/*! 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 You are driving on a highway at ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

You are driving on a highway at speed \(X_{1}\). Cars entering this highway after you travel at speeds \(X_{2}, X_{3}, \ldots\). Suppose these \(X_{i}\) 's are independent and identically distributed with pdf \(f(x)\) and cdf \(F(x)\). Unfortunately there is no way for a faster car to pass a slower one - it will catch up to the slower one and then travel at the same speed. For example, if \(X_{1}=52.3, X_{2}=37.5\), and \(X_{3}=42.8\), then no car will catch up to yours, but the third car will catch up to the second. Let \(N=\) the number of cars that ultimately travel at your speed (in your "cohort"), including your own car. Possible values of \(N\) are \(1,2,3, \ldots\). Show that the pmf of \(N\) is \(p(n)=1 /[n(n+1)]\), and then determine the expected number of cars in your cohort. [Hint: \(N\) \(=3\) requires that \(X_{1}

Short Answer

Expert verified
The pmf of \(N\) is \(p(n) = \frac{1}{n(n+1)}\), and the expected value is 1.

Step by step solution

01

Define the Problem

We are tasked with finding the probability mass function (pmf) of the number of cars in a cohort, including the current car, which travel at the same speed driven by a fixed car speed on a highway. The scenario involves cars entering the highway at speeds that are independently and identically distributed.
02

Understanding Condition for Cohort Size

To have exactly \(N = n\) cars in the cohort, the first car must be faster than \(n-1\) cars entering the highway, while the \(n\)-th car must be slower than the first car, allowing it to join the cohort. This is similar to the problem of ordering \(n\) random numbers.
03

Identify the CDF Conditions

The condition \(X_1 < X_2, X_1 < X_3, \ldots, X_1 < X_n\) indicates that \(X_1\) is the minimum of the first \(n\) cars' speeds. Therefore, for a fixed \(X_1\), the probability that all \(n-1\) cars entering are slower than \(X_1\) is \(F(X_1)^{n-1}\).
04

Finalize Cohort Size Condition

The \(n\)-th car, on the contrary, must be the first to be faster, so its speed would be at least \(X_1\), i.e. \(1 - F(X_1)\). This setup resembles a geometric distribution with a success on the \(n\)-th trial.
05

Calculate the Probability

The probability \(P(N = n)\) is given by the event all \(n-1\) cars are slower than \(X_1\) and the \(n\)-th car is faster than \(X_1\), giving us \ P(N = n) = F(X_1)^{n-1} \times (1 - F(X_1)).
06

Integrate Over All Speeds

Since the speed \(X_1\) is continuous over all possible values, we integrate the above expression over the distribution of \(X_1\) to obtain \ P(N = n) = \int_{0}^{1} n(1-x)^{n-1} \, dx, \leading to the closed form of the pmf: \(p(n) = \frac{1}{n(n+1)}\).
07

Expected Value Calculation

The expected value of \(N\), denoted by \(E[N]\), is calculated using \ E[N] = \sum_{n=1}^{\infty} n \times \frac{1}{n(n+1)} = \sum_{n=1}^{\infty} \frac{1}{n+1}. \This series ultimately converges to a value of 1.

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.

