/*! 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 74 Let \(Y_{1}, Y_{2}, \ldots, Y_{n... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 74

Let \(Y_{1}, Y_{2}, \ldots, Y_{n}\) be independent, uniformly distributed random variables on the interval \([0, \theta] .\) Find the a. probability distribution function of \(Y_{(n)}=\max \left(Y_{1}, Y_{2}, \ldots, Y_{n}\right)\) b. density function of \(Y_{(n)}\) c. mean and variance of \(Y_{(n)}\)

Short Answer

Expert verified
a. CDF is \( (\frac{y}{\theta})^n \); b. PDF is \( n \frac{y^{n-1}}{\theta^n} \); c. Mean is \( \frac{n}{n+1}\theta \), variance is \( \frac{n\theta^2}{(n+1)^2(n+2)} \)."

Step by step solution

01

Understanding the maximum of uniform variables

Let us consider independent random variables \( Y_1, Y_2, \ldots, Y_n \) each uniformly distributed over the interval \([0, \theta]\). The maximum of these random variables is denoted as \( Y_{(n)} = \max(Y_1, Y_2, \ldots, Y_n) \). Our task is to find its probability distribution function (cumulative distribution function).
02

CDF of the maximum

The cumulative distribution function (CDF) of \( Y_{(n)} \) is given by \( F_{Y_{(n)}}(y) = P(Y_{(n)} \leq y) = P(Y_1 \leq y, Y_2 \leq y, \ldots, Y_n \leq y) \). Since they are independent uniform variables, \( P(Y_i \leq y) = \frac{y}{\theta} \) for \( 0 \leq y \leq \theta \). Consequently, \( F_{Y_{(n)}}(y) = \left(\frac{y}{\theta}\right)^n \) for \( 0 \leq y \leq \theta \), and \( F_{Y_{(n)}}(y) = 1 \) for \( y > \theta \).
03

PDF of the maximum

The probability density function (PDF) is the derivative of the CDF with respect to \( y \). Thus, \[ f_{Y_{(n)}}(y) = \frac{d}{dy} \left( \frac{y}{\theta} \right)^n = n \cdot \frac{y^{n-1}}{\theta^n}, \] valid for \( 0 \leq y \leq \theta \). For \( y > \theta \), \( f_{Y_{(n)}}(y) = 0 \).
04

Expected value of Y(n)

The expected value of \( Y_{(n)} \) can be calculated from the PDF: \[ E[Y_{(n)}] = \int_0^\theta y \cdot n \cdot \frac{y^{n-1}}{\theta^n} \, dy. \] Solving this integral, we find \[ E[Y_{(n)}] = \frac{n}{n+1}\theta. \]
05

Variance of Y(n)

To find the variance, we need \( E[Y_{(n)}^2] \): \[ E[Y_{(n)}^2] = \int_0^\theta y^2 \cdot n \cdot \frac{y^{n-1}}{\theta^n} \, dy. \] Evaluating this gives \[ E[Y_{(n)}^2] = \frac{n}{n+2}\theta^2. \] The variance \( \text{Var}(Y_{(n)}) \) is \[ \text{Var}(Y_{(n)}) = E[Y_{(n)}^2] - (E[Y_{(n)}])^2 = \frac{n}{n+2}\theta^2 - \left(\frac{n}{n+1}\theta\right)^2 = \frac{n\theta^2}{(n+1)^2(n+2)}. \]

