/*! 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 3 Suppose that the distribution fu... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose that the distribution function of a discrete random variable \(X\) is given by $$ F(a)= \begin{cases}0 & \text { for } a<0 \\ \frac{1}{3} & \text { for } 0 \leq a<\frac{1}{2} \\ \frac{1}{2} & \text { for } \frac{1}{2} \leq a<\frac{3}{4} \\ 1 & \text { for } a \geq \frac{3}{4}\end{cases} $$ Determine the probability mass function of \(X\).

Short Answer

Expert verified
The PMF of X is: \( p(x) = 1/3 \) at \( x=0 \), \( 1/6 \) at \( x=1/2 \), \( 1/2 \) at \( x=3/4 \).

Step by step solution

01

Understand the Distribution Function

The problem provides a cumulative distribution function (CDF) \( F(a) \) for a discrete random variable \( X \). It describes how probabilities accumulate over certain intervals of \( a \). The CDF is piecewise, providing different probability values for different ranges of \( a \).
02

Identify the Jumps in the CDF

The jumps in the CDF indicate where the probability mass function (PMF) has non-zero values. The function jumps at \( a = 0, \frac{1}{2}, \) and \( \frac{3}{4} \). These jumps are where the probability mass is located.
03

Calculate the Probability Mass

The probability mass at each point is computed by subtracting the CDF values at the edges of the intervals:- For \( X = 0 \), the change is \( F(0) - F(-0^+) = \frac{1}{3} - 0 = \frac{1}{3} \).- For \( X = \frac{1}{2} \), the change is \( F(\frac{1}{2}) - F(0^+) = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} \).- For \( X = \frac{3}{4} \), the change is \( F(\frac{3}{4}) - F(\frac{1/2}^+) = 1 - \frac{1}{2} = \frac{1}{2} \).
04

Write the Probability Mass Function

The probability mass function \( p(x) \) is given by the masses calculated in the previous step:\[ p(x) = \begin{cases} \frac{1}{3} & \text{if } x = 0 \ \frac{1}{6} & \text{if } x = \frac{1}{2} \ \frac{1}{2} & \text{if } x = \frac{3}{4} \ 0 & \text{otherwise} \end{cases} \]

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.

Discrete Random Variable
A discrete random variable is a type of random variable that can take on a finite or countably infinite number of possible outcomes. In other words, it can only assume distinct and separate values. These values usually come from a countable set. For example, the roll of a die results in a discrete random variable, with possible outcomes of 1 through 6.

The key characteristic of discrete random variables is that you can list all the possible values they can take. This makes them different from continuous random variables, which can assume any value within a given range or interval.

The problem we are dealing with features a discrete random variable, denoted by \(X\). This random variable assumes values at specific points where the cumulative distribution function (CDF) jumps. These values represent the points where the probability mass is concentrated. Understanding this helps us analyze how likely certain outcomes are, which is crucial for calculating probabilities with these types of variables.
Cumulative Distribution Function
The cumulative distribution function (CDF) of a random variable tells us the probability that the variable takes on a value less than or equal to a certain level. Specifically, for a discrete random variable \(X\), the CDF \(F(a)\) gives the cumulative probability up to value \(a\). This function is often piecewise, with different rules or expressions for different intervals of \(a\).

In our example, the CDF is given for a discrete random variable \(X\):
  • \(F(a) = 0\) for \(a < 0\)
  • \(F(a) = \frac{1}{3}\) for \(0 \leq a < \frac{1}{2}\)
  • \(F(a) = \frac{1}{2}\) for \(\frac{1}{2} \leq a < \frac{3}{4}\)
  • \(F(a) = 1\) for \(a \geq \frac{3}{4}\)
These steps indicate the cumulative nature of the function: moving from smaller to larger \(a\) captures more probability. Each jump in the CDF occurs at a point where the PMF, or probability mass function, holds non-zero values. By examining these jumps, we can extract important information about the main discrete values of \(X\) and their associated probabilities.
Calculating Probability
Calculating probabilities for discrete random variables often involves analyzing its probability mass function (PMF). The PMF gives the probability that a discrete random variable takes exactly a specific value. This is important because these values define how probability is distributed across different possible outcomes for \(X\).

To determine the PMF from a given CDF, we look at the differences between cumulative probabilities across intervals. This is because, in a CDF, the increase between intervals represents the allocation of probability mass at specific discrete points.

