/*! 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 70 Suppose that \(X\) has a uniform... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 70

Suppose that \(X\) has a uniform probability distribution $$ f_{X}(x)=1, \quad 0 \leq x \leq 1 $$ Show that the probability distribution of the random variable \(Y=-2 \ln X\) is chi-squared with two degrees of freedom.

Short Answer

Expert verified
The distribution of \(Y = -2 \ln X\) is chi-squared with 2 degrees of freedom.

Step by step solution

01

Understand the Problem

We are given a random variable \(X\) that follows a uniform distribution on the interval [0, 1]. We need to find the distribution of \(Y = -2 \ln X\). This requires applying transformations to probability distributions.
02

Use Transformation Technique

To find the distribution of \(Y\), we'll use the transformation technique for continuous random variables. This involves using the change of variables method and the relationship \(Y = g(X)\) where \(g(X) = -2 \ln X\).
03

Find the Inverse Transformation

To apply the transformation, solve for \(X\) in terms of \(Y\): \[-2 \ln X = Y \Rightarrow \ln X = -\frac{Y}{2} \Rightarrow X = e^{-Y/2}.\]
04

Differentiate the Transformation

Differentiate \(X = e^{-Y/2}\) with respect to \(Y\) to find the Jacobian:\[\frac{dX}{dY} = \frac{d}{dY}(e^{-Y/2}) = -\frac{1}{2}e^{-Y/2}.\]
05

Determine the Support of Y

Since \(X\) is in the interval \([0, 1]\), the transformation \(-2 \ln X\) means \(0 \leq Y < \infty\). Thus, \(Y\) is supported on \([0, \infty)\).
06

Find the PDF of Y

The PDF of \(Y\) can be found using the formula for transformation of variables: \[ f_Y(y) = f_X(x) \left|\frac{dX}{dY}\right| = 1 \times \left| -\frac{1}{2} e^{-Y/2} \right| = \frac{1}{2} e^{-Y/2}, \text{ for } y \geq 0. \]
07

Recognize the Chi-Squared Distribution

The derived PDF \(\frac{1}{2} e^{-Y/2}\) matches the chi-squared distribution with 2 degrees of freedom: \[f_Y(y) = \frac{1}{2} e^{-y/2}, \quad y \geq 0,\] which is the PDF for \(\chi^2\) with 2 degrees of freedom.

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.

Uniform Distribution
The uniform distribution is a type of probability distribution where every outcome in a certain range is equally likely. This makes it quite unique compared to other distributions.
  • The probability density function (PDF) for a uniform distribution on the interval \[0, 1\] is given by \( f_X(x) = 1 \) for \( 0 \leq x \leq 1 \).
  • This means every value of \(X\) between 0 and 1 has an equal probability of occurring.
In our exercise, \(X\) is a continuous random variable uniformly distributed over the interval \[0, 1\]. This is the starting point for exploring how transforming this distribution can lead us to a different distribution, the chi-squared distribution.
Probability Distribution Transformation
Transforming a probability distribution involves changing a random variable using a mathematical function. This process helps us find new probability distributions. Here鈥檚 how it works:
  • Given a random variable \(X\) with a known distribution, we find \(Y = g(X)\), where \(g\) is some function applied to \(X\).
  • In our case, \(Y = -2 \ln X\), transforming the uniform distribution of \(X\) into a new distribution for \(Y\).
  • The challenge lies in finding the PDF of \(Y\), which involves using the change of variables method or transformation technique to ensure the probability distribution is correctly adjusted.
Transformation helps us derive new insights from already known distributions, as demonstrated by converting a uniform distribution into a chi-squared distribution.
Continuous Random Variables
Continuous random variables can take on an infinite number of possible values within a given interval. Unlike discrete random variables, which can only take specific, separate values (like whole numbers), continuous variables fill every possible value in a range.
  • In our case, the random variable \(X\) is continuous and can take any real value between 0 and 1.
  • The continuity of \(X\) allows us to apply calculus-based techniques, like differentiation and integration, to find the transformed PDF of \(Y\).
Understanding continuous random variables is critical in deriving distributions through transformations, as it involves precise and careful calculations in continuous ranges.
Transformation Technique
The transformation technique is a mathematical method used to find the distribution of a new variable \(Y = g(X)\) from a given variable \(X\). Here are the key steps involved:
  • Find the inverse transformation \(X\) in terms of \(Y\). For our scenario, \(Y = -2 \ln X\) was transformed into \(X = e^{-Y/2}\).
  • Differentiate this inverse transformation with respect to \(Y\) to determine the Jacobian, which adjusts the probability distribution. The derivative \(\frac{dX}{dY} = -\frac{1}{2} e^{-Y/2}\) plays a crucial role in deriving the final PDF of \(Y\).
  • Apply the transformation formula: \(f_Y(y) = f_X(x) \left|\frac{dX}{dY}\right|\) to find the new distribution.
This technique elegantly transforms uniform distributions into more complex distributions like the chi-squared, unfolding surprising relationships between disparate statistical concepts.

