/*! 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 15 A prisoner is told that he will ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A prisoner is told that he will be released at a time chosen uniformly at random within the next 24 hours. Let \(T\) denote the time that he is released. What is the hazard function for \(T ?\) For what values of \(t\) is it smallest and largest? If he has been waiting for 5 hours, is it more likely that he will be released in the next few minutes than if he has been waiting for 1 hour?

Short Answer

Expert verified
The hazard function is smallest at \(t=0\) and largest as \(t\) approaches 24. It is more likely he will be released soon after waiting 5 hours than after 1 hour.

Step by step solution

01

Define the Uniform Distribution

The prisoner will be released at a time chosen uniformly within 24 hours, which means the time \( T \) is uniformly distributed between 0 and 24 hours. The probability density function (PDF) for a continuous uniform distribution on the interval \([a, b]\) is given by \(f(t) = \frac{1}{b-a}\). Here, \(a = 0\) and \(b = 24\), so the PDF is \(f(t) = \frac{1}{24}\) for \(0 \le t \le 24\).
02

Define the Cumulative Distribution Function (CDF)

The CDF \(F(t)\) gives the probability that \(T\) is less than or equal to \(t\). For a uniform distribution, it is calculated as \(F(t) = \frac{t-a}{b-a}\). Here, it becomes \(F(t) = \frac{t}{24}\) for \(0 \le t \le 24\).
03

Derive the Survivor Function

The survivor function \(S(t)\) describes the probability that the time \(T\) is greater than \(t\). It is given by \(S(t) = 1 - F(t)\). For this uniform distribution, \(S(t) = 1 - \frac{t}{24}\) for \(0 \le t \le 24\).
04

Compute the Hazard Function

The hazard function \(h(t)\) is the PDF divided by the survivor function: \(h(t) = \frac{f(t)}{S(t)}\). Substituting in the expressions for \(f(t)\) and \(S(t)\), we get:\[ h(t) = \frac{\frac{1}{24}}{1 - \frac{t}{24}} = \frac{1}{24 - t} \]\ for \(0 \le t < 24\).
05

Analyze Values for Smallest and Largest Hazard Function

The hazard function \(h(t) = \frac{1}{24 - t}\) reaches its smallest value at \(t = 0\), which is \(\frac{1}{24}\), and it increases as \(t\) approaches 24. It is largest as \(t\) approaches 24.
06

Answer Whether Waiting Changes Likelihood

The hazard function \(h(t)\) increases with time, which means as time goes by, the rate of the event (release) per unit time increases. Therefore, it is more likely the prisoner will be released in the next few minutes after waiting for 5 hours compared to 1 hour.

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.

Understanding Uniform Distribution
In the context of probability and statistics, a uniform distribution is a type of probability distribution where all outcomes are equally likely. For instance, if you roll a fair six-sided die, each face (1 through 6) has an equal chance of landing face up. Similarly, in the exercise, the time at which the prisoner will be released is uniformly distributed over the 24-hour period. This means that any given moment within that time frame is just as likely as any other. This uniform distribution is characterized mathematically by its probability density function (PDF), which we will explore in the next section.
Probability Density Function (PDF)
The probability density function (PDF) is a fundamental concept in understanding how probabilities are distributed over a continuous range of values. For a uniform distribution between two boundaries, say \(a\) and \(b\), the PDF is given by \(f(t) = \frac{1}{b-a}\). This formula reflects that each outcome in this interval has an equal chance of occurring.

In our exercise, since the release time \(T\) is uniformly distributed between 0 and 24 hours, the PDF is \(f(t) = \frac{1}{24}\) for any \(0 \leq t \leq 24\). This constant value indicates that the probability of release does not favor any specific hour within the 24-hour period. It is crucial in calculating other functions such as the cumulative distribution and hazard functions.
Cumulative Distribution Function (CDF)
The cumulative distribution function (CDF) is a useful tool to determine the probability that a random variable will take a value less than or equal to a particular number. For a uniform distribution, the CDF is determined by the formula \(F(t) = \frac{t-a}{b-a}\).