For our example, from the CDF:
  • The probability for \(X = 0\) is \(\frac{1}{3} - 0 = \frac{1}{3}\).
  • For \(X = \frac{1}{2}\), it's \(\frac{1}{2} - \frac{1}{3} = \frac{1}{6}\).
  • And for \(X = \frac{3}{4}\), it's \(1 - \frac{1}{2} = \frac{1}{2}\).
Thus, the PMF can be written as:
  • \(p(x) = \frac{1}{3}\) if \(x = 0\)
  • \(p(x) = \frac{1}{6}\) if \(x = \frac{1}{2}\)
  • \(p(x) = \frac{1}{2}\) if \(x = \frac{3}{4}\)
  • \(p(x) = 0\) otherwise
By following these calculations, you learn how the discrete random variable's behavior is quantified, allowing you to predict outcomes with defined probabilities.

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

A shop receives a batch of 1000 cheap lamps. The odds that a lamp is defective are \(0.1 \%\). Let \(X\) be the number of defective lamps in the batch. a. What kind of distribution does \(X\) have? What is/are the value(s) of parameter(s) of this distribution? b. What is the probability that the batch contains no defective lamps? One defective lamp? More than two defective ones?

You and a friend want to go to a concert, but unfortunately only one ticket is still available. The man who sells the tickets decides to toss a coin until heads appears. In each toss heads appears with probability \(p\), where \(0

We throw a coin until a head turns up for the second time, where \(p\) is the probability that a throw results in a head and we assume that the outcome of each throw is independent of the previous outcomes. Let \(X\) be the number of times we have thrown the coin. a. Determine \(\mathrm{P}(X=2), \mathrm{P}(X=3)\), and \(\mathrm{P}(X=4)\). b. Show that \(\mathrm{P}(X=n)=(n-1) p^{2}(1-p)^{n-2}\) for \(n \geq 2\).

Let \(X\) be a discrete random variable with probability mass function \(p\) given by: \begin{tabular}{ccccc} \(a\) & \(-1\) & 0 & 1 & 2 \\ \hline\(p(a)\) & \(\frac{1}{4}\) & \(\frac{1}{8}\) & \(\frac{1}{8}\) & \(\frac{1}{2}\) \end{tabular} and \(p(a)=0\) for all other \(a\). a. Let the random variable \(Y\) be defined by \(Y=X^{2}\), i.e., if \(X=2\), then \(Y=4\). Calculate the probability mass function of \(Y\). b. Calculate the value of the distribution functions of \(X\) and \(Y\) in \(a=1\), \(a=3 / 4\), and \(a=\pi-3 .\)

A box contains an unknown number \(N\) of identical bolts. In order to get an idea of the size \(N\), we randomly mark one of the bolts from the box. Next we select at random a bolt from the box. If this is the marked bolt we stop, otherwise we return the bolt to the box, and we randomly select a second one, etc. We stop when the selected bolt is the marked one. Let \(X\) be the number of times a bolt was selected. Later (in Exercise 21.11) we will try to find an estimate of \(N\). Here we look at the probability distribution of \(X\). a. What is the probability distribution of \(X ?\) Specify its parameter(s)! b. The drawback of this approach is that \(X\) can attain any of the values \(1,2,3, \ldots\), so that if \(N\) is large we might be sampling from the box for quite a long time. We decide to sample from the box in a slightly different way: after we have randomly marked one of the bolts in the box, we select at random a bolt from the box. If this is the marked one, we stop, otherwise we randomly select a second bolt (we do not return the selected bolt). We stop when we select the marked bolt. Let \(Y\) be the number of times a bolt was selected. Show that \(\mathrm{P}(Y=k)=1 / N\) for \(k=1,2, \ldots, N\) ( \(Y\) has a so-called discrete uniform distribution). c. Instead of randomly marking one bolt in the box, we mark \(m\) bolts, with \(m\) smaller than \(N\). Next, we randomly select \(r\) bolts; \(Z\) is the number of marked bolts in the sample. Show that $$ \mathrm{P}(Z=k)=\frac{\left(\begin{array}{c} m \\ k \end{array}\right)\left(\begin{array}{c} N-m \\ r-k \end{array}\right)}{\left(\begin{array}{c} N \\ r \end{array}\right)}, \quad \text { for } \quad k=0,1,2, \ldots, r $$ ( \(Z\) has a so-called hypergeometric distribution, with parameters \(m, N\), and \(r .\) )

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.