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

Show that the following function satisfies the properties of a joint probability mass function. $$ \begin{array}{ccc} \hline x & y & f_{X Y}(x, y) \\ \hline-1 & -2 & 1 / 8 \\ -0.5 & -1 & 1 / 4 \\ 0.5 & 1 & 1 / 2 \\ 1 & 2 & 1 / 8 \\ \hline \end{array} $$ Determine the following: (a) \(P(X < 0.5, Y < 1.5)\) (b) \(P(X < 0.5)\) (c) \(P(Y < 1.5)\) (d) \(P(X > 0.25, Y < 4.5)\) (e) \(E(X), E(Y), V(X),\) and \(V(Y)\) (f) Marginal probability distribution of the random variable \(X\) (g) Conditional probability distribution of \(Y\) given that \(X=1\) (h) Conditional probability distribution of \(X\) given that \(Y=1\) (i) \(E(X \mid y=1)\) (j) Are \(X\) and \(Y\) independent?

The permeability of a membrane used as a moisture barrier in a biological application depends on the thickness of two integrated layers. The layers are normally distributed with means of 0.5 and 1 millimeters, respectively. The standard deviations of layer thickness are 0.1 and 0.2 millimeters, respectively. The correlation between layers is \(0.7 .\) (a) Determine the mean and variance of the total thickness of the two layers. (b) What is the probability that the total thickness is less than 1 millimeter? (c) Let \(X_{1}\) and \(X_{2}\) denote the thickness of layers 1 and \(2,\) respectively. A measure of performance of the membrane is a function \(2 X_{1}+3 X_{2}\) of the thickness. Determine the mean and variance of this performance measure.

$$ \begin{aligned} \text { If } f_{X Y}(x, y)=\frac{1}{1.2 \pi} \exp \left\\{\frac{-1}{0.72}\left[(x-1)^{2}\right.\right.\\\ &\left.\left.-1.6(x-1)(y-2)+(y-2)^{2}\right]\right\\} \end{aligned} $$ determine \(E(X), E(Y), V(X), V(Y),\) and \(\rho\) by reorganizing the parameters in the joint probability density function.

A manufacturer of electroluminescent lamps knows that the amount of luminescent ink deposited on one of its products is normally distributed with a mean of 1.2 grams and a standard deviation of 0.03 gram. Any lamp with less than 1.14 grams of luminescent ink will fail to meet customers' specifications. A random sample of 25 lamps is collected and the mass of luminescent ink on each is measured. (a) What is the probability that at least one lamp fails to meet specifications? (b) What is the probability that five lamps or fewer fail to meet specifications? (c) What is the probability that all lamps conform to specifications? (d) Why is the joint probability distribution of the 25 lamps not needed to answer the previous questions?

A Web site uses ads to route visitors to one of four landing pages. The probabilities for each landing page are equal. Consider 20 independent visitors and let the random variables \(W\), \(X, Y,\) and \(Z\) denote the number of visitors routed to each page. Calculate the following: 5-44. A Web site uses ads to route visitors to one of four landing pages. The probabilities for each landing page are equal. Consider 20 independent visitors and let the random variables \(W\) \(X, Y,\) and \(Z\) denote the number of visitors routed to each page. Calculate the following: (a) \(P(W=5, X=5, Y=5, Z=5)\) (b) \(P(W=5, X=5, Y=5)\) (c) \(P(W=7, X=7, Y=6 \mid Z=3)\) (d) \(P(W=7, X=7, Y=3 \mid Z=3)\) (e) \(P(W \leq 2)\) (f) \(E(W)\) (g) \(P(W=5, X=5)\) (h) \(P(W=5 \mid X=5)\)
See all solutions

Recommended explanations on Math Textbooks

Pure Maths

Read Explanation

Geometry

Read Explanation

Calculus

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Logic and Functions

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.