Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

. In the proof of \(11.1 .1\), we considered the case in which \(p_{3}p_{1}+p_{2}\) and create a Dutch book for him. Hint: Let \(d=p_{3}-\left(p_{1}+p_{2}\right)\). The gambler buys from the person \(C_{1}\) at a premium price of \(p_{1}+d / 4\) and \(C_{2}\) for \(p_{2}+d / 4\). Then the gambler sells \(C_{1} \cup C_{2}\) to that person at a discount of \(p_{3}-d / 4\). All those are good deals for that person who believes that \(p_{1}, p_{2}, p_{3}\) are correct with \(p_{3}>p_{1}+p_{2} .\) Show that the person has a Dutch book.

Short Answer

Expert verified
Yes, the person has a Dutch book because the expected return from the bets is greater than the total cost incurred, resulting in a sure win.

Step by step solution

01

Define the Key Components

For the existence of a Dutch book, the total cost paid should be lower than the expected value of the bet. The gambler buys \(C_{1}\) for \(p_{1}+d / 4\) and \(C_{2}\) for \(p_{2}+d / 4\), and sells \(C_{1} \cup C_{2}\) for \(p_{3}-d / 4\). Here, \(d = p_{3}-(p_{1}+p_{2})\).
02

Calculate the Total Cost of the Bets

The total cost spent by the gambler is \(p_{1}+d / 4 + p_{2}+d / 4 + p_{3}-d / 4 = p_{1}+p_{2}+p_{3}=1\). This can be derived from the fact that the probabilities should add up to 1 in a probability space.
03

Calculate the Expected Return

For \(C_{1}\), the minimum return is 1, and for \(C_{2}\), the minimum return is still 1, regardless of whether the gambler wins or loses. Thus, the total return is \(1+1 =2\).
04

Compare the Total Cost and Expected Return

It is clear from the prior calculations that the expected return (2) is more than the total cost (1). Therefore, this set of bets leads to a sure win for the gambler hence creating a Dutch book for the person.

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 Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Probability Theory
Probability theory is a branch of mathematics that deals with calculating the likelihood of different events happening. It involves the use of numbers to represent how likely an event is to occur. The basic idea is that probabilities are numbers between 0 and 1, where 0 means an event will not happen, and 1 means it definitely will.

For example, when flipping a fair coin, there are two possible outcomes: heads or tails. Since each outcome is equally likely, the probability of obtaining heads is 0.5. In our exercise, the probabilities don't just stop at simple events but form the core of proving or disproving complex theories like the Dutch Book Theorem.

In the context of a Dutch Book, the concept of probability plays a crucial role. It ensures that the probabilities of all outcomes should add up to 1. Suppose one believes that the sum of probabilities is less than or more than 1, this opens up possibilities for a Dutch Book situation, where a series of bets is set up to guarantee a profit for someone at the expense of the bettor.
Exploring Gambling Strategies
Gambling strategies often involve making decisions based on the perceived probabilities and potential payouts. These strategies can be as diverse as the kinds of bets being placed. A popular strategy in probability and gambling discussions is avoiding a Dutch Book scenario.

A **Dutch Book** is a set of bets that ensures a win, regardless of the outcome of events. It's named after Dutch bookmakers who were supposedly able to construct such bets. The strategy arises when a bettor's belief in probabilities is inconsistent, allowing someone to exploit the situation by constructing a sequence of favorable bets.
  • The gambler buys and sells bets in a way where, due to misestimated probabilities, they are guaranteed to profit.
  • This involves transactions like buying a series of bets or selling combined ones, resulting in assured gains.

In the given exercise, the suggested gambling strategy illustrates how the Dutch Book can work: by believing incorrect estimates of probability, the person ends up making decisions that a strategic gambler can exploit. The gambler buys and sells bets at prices driven by the person's belief in flawed probabilities, ensuring a profit.
Calculating Expected Value
Expected value calculations are a fundamental concept in both probability theory and gambling. The expected value is essentially the average outcome one can anticipate if they were to repeat an action many times under the same conditions.

**Expected Value Formula**: The expected value \(EV\) of a random variable is calculated by multiplying each possible outcome by its probability and then summing all these products:\(EV = \displaystyle \sum X_i \times P(X_i)\)

Where \(X_i\) represents each outcome and \(P(X_i)\) its probability.

In the context of the Dutch Book exercise:
  • The gambler's total cost of bets equals 1 because each component, when combined, matches the entire probability space.
  • The expected return, however, amounts to 2, given the minimum returns irrespective of outcomes, leading to a certain profit.

