used to transform the PDF of one random variable into another, especially when dealing with variables connected non-linearly. This transformation is essential for finding the PDF of a new random variable when a transformation is applied.
In the exercise, we transform the random variable X to Y through the relationship: \( Y = \sqrt{X / n} \). Correspondingly, we can rewrite X as \( X = nY^2 \). To apply the transformation, we need the Jacobian determinant, calculated by: \[\frac{dX}{dY} = \frac{d(nY^2)}{dY} = 2nY.\]This Jacobian factor accounts for the change in variable and must be included in the transformed PDF.
Using the Jacobian, the PDF of Y from the PDF of X is adjusted as:\[f_Y(y) = f_X(nY^2) \, \big|2nY\big|.\]
Random Variables
Random variables (RVs) are fundamental components of probability theory. They provide a numerical value to outcomes of a random phenomenon. RVs can be discrete (taking specific values) or continuous (taking any value within a range).
In the given exercise, we start with X, a chi-square random variable, and aim to find the PDF of the transformed variable Y. The transformation relationship \( Y = \sqrt{X / n} \) helps in understanding how different outcomes of X influence Y.
The final step in the exercise brings us to the new PDF of Y, ensuring it considers the transformation and domain conditions:
\[ f_Y(y) = \frac{n^{n/2} y^{n-1}}{2^{n/2-1} \Gamma(n/2)} e^{-ny^2/2}, \text{for } y > 0.\]This final PDF combines the original characteristics of X and the transformation applied, providing a clear and adjusted probability description for Y.

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

has the chi-square distribution with \(2 k\) degrees of freedom. Consider a nonlinear amplifier whose input \(X\) and the output \(Y\) are related by its transfer characteristic: $$ Y= \begin{cases}X^{1 / 2}, & X \geq 0 \\ -|X|^{1 / 2}, & X<0\end{cases} $$ Assuming that \(X \sim N(0,1)\) compute the pdf of \(Y\) and plot your result.

A system with three independent components works correctly if at least one component is functioning properly. Failure rates of the individual components are \(\lambda_1=0.0001, \lambda_2=0.0002\), and \(\lambda_3=0.0004\) (assume exponential lifetime distributions). (a) Determine the probability that the system will work for \(1000 \mathrm{~h}\). (b) Determine the density function of the lifetime \(X\) of the system.

Consider a normalized floating-point number in base (or radix) \(\beta\) so that the mantissa \(X\) satisfies the condition \(1 / \beta \leq X<1\). Assume that the density of the continuous random variable \(X\) is \(1 /(x \ln \beta)\). (a) Show that a random deviate of \(X\) is given by the formula, \(\beta^{u-1}\) where \(u\) is a random number. (b) Determine the pdf of the normalized reciprocal \(Y=1 /(\beta X)\).

The test effort during the software testing phase, such as human resources, the number of test cases, and CPU time, can be measured by the cumulative amount of testing effort during the time interval \((0, t]\). Yamada et al. [YAMA 1986] proposed a formula for \(W(t): d W(t) / d t=g(t)(1-W(t))\) where \(g(t)\) is the instantaneous consumption rate of the testing effort expenditures. \(W(t)\) is defined as \(W(t)=\int_0^t w(t) d t\) where \(w(t)\) is the testing-effort consumption rate at time \(t\). Find an explicit expression for \(W(t)\) in terms of \(g(t)\) and show that \(W(t)\) is the Rayleigh distribution.

Consider a series system consisting of \(n\) independent components. Assuming that the lifetime of the ith component is Weibull distributed with parameter \(\lambda_i\) and \(\alpha\), show that the system lifetime also has a Weibull distribution. As a concrete example, consider a liquid cooling cartridge system that is used in enterprise-class servers made by Sun Microsystems [KOSL 2001]. The series system consists of a blower, a water pump and a compressor. The following table gives the Weibull data for the three components. \begin{tabular}{crc} \hline Component & L10 \((\mathrm{h})\) & Shape parameter \((\alpha)\) \\ \hline Blower & 70,000 & \(3.0\) \\ Water pump & 100,000 & \(3.0\) \\ Compressor & 100,000 & \(3.0\) \\ \hline \end{tabular} L10 is the rating life of the component, which is the time at which \(10 \%\) of the components are expected to have failed or \(R(\mathrm{~L} 10)=0.9\). Derive the system reliability expression.
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation

Probability and Statistics

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.