In the scenario of the prisoner's release, it becomes \(F(t) = \frac{t}{24}\) for \(0 \leq t \leq 24\). This function increases linearly from 0 to 1 as \(t\) moves from 0 to 24. Essentially, the CDF tells us the probability of the prisoner being released by any given time, providing an accumulation of probabilities up to \(t\). As \(t\) approaches the upper limit of the interval, the probability of a release increases, reaching 100% by 24 hours.
Survivor Function
The survivor function provides a different perspective by showing the probability that the event of interest has not occurred by a certain time. It complements the cumulative distribution function, as it represents the probability that the time of release \(T\) is greater than some particular value \(t\).

The survivor function is calculated as \(S(t) = 1 - F(t)\), which in our example becomes \(S(t) = 1 - \frac{t}{24}\). This function decreases from 1 to 0 over the interval \(0 \leq t \leq 24\). Practically, it means that at the start, there is a 100% chance the prisoner has not been released, and as time progresses, this chance decreases, indicating an increasing likelihood of release. Understanding this function helps visualize and compute the hazard function, which provides deeper insights into the timing of events within this 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 \(X_{1}, \ldots, X_{n}\) be a sample (i.i.d.) from a distribution function, \(F,\) and let \(F_{n}\) denote the ecdf. Show that $$ \operatorname{Cov}\left[F_{n}(u), F_{n}(v)\right]=\frac{1}{n}[F(m)-F(u) F(v)] $$ where \(m=\min (u, v) .\) Conclude that \(F_{n}(u)\) and \(F_{n}(v)\) are positively correlated: If \(F_{n}(u)\) overshoots \(F(u),\) then \(F_{n}(v)\) will tend to overshoot \(F(v)\)

Barlow, Toland, and Frecman (1984) studicd the lifetimes of Kevlar 49/cpoxy strands subjected to sustained stress. (The space shuttle uses Kevlar/epoxy spherical vessels in an environment of sustained pressure.) The files kevlar?0, kevlar80, and kevlar90 contain the times to failure (in hours) of strands tested at \(70 \%, 80 \%,\) and \(90 \%\) stress levels. What do these data indicate about the nature of the distribution of lifetimes and the effect of increasing stress?

The 2000 U.S. Presidential election was very close and hotly contested. George W. Bush was ultimately appointed to the Presidency by the U.S. Supreme Court. Among the issues was a confusing ballot in Palm Beach County, Florida, the so- called Butterfly Ballot, shown in the following figure. Notice that on this ballot, although the Democrats are listed in the second row on the left, a voter wishing to specify them would have to punch the third hole-punching the second hole would result in a vote for the Reform Party (Pat Buchanan). After the election, many distraught Democratic voters claimed that they had inadvertently voted for Buchanan, a right-wing candidate. The file PalmBeach contains relevant data: vote counts by county in Florida for Buchanan and for four other presidential candidates in \(2000,\) the total vote counts in \(2000,\) the presidential vote counts for three presidential candidates in \(1996,\) the vote count for Buchanan in the 1996 Republican primary, the registration in Buchanan's Reform Party, and the total registration in the county. Does this data support voters' claims that they were misled by the form of the ballot? Start by making two scatterplots: a plot of Buchanan's votes versus Bush's votes in \(2000,\) and a plot of Buchanan's votes in 2000 versus his votes in the 1996 primary.

Consider a sample of size 100 from an exponential distribution with parameter \(\lambda=1\) a. Sketch the approximate standard deviation of the empirical log survival function, \(\log S_{n}(t),\) as a function of \(t\) b. Generate several such samples of size 100 on a computer and for each sample plot the empirical log survival function. Relate the plots to your answer to (a).

Give an example of a probability distribution with increasing failure rate.

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.