/*! 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 86 Suppose that \(Z\) is a standard... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 86

Suppose that \(Z\) is a standard normal random variable and that \(Y_{1}\) and \(Y_{2}\) are \(\chi_{2}\) -distributed random variables with \(\nu_{1}\) and \(\nu^{2}\) degrees of freedom, respectively. Further, assume that \(Z, Y_{1},\) and \(Y_{2}\) are independent. a. Define \(W=Z / \sqrt{Y_{1}} .\) Find \(E(W)\) and \(V(W) .\) What assumptions do you need about the value of $$\nu_{1} ?$$

Short Answer

Expert verified
\(E(W) = 0, V(W) = \frac{\nu_1}{\nu_1 - 2}\) with \(\nu_1 > 2\).

Step by step solution

01

Define the Variables and Distribution

- The variables presented are as follows: - \(Z\) is a standard normal random variable \(Z \sim N(0,1)\). - \(Y_1\) is a \(\chi^2\)-distributed random variable with \(u_1\) degrees of freedom \(Y_1 \sim \chi^2_{u_1}\). - \(W = \frac{Z}{\sqrt{Y_1}}\) is defined.
02

Identify the Distribution of W

- Since \(Z\) is standard normal and \(Y_1\) is chi-squared, the distribution of \(W\) is a t-distribution.- \(W\) follows a t-distribution with \(u_1\) degrees of freedom, i.e., \(W \sim t_{u_1}\).
03

Calculate the Expectation E(W)

- For a t-distribution with \(u_1 > 1\), the mean or expected value is 0.- Thus, \(E(W) = 0\) for \(u_1 > 1\).
04

Calculate the Variance V(W)

- For \(u_1 > 2\), the variance of a t-distribution is given by: \[V(W) = \frac{u_1}{u_1 - 2}\].
05

State the Assumptions

- To find both the expectation and variance, \(u_1\) must meet the following conditions: - \(u_1 > 1\) for the expectation to exist. - \(u_1 > 2\) for the variance to exist.

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.

t-distribution
The t-distribution is a fundamental concept in statistics, especially useful when dealing with small sample sizes or when the population standard deviation is unknown. It arises when you standardize a normally distributed variable by an independent estimate of its standard deviation, typically in the form of the square root of a chi-squared distribution divided by its degrees of freedom.
  • The t-distribution is symmetric similar to the normal distribution but slightly different in that it has heavier tails, which means it allows for a higher probability of extreme values.
  • As the sample size increases, the t-distribution approaches a normal distribution.
In the original exercise, the variable \( W \) is shown to follow a t-distribution because it is the result of dividing a standard normal random variable \( Z \) by the square root of a chi-squared variable \( Y_1 \). This relationship highlights how the t-distribution naturally emerges from this setup, allowing it to be a robust tool in statistical inference. When the degrees of freedom \( u_1 \) is greater than 1, the mean of the distribution is 0, and when \( u_1 \) exceeds 2, the variance is defined by \( \frac{u_1}{u_1 - 2} \).
chi-squared distribution
The chi-squared distribution is derived from the sum of the squares of \( k \) independent standard normal random variables. It is a special case of the gamma distribution and has several important properties and applications.
  • This distribution is primarily used in hypothesis testing, particularly in tests of independence and goodness-of-fit tests.
  • The distribution is positively skewed and becomes more symmetrical as the degrees of freedom increase.
The original exercise involves chi-squared distributed random variables, namely \( Y_1 \) and \( Y_2 \), each characterized by different degrees of freedom, \( u_1 \) and \( u_2 \). The degrees of freedom in a chi-squared distribution represent the number of independent pieces of information from the data used to estimate a parameter. Chi-squared distributions are valuable in forming ratios with standard normal variables, leading directly to distributions like the F and t-distributions, as demonstrated in the provided problem.
degrees of freedom
Degrees of freedom (DOF) are a critical concept in statistics, influencing the shape and characteristics of various statistical distributions. The degrees of freedom are the number of independent values or quantities which can be assigned to a statistical distribution.
  • In the context of the t-distribution, degrees of freedom are typically thought of as a function of the sample size; specifically, they are often calculated as \( n - 1 \) where \( n \) is the sample size.
  • In a chi-squared distribution, DOF are integral to determining the distribution's shape, often equating to the number of variables whose squares are summed to form a chi-square variable.
In the scenario posed by the exercise, \( W = \frac{Z}{\sqrt{Y_1}} \) involves degrees of freedom \( u_1 \) because \( Z \) and \( Y_1 \) are independent random variables. The degrees of freedom determine if certain statistical characteristics, like the mean and variance of \( W \), can be defined. Specifically, \( u_1 \) must be greater than 1 for the mean to be zero and greater than 2 for calculating the variance via the formula \( \frac{u_1}{u_1 - 2} \).