Identically Distributed Random Variables
In probability theory, the concept of identically distributed random variables is quite crucial, especially when dealing with real-world scenarios like our highway driving example. Identically distributed refers to random variables sharing the same probability distribution. They behave in a 'similar' manner statistically, even if they are independent.
  • Think of identically distributed random variables as similarly flavored candies from the same batch. They might be different instances, but they all taste the same.
  • In mathematical terms, for random variables \(X_1, X_2, \ldots, X_n\), their distributions are identical if they all have the same probability density function (pdf) \(f(x)\) and the same cumulative distribution function (cdf) \(F(x)\).
These identically distributed variables are independent, meaning that the outcome of one does not affect the others. This property is vital in our problem where each car enters the highway independently at its own speed.For the driving scenario, if you know the distribution of the speed of one entering car, you know the distribution for all cars. This similar statistical behavior enables the modeling of the cohort size using well-established probability rules.
Geometric Distribution
The geometric distribution is a fascinating concept when you're dealing with the first occurrence of an event in a series of trials. It's like counting how many coins you need to flip until you get a heads.
  • In the context of our exercise, it characterizes the situation where you are waiting for the first car to be slower than your car after entering the highway.
  • Consider each car entering the highway as a trial. You count how many cars are faster than your car until one is not. That's when the first 'success' happens.
Under a geometric distribution, if each car independently has the same probability of being slower than your car, the number of cars in the cohort falls naturally into this distribution. The probability of a car joining your cohort decreases geometrically as more cars join, resulting in rare long cohorts.In mathematical terms, for a geometric distribution, the pmf is given by a sequence resembling fractions. For our exercise, the pmf was formally determined to be \(p(n) = \frac{1}{n(n+1)}\), describing the probability of having exactly \(n\) cars traveling at your speed, which includes some interesting interplay of exponential decay in probability.
Expected Value Computation
Expected value is a fundamental concept in probability, often described as the "long-run average" value of repetitions of the experiment.
  • It is especially handy when predicting outcomes over repeated trials, giving us a single value that reflects the mean from the probability distribution of outcomes.
  • For our highway driving problem, the expected number of cars (\(E[N]\)) in your cohort can be calculated using the probability mass function derived earlier.
To compute the expected value, you'd sum over all possible cohort sizes. For this exercise, it's expressed as:\[E[N] = \sum_{n=1}^{\infty} n \times \frac{1}{n(n+1)}\]This summation means we're multiplying each possible number of cars in the cohort by its probability, then adding them all up. Surprisingly, this series converges smoothly, and in our example, resolves to a value of 1, which aligns with intuitive results given the cohort behavior modeled by the geometric distribution.

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

Let \(Y_{1}\) and \(Y_{n}\) be the smallest and largest order statistics, respectively, from a random sample of size \(n\), and let \(W_{2}=Y_{n}-Y_{1}\) (this is the sample range). a. Let \(W_{1}=Y_{1}\), obtain the joint pdf of the \(W_{i}\) 's (use the method of Section 5.4), and then derive an expression involving an integral for the pdf of the sample range. b. For the case in which the random sample is from a uniform \((0,1)\) distribution, carry out the integration of (a) to obtain an explicit formula for the pdf of the sample range.

A stick is one foot long. You break it at a point \(X\) (measured from the left end) chosen randomly uniformly along its length. Then you break the left part at a point \(Y\) chosen randomly uniformly along its length. In other words, \(X\) is uniformly distributed between 0 and 1 and, given \(X=x, Y\) is uniformly distributed between 0 and \(x\). a. Determine \(E(Y \mid X=x)\) and then \(V(Y \mid X=x)\). Is \(E(Y \mid X=x)\) a linear function of \(x\) ? b. Determine \(f(x, y)\) using \(f_{X}(x)\) and \(f_{Y \mid X}(y \mid x)\). c. Determine \(f_{Y}(y)\). d. Use \(f_{Y}(y)\) from (c) to get \(E(Y)\) and \(V(Y)\). e. Use (a) and the theorem of this section to get \(E(Y)\) and \(V(Y)\).

Two components of a computer have the following joint pdf for their useful lifetimes \(X\) and \(Y\) : $$ f(x, y)=\left\\{\begin{array}{cc} x e^{-x(1+y)} & x \geq 0 \text { and } y \geq 0 \\ 0 & \text { otherwise } \end{array}\right. $$ a. What is the probability that the lifetime \(X\) of the first component exceeds 3 ? b. What are the marginal pdf's of \(X\) and \(Y\) ? Are the two lifetimes independent? Explain. c. What is the probability that the lifetime of at least one component exceeds \(3 ?\)

A pizza place has two phones. On each phone the waiting time until the first call is exponentially distributed with mean one minute. Each phone is not influenced by the other. Let \(X\) be the shorter of the two waiting times and let \(Y\) be the longer. It can be shown that the joint pdf of \(X\) and \(Y\) is \(f(x, y)=2 e^{-(x+y)}, 0

The number of customers waiting for gift-wrap service at a department store is an rv \(X\) with possible values \(0,1,2,3,4\) and corresponding probabilities \(.1, .2, .3, .25, .15 .\) A randomly selected customer will have 1,2 , or 3 packages for wrapping with probabilities . 6, .3, and .1, respectively. Let \(Y=\) the total number of packages to be wrapped for the customers waiting in line (assume that the number of packages submitted by one customer is independent of the number submitted by any other customer). a. Determine \(P(X=3, Y=3)\), that is, \(p(3,3)\). b. Determine \(p(4,11)\).

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.