This mismatch between the cost paid and the expected value is what guarantees a profit, highlighting the power of expected value in assessing the potential benefits of different strategies in gambling or betting scenarios.

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

. Consider the Bayes model $$ \begin{aligned} X_{i} \mid \theta & \sim \operatorname{iid} \Gamma\left(1, \frac{1}{\theta}\right) \\ \Theta \mid \beta & \sim \Gamma(2, \beta) \end{aligned} $$ By performing the following steps obtain the empirical Bayes estimate of \(\theta\). (a) Obtain the likelihood function $$ m(\mathbf{x} \mid \beta)=\int_{0}^{\infty} f(\mathbf{x} \mid \theta) h(\theta \mid \beta) d \theta $$ (b) Obtain the mle \(\widehat{\beta}\) of \(\beta\) for the likelihood \(m(\mathbf{x} \mid \beta)\). (c) Show that the posterior distribution of \(\Theta\) given \(\mathbf{x}\) and \(\widehat{\beta}\) is a gamma distribution.

Consider the Bayes model \(X_{i} \mid \theta, i=1,2, \ldots n \sim\) iid with distribution \(\Gamma(1, \theta), \theta>0\) $$ \Theta \sim h(\theta) \propto \frac{1}{\theta} $$ (a) Show that \(h(\theta)\) is in the class of Jeffrys priors. (b) Show that the posterior pdf is $$ h(\theta \mid y) \propto\left(\frac{1}{\theta}\right)^{n+2-1} e^{-y / \theta} $$ where \(y=\sum_{i=1}^{n} x_{i}\) (c) Show that if \(\tau=\theta^{-1}\) then the posterior \(k(\tau \mid y)\) is the pdf of a \(\Gamma(n, 1 / y)\) distribution. (d) Determine the the posterior pdf of \(2 y \tau\). Use it to obtain a \((1-\alpha) 100 \%\) credible interval for \(\theta\). (e) Use the posterior pdf in Part (d) to determine a Bayesian test for the hypotheses \(H_{0}: \theta \geq \theta_{0}\) versus \(H_{1}: \theta<\theta_{0}\), where \(\theta_{0}\) is specified.

Let \(X_{1}, X_{2}, \ldots, X_{10}\) be a random sample of size \(n=10\) from a gamma distribution with \(\alpha=3\) and \(\beta=1 / \theta\). Suppose we believe that \(\theta\) has a gamma distribution with \(\alpha=10\) and \(\beta=2\) (a) Find the posterior distribution of \(\theta\). (b) If the observed \(\bar{x}=18.2\), what is the Bayes point estimate associated with square error loss function. (c) What is the Bayes point estimate using the mode of the posterior distribution? (d) Comment on an HDR interval estimate for \(\theta\). Would it be easier to find one having equal tail probabilities? Hint: Can the posterior distribution be related to a chi-square distribution?

Consider the Bayes model \(X_{i} \mid \theta, i=1,2, \ldots n \sim\) iid with distribution Poisson \((\theta), \theta>0\) $$ \Theta \sim h(\theta) \propto \theta^{-1 / 2} $$ (a) Show that \(h(\theta)\) is in the class of Jeffrys priors. (b) Show that the posterior pdf of \(2 n \theta\) is the pdf of a \(\chi^{2}(2 y+1)\) distribution, where \(y=\sum_{i=1}^{n} x_{i}\) (c) Use the posterior pdf of Part (b) to obtain a \((1-\alpha) 100 \%\) credible interval for \(\theta\) (d) Use the posterior pdf in Part (d) to determine a Bayesian test for the hypotheses \(H_{0}: \theta \geq \theta_{0}\) versus \(H_{1}: \theta<\theta_{0}\), where \(\theta_{0}\) is specified.

In Example 11.2.2 let \(n=30, \alpha=10\), and \(\beta=5\) so that \(\delta(y)=(10+y) / 45\) is the Bayes' estimate of \(\theta\). (a) If \(Y\) has a binomial distribution \(b(30, \theta)\), compute the risk \(E\left\\{[\theta-\delta(Y)]^{2}\right\\}\). (b) Find values of \(\theta\) for which the risk of Part (a) is less than \(\theta(1-\theta) / 30\), the risk associated with the maximum likelihood estimator \(Y / n\) of \(\theta\).

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.

Sign-up for free