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 forester studying diseased pine trees models the number of diseased trees per acre, \(Y\), as a Poisson random variable with mean \(\lambda\). However, \(\lambda\) changes from area to area, and its random behavior is modeled by a gamma distribution. That is, for some integer \(\alpha\) $$f(\lambda)=\left\\{\begin{array}{ll} \frac{1}{\Gamma(\alpha) \beta^{\alpha}} \lambda^{\alpha-1} e^{-\lambda / \beta}, & \lambda>0 \\ 0, & \text { elsewhere } \end{array}\right.$$ Find the unconditional probability distribution for \(Y\)

In Exercise \(5.3,\) we determined that the joint probability distribution of \(Y_{1}\), the number of married executives, and \(Y_{2},\) the number of never- married executives, is given by $$p\left(y_{1}, y_{2}\right)=\frac{\left(\begin{array}{c}4 \\\y_{1}\end{array}\right)\left(\begin{array}{c}3 \\ y_{2}\end{array}\right)\left(\begin{array}{c}2 \\\3-y_{1}-y_{2}\end{array}\right)}{\left(\begin{array}{l}9 \\\3 \end{array}\right)}$$ where \(y_{1}\) and \(y_{2}\) are integers, \(0 \leq y_{1} \leq 3,0 \leq y_{2} \leq 3,\) and \(1 \leq y_{1}+y_{2} \leq 3 .\) Find \(\operatorname{Cov}\left(Y_{1}, Y_{2}\right)\).

In the production of a certain type of copper, two types of copper powder (types A and B) are mixed together and sintered (heated) for a certain length of time. For a fixed volume of sintered copper, the producer measures the proportion \(Y_{1}\) of the volume due to solid copper (some pores will have to be filled with air) and the proportion \(Y_{2}\) of the solid mass due to type A crystals. Assume that appropriate probability densities for \(Y_{1}\) and \(Y_{2}\) are $$\begin{array}{l} f_{1}\left(y_{1}\right)=\left\\{\begin{array}{ll} 6 y_{1}\left(1-y_{1}\right), & 0 \leq y_{1} \leq 1 \\ 0, & \text { elsewhere } \end{array}\right. \\ f_{2}\left(y_{2}\right)=\left\\{\begin{array}{ll} 3 y_{2}^{2}, & 0 \leq y_{2} \leq 1 \\ 0, & \text { elsewhere } \end{array}\right. \end{array}$$ The proportion of the sample volume due to type A crystals is then \(Y_{1} Y_{2} .\) Assuming that \(Y_{1}\) and \(Y_{2}\) are independent, find \(P\left(Y_{1} Y_{2} \leq .5\right)\)

A bus arrives at a bus stop at a uniformly distributed time over the interval 0 to 1 hour. A passenger also arrives at the bus stop at a uniformly distributed time over the interval 0 to 1 hour. Assume that the arrival times of the bus and passenger are independent of one another and that the passenger will wait for up to \(1 / 4\) hour for the bus to arrive. What is the probability that the passenger will catch the bus? [Hint: Let \(Y_{1}\) denote the bus arrival time and \(Y_{2}\) the passenger arrival time; determine the joint density of \(\left.Y_{1} \text { and } Y_{2} \text { and find } P\left(Y_{2} \leq Y_{1} \leq Y_{2}+1 / 4\right) .\right]\).

Suppose that the random variables \(Y_{1}\) and \(Y_{2}\) have joint probability density function, \(f\left(y_{1}, y_{2}\right)\) given by (see Exercises 5.14 and 5.32 ) $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll}6 y_{1}^{2} y_{2}, & 0 \leq y_{1} \leq y_{2}, y_{1}+y_{2} \leq 2 \\\0, & \text { elsewhere }\end{array}\right.$$ Show that \(Y_{1}\) and \(Y_{2}\) are dependent random variables.
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Pure Maths

Read Explanation

Statistics

Read Explanation

Probability and Statistics

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.