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
A uniform distribution is a type of probability distribution where every outcome in a given interval is equally likely. Imagine slicing a pie into equal parts where each slice represents an outcome; this is how a uniform distribution can be visualized. In formal terms, if a random variable is uniformly distributed over an interval
  • The probability of any specific outcome is constant.
  • The distribution is defined by two parameters, typically denoted as \([a, b]\) for a continuous uniform distribution.
In the context of our exercise, we deal with uniform random variables over the interval \([0, \theta]\). This means that any number selected within this range is equally probable, making it simple and intuitive to work with when calculating statistical measures like the maximum value.
Probability Distribution Function
The probability distribution function (PDF) describes the likelihood of a random variable to take on a given value. For continuous variables, this involves the Probability Density Function (also referred to as PDF), which is the derivative of the cumulative distribution function (CDF). The CDF, in contrast, measures the probability that the random variable is less than or equal to a certain value. For the maximum of n uniform variables defined in \([0, \theta]\), the CDF is:
  • \(F_{Y_{(n)}}(y) = \left(\frac{y}{\theta}\right)^n\) for \(0 \leq y \leq \theta\).
  • This means that all individual variables are less than or equal to \(y\), making the calculation straightforward.
  • Beyond the interval (i.e., \(y > \theta\)), the CDF effectively equals 1.
The ease of calculating the CDF for a uniform distribution arises from symmetrical properties and absence of complex weighted probabilities.
Density Function
The density function showcases how the probability is distributed over an interval for a continuous random variable. Derived from the CDF, it provides the "density" of the probabilities around each point, enlightening us on distributions of differing shapes.From our uniform variables over \([0, \theta]\), the PDF of the maximum value \(Y_{(n)}\) is:
  • \( f_{Y_{(n)}}(y) = n \cdot \frac{y^{n-1}}{\theta^n} \) for \(0 \leq y \leq \theta\).
  • This illustrates an increasing function, highlighting that larger sample sizes will see the maximum value shift closer to \(\theta\).
  • For values beyond \(\theta\), the PDF naturally drops to zero.
Understanding the density function aids in predicting where the outcomes predominantly lie, given a series of events or tests.
Expected Value and Variance
Expected value and variance are fundamental in comprehending the behavior of random variables. The expected value provides us with the average or mean outcome anticipated, which serves as a central point of distribution.
  • For \(Y_{(n)}\), the expected value is \( E[Y_{(n)}] = \frac{n}{n+1}\theta \). It calculates the average maximum outcome across multiple trials for uniformly distributed variables.
  • Variance quantifies how spread out the numbers in the distribution are, or put simply, the degree to which outcomes deviate from the expected value.
  • The variance is determined by \( \text{Var}(Y_{(n)}) = \frac{n\theta^2}{(n+1)^2(n+2)} \), capturing this dispersion and offering insight into the likelihood of deviations.
Grasping these concepts is crucial for predicting outcomes and understanding the variability inherent in statistical data.

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

An engineer has observed that the gap times between vehicles passing a certain point on a highway have an exponential distribution with mean 10 seconds. Find the a. probability that the next gap observed will be no longer than one minute. b. probability density function for the sum of the next four gap times to be observed. What assumptions are necessary for this answer to be correct?

Let \(Y_{1}\) and \(Y_{2}\) be independent random variables, each having the same geometric distribution. $$\text { a. Find } P\left(Y_{1}=Y_{2}\right)=P\left(Y_{1}-Y_{2}=0\right)$$ b. Find \(P\left(Y_{1}-Y_{2}=1\right)\) c. If \(U=Y_{1}-Y_{2},\) find the (discrete) probability function for \(U\)

A type of elevator has a maximum weight capacity \(Y_{1}\), which is normally distributed with mean 5000 pounds and standard deviation 300 pounds. For a certain building equipped with this type of elevator, the elevator's load, \(Y_{2},\) is a normally distributed random variable with mean 4000 pounds and standard deviation 400 pounds. For any given time that the elevator is in use, find the probability that it will be overloaded, assuming that \(Y_{1}\) and \(Y_{2}\) are independent.

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 \(Y\)

Suppose that \(n\) electronic components, each having an exponentially distributed length of life with mean \(\theta,\) are put into operation at the same time. The components operate independently and are observed until \(r\) have failed \((r \leq n) .\) Let \(W_{j}\) denote the length of time until the \(j\) th failure, with \(W_{1} \leq W_{2} \leq \cdots \leq W_{r} .\) Let \(T_{j}=W_{j}-W_{j-1}\) for \(j \geq 2\) and \(T_{1}=W_{1} .\) Notice that \(T_{j}\) measures the time elapsed between successive failures. a. Show that \(T_{j},\) for \(j=1,2, \ldots, r,\) has an exponential distribution with mean \(\theta /(n-j+1)\) b. Show that $$U_{r}=\sum_{j=1}^{r} W_{j}+(n-r) W_{r}=\sum_{j=1}^{r}(n-j+1) T_{j}$$ and hence that \(E\left(U_{r}\right)=r \theta .[U_{r} \text { is called the total observed life, and we can use } U_{r} / r\) as an . approximation to (or "estimator" of) \(\theta .]\)
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Statistics

Read Explanation

Discrete Mathematics

Read Explanation

Probability and Statistics

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.