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Chapter 1: Problem 6

If the sample space is \(\mathcal{C}=\\{c:-\infty

Short Answer

Expert verified
The given set function is not a probability set function since its integration over the entire sample space gives 2, not 1. We need to multiply the integrand by the constant \(0.5\) or \(1/2\) in order to make it a probability set function.

Step by step solution

01

Check whether the given function is a probability set function

We integrate the given function \( e^{-|x|} \) over the entire sample space \(\mathcal{C}\). This is done using two separate integrals due to the absolute value operation, resulting in the below expressions: \[\int_{-\infty}^{0} e^{x} dx + \int_{0}^{\infty} e^{-x} dx\]Now we can proceed with the integration.
02

Calculate the integrals

First we solve the integrals separately. This gives: \[\[-e^x]_{-\infty}^{0} + [-e^{-x}]_{0}^{\infty}\] = 1 + 1 = 2\]
03

Conclude whether given set function is a probability function

Since the integral of the absolute function over the entire sample space results in 2, not 1, it can be concluded that the given set function is not a probability set function.
04

Find the constant to make it a probability set function

To turn the original function into a probability set function, we need to find a constant that makes the total integral equal to 1. Since we know the total integral was 2 for the unadjusted function, the necessary constant is simply the reciprocal of 2, giving a constant of \(0.5\) or \(1/2\).

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Key Concepts

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

Mathematical Statistics
Mathematical statistics is a branch of mathematics that deals with the collection, analysis, interpretation, and presentation of data. It uses probability theory as a foundational tool to make inferences about a population based on a sample. One of the essential parts of mathematical statistics is the concept of a probability function, which assigns probabilities to events in a sample space.

For the function to be a probability set function, it must satisfy specific properties, including non-negativity, normalization, and additivity. In the context of our exercise where we are evaluating the integral of the function over a set, we're essentially verifying whether the function adheres to these properties. The integral across the entire sample space should equal one (normalization property), and since it doesn't, the given function cannot be considered a probability set function without further modification.
Integration in Statistics
Integration in statistics is pivotal in calculating probabilities and expected values, particularly when dealing with continuous random variables. Integrals are mathematically formal ways to sum up infinitely small pieces across a continuum, which for probability distributions translates to finding the total probability across a range of outcomes.

In the given problem, the technique of integration is used to evaluate the area under the curve of the function representing the event probabilities across the continuum of values. However, the result of the integration process shows a sum greater than one, indicating that the function, as it stands, isn't a legitimate probability density function. This is because, in probability theory, the total probability for all possible outcomes must equate to one (a requirement known as the normalization condition).
Normalization Constant
A normalization constant is a factor used in mathematical statistics to adjust a function so the area under its curve equals one, which is a prerequisite for a probability density function. The concept is grounded in the need for the integral of a probability density function over its entire range to be exactly 1, confirming that the sum of probabilities of all possible outcomes is certain.

To normalize a function, as in our exercise, we divide the function by the integral's result when integrated over the entire range of possible outcomes. Here, because the original integral equated to 2, the normalization constant is half (\(1/2\) or 0.5). Multiplying the integrand by this constant adjusts the outcome of the integral to 1, thereby transforming the set function into a proper probability set function, fulfilling the integral property of total probability.

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Most popular questions from this chapter

Players \(A\) and \(B\) play a sequence of independent games. Player \(A\) throws a die first and wins on a "six." If he fails, \(B\) throws and wins on a "five" or "six." If he fails, \(A\) throws and wins on a "four," "five," or "six." And so on. Find the probability of each player winning the sequence.

The distribution of the random variable \(X\) in Example \(1.7 .3\) is a member of the log- \(F\) familily. Another member has the cdf $$ F(x)=\left[1+\frac{2}{3} e^{-x}\right]^{-5 / 2}, \quad-\infty

Suppose there are three curtains. Behind one curtain there is a nice prize, while behind the other two there are worthless prizes. A contestant selects one curtain at random, and then Monte Hall opens one of the other two curtains to reveal a worthless prize. Hall then expresses the willingness to trade the curtain that the contestant has chosen for the other curtain that has not been opened. Should the contestant switch curtains or stick with the one that she has? To answer the question, determine the probability that she wins the prize if she switches.

To join a certain club, a person must be either a statistician or a mathematician or both. Of the 25 members in this club, 19 are statisticians and 16 are mathematicians. How many persons in the club are both a statistician and a mathematician?

Let \(C_{1}\) and \(C_{2}\) be independent events with \(P\left(C_{1}\right)=0.6\) and \(P\left(C_{2}\right)=0.3\). Compute (a) \(P\left(C_{1} \cap C_{2}\right)\), (b) \(P\left(C_{1} \cup C_{2}\right)\), and (c) \(P\left(C_{1} \cup C_{2}^{c}\right)\).
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