/*! 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 174 . Find the joint cdf of the inde... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

. Find the joint cdf of the independent random variables \(X\) and \(Y\), where \(f_{X}(x)=\frac{x}{2}, 0 \leq x \leq 2\), and \(f_{Y}(y)=2 y, 0 \leq y \leq 1 .\)

Short Answer

Expert verified
To find the joint cumulative distribution function (CDF) of the two independent random variables \(X\) and \(Y\), we first compute their individual CDFs given by their specific probability density functions, and then calculate the joint CDF by simply taking the product of the two individual CDFs.

Step by step solution

01

Compute the CDF of X

We first need to calculate the cumulative distribution function (CDF) of \(X\). \[F_{X}(x)=\int_{0}^{x} f_{X}(t) dt \]Substitute \(f_{X}(t) = \frac{t}{2}\) into the equation, the value of \(F_{X}(x)\) is then calculated by integrating \(\frac{t}{2}\) from 0 to \(x\).
02

Compute the CDF of Y

Next, we compute the cumulative distribution function (CDF) of \(Y\). \[F_{Y}(y)=\int_{0}^{y} f_{Y}(s) ds \]Substitute \(f_{Y}(s) = 2s\) into the equation, the value of \(F_{Y}(y)\) is then computed by integrating \(2s\) from 0 to \(y\).
03

Compute the Joint CDF

Since \(X\) and \(Y\) are independent, the joint cumulative distribution function \(F(x,y)\) is the product of \(F_{X}(x)\) and \(F_{Y}(y)\):\[ F(x,y) = F_{X}(x)F_{Y}(y)\]

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.

CDF of independent random variables
Understanding the cumulative distribution function (CDF) of independent random variables is crucial in the field of probability and statistics. If we have two independent random variables, let's call them 'X' and 'Y', their joint CDF can be expressed as the product of their individual CDFs. This is a direct consequence of their independence, which implies that the occurrence of an event for 'X' does not affect the probability of an event for 'Y' and vice versa.

For example, let's consider a step-by-step solution where we are given two independent random variables 'X' and 'Y' with their corresponding probability density functions (PDFs). To find their joint CDF, we would first determine the CDF of each variable separately by integrating their PDFs. Then, because of their independence, we multiply these individual CDFs together to get the joint CDF, which is the probability that 'X' is less than or equal to some value 'x' and 'Y' is less than or equal to some value 'y'. Such straightforward methods greatly simplify the process of understanding multivariate distributions for students.
Probability density function
A probability density function, often abbreviated as PDF, is a function that describes the likelihood of a random variable to take on a certain value. For continuous random variables, the PDF provides the probability that the variable falls within a particular range of values. The fundamental property of a PDF is that the integral over the entire space is equal to 1, representing the certainty that the random variable will take on a value within its range.

The exercise given involves finding the PDFs for the random variables 'X' and 'Y', which are denoted as 'f_X(x)' and 'f_Y(y)', respectively. The PDFs are critical for calculating the CDFs later on. Adequate comprehension of PDFs is essential for students tackling probability theory, as they form the basis for interpreting and computing various statistical measures. It is the cornerstone for building up to more complex concepts, such as the calculation of expected values, variances, and even the application of Bayes' theorem.
Integration in probability
Integration is a powerful tool in probability theory which is often used to calculate cumulative distribution functions (CDFs) from probability density functions (PDFs). Integration can be thought of as a way to add up all probabilities up to a certain point, providing us with the cumulative probability.

For instance, in our exercise, to find the CDF of 'X', we integrate its PDF from the lower limit of the variable up to 'x'. Similarly, to find the CDF of 'Y', we integrate its PDF from zero to 'y'. This process essentially sums up the probabilities of 'X' being less than or equal to 'x' and 'Y' being less than or equal to 'y'. Especially for students, grasping the concept of integration within the context of probability is fundamental, as it is a frequent operation in many probabilistic applications. It enables them to transition from the density function, which shows the probability per unit, to the cumulative function, which illustrates the accumulated probability.

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 gambler plays \(n\) hands of poker. If he wins the \(k\) th hand, he collects \(k\) dollars; if he loses the \(k\) th hand, he collects nothing. Let \(T\) denote his total winnings in \(n\) hands. Assuming that his chances of winning each hand are constant and independent of his success or failure at any other hand, find \(E(T)\) and \(\operatorname{Var}(T)\).

Consider the experiment of tossing a fair coin three times. Let \(X\) denote the number of heads on the last flip, and let \(Y\) denote the total number of heads on the three flips. Find \(p_{X, Y}(x, y)\).

Frankie and Johnny play the following game. Frankie selects a number at random from the interval \([a, b]\). Johnny, not knowing Frankie's number, is to pick a second number from that same interval and pay Frankie an amount, \(W\), equal to the squared difference between

Suppose a life insurance company sells a \(\$ 50,000\), five-year term policy to a twenty-five-year-old woman. At the beginning of each year the woman is alive, the company collects a premium of \(\$ P\). The probability that the woman dies and the company pays the \(\$ 50,000\) is given in the table below. So, for example, in Year 3 , the company loses \(\$ 50,000-\$ P\) with probability \(0.00054\) and gains \(\$ P\) with probability \(1-0.00054=0.99946\). If the company expects to make \(\$ 1000\) on this policy, what should \(P\) be? \begin{tabular}{cc} \hline Year & Probability of Payoff \\ \hline 1 & \(0.00051\) \\ 2 & \(0.00052\) \\ 3 & \(0.00054\) \\ 4 & \(0.00056\) \\ 5 & \(0.00059\) \\ \hline \end{tabular}

A certain brand of fluorescent bulbs will last, on the average, one thousand hours. Suppose that four of these bulbs are installed in an office. What is the probability that all four are still functioning after one thousand fifty hours? If \(X_{i}\) denotes the \(i\) th bulb's life, assume that $$ f_{X_{1}, X_{2}, X_{3}, X_{4}}\left(x_{1}, x_{2}, x_{3}, x_{4}\right)=\prod_{i=1}^{4}\left(\frac{1}{1000}\right) e^{-x / 1000} $$

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.