/*! 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 59 Let \(X_{1}\) denote the time (h... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(X_{1}\) denote the time (hr) it takes to perform a first task and \(X_{2}\) denote the time it takes to perform a second one. The second take always takes at least as long to perform as the first task. The joint pdf of these variables is \(f\left(x_{1}, x_{2}\right)=\left\\{\begin{array}{cl}2\left(x_{1}+x_{2}\right) & 0 \leq x_{1} \leq x_{2} \leq 1 \\ 0 & \text { otherwise }\end{array}\right.\) a. Obtain the pdf of the total completion time for the two tasks. b. Obtain the pdf of the difference \(X_{2}-X_{1}\) between the longer completion time and the shorter time.

Short Answer

Expert verified
a. Total time: piecewise pdf with segments \(y\) and \(2-y\). b. Difference: pdf \(2(1-z)z\).

Step by step solution

01

Define the Transformation for Total Time

To find the pdf of the total completion time for both tasks, we define the new variable \(Y = X_1 + X_2\). We then need to determine the range over which this new variable, \(Y\), is defined. Since \(0 \leq x_1 \leq x_2 \leq 1\), it follows that \(0 \leq x_1 \leq 1\) and \(0 \leq x_2 \leq 1\). Therefore, \(0 \leq x_1 + x_2 \leq 2\). Thus, the range for \(Y\) is \(0 \leq Y \leq 2\).
02

Compute the Joint Cumulative Distribution Function (CDF) for Y

To find the pdf of \(Y\), we first find the CDF, \(F_Y(y) = P(Y \leq y)\). For each fixed \(y\), the integration limits for \(x_1\) and \(x_2\) are from \(0\) to \(\frac{y}{2}\) since \(x_1 + x_2 = y\). The limits must also satisfy \(0 \leq x_1 \leq x_2\). So the integration is performed under the valid intersection of these conditions.
03

Find the CDF Expression for Y

We integrate the joint pdf over the valid range for \(x_1\). The CDF \(F_Y(y)\) can be expressed by the integral:\[F_Y(y) = \int_{0}^{\min(1, y/2)} \int_{x_1}^{\min(1, y-x_1)} 2(x_1+x_2) \ dx_2 \ dx_1\] where the inner integral computes for \(x_2\) from \(x_1\) to \(\min(1,y-x_1)\). Simplify this expression and these boundaries considering the constraints.
04

Differentiate to Find the pdf of Y

Differentiate the CDF \(F_Y(y)\) with respect to \(y\) to get the pdf \(f_Y(y)\). The PDF turns out to be: \[f_Y(y) = \begin{cases} y, & 0 \leq y \leq 1 \ 2-y, & 1 < y \leq 2 \ 0, & \text{otherwise} \end{cases}\]
05

Define the Transformation for Time Difference

To find the pdf of the difference \(Z = X_2 - X_1\), consider the transformation with this new variable. The possible range for \(Z\) is from \( 0 \) to \( 1 \) because the difference must satisfy \(0 \leq x_1 \leq x_2\) and \(x_2 = x_1 + Z\). This means \(0 \leq Z \leq 1\).
06

Compute the Joint CDF for Z

To find the CDF \(F_Z(z) = P(Z \leq z)\), we integrate the joint pdf with respect to \(x_1\) and substitute \(x_2 = x_1 + z\).\[F_Z(z) = \int_{0}^{1} \int_{x_1}^{x_1 + z} 2(x_1+x_2) \, dx_2 \, dx_1\] and simplify this expression under the defined boundaries.
07

Differentiate to Find the pdf of Z

Differentiate \(F_Z(z)\) to get the pdf of \(Z\).\[f_Z(z) = \begin{cases} 2(1-z)z, & 0 \leq z \leq 1 \ 0, & \text{otherwise} \end{cases}\]This gives us the distribution of the difference between the task times.

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Key Concepts

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

Joint Probability Distribution
A joint probability distribution describes the probability of two or more random variables occurring at the same time. In the context of continuous random variables, it specifies a function called a joint probability density function (pdf), denoted as \(f(x_1, x_2)\). This function gives the likelihood of \(X_1\) and \(X_2\) taking on particular values within a specified range.
In the given problem, \(X_1\) and \(X_2\) represent the times taken to complete two tasks. The joint pdf is defined as:
  • \(f(x_1, x_2) = 2(x_1 + x_2)\) for \(0 \leq x_1 \leq x_2 \leq 1\)
  • And \(f(x_1, x_2) = 0\) otherwise.
This means that within the constraint \(0 \leq x_1 \leq x_2 \leq 1\), the probability density is determined by \(2(x_1 + x_2)\). It captures both the individual behaviors of \(X_1\) and \(X_2\), as well as their dependence on each other, since the time for the second task (\(X_2\)) is always greater than or equal to the time for the first task (\(X_1\)). Understanding this relationship is crucial for solving problems that involve the transformation of variables.
Transformation of Variables
Transformation of Variables is a powerful technique in probability theory. It involves changing variables to simplify the analysis of a problem. In our case, transformations help us find the pdf of new variables like the total completion time and the time difference.
For the total completion time of the two tasks, we define a new variable \(Y = X_1 + X_2\). This transformation allows us to describe the combined effect of \(X_1\) and \(X_2\) as a single random variable. Similarly, we define \(Z = X_2 - X_1\), representing the difference in times, which helps to understand the variability between the tasks.
The transformation process involves setting up equations for new variables and finding their ranges. In the exercise:
  • For \(Y\), the total time ranges from 0 to 2 because \(X_1\) and \(X_2\) can independently vary from 0 to 1.
  • For \(Z\), its range is from 0 to 1 given that the second task cannot be completed in less time than the first.
Performing such transformations simplifies the calculations required to find new pdfs. It helps recast complex relationships into manageable forms.
Probability Density Function
The Probability Density Function (pdf) is a critical concept when dealing with continuous random variables. It represents the density of probability at any given point for a random variable. For a function \(f(x)\) to be a probability density function, it must satisfy the condition \(\int_{-\infty}^{\infty} f(x)\,dx = 1\).
Occupying center stage in our analysis is the derivation of pdfs for the transformed variables \(Y\) and \(Z\). By performing transformations:
  • For \(Y = X_1 + X_2\), its pdf is derived by differentiating the cumulative distribution function (CDF) \(F_Y(y)\). The resulting pdf is: \[f_Y(y) = \begin{cases} y, & 0 \leq y \leq 1 \ 2-y, & 1 < y \leq 2 \ 0, & \text{otherwise} \end{cases}\]
  • For the difference \(Z = X_2 - X_1\), the pdf is found similarly with: \[f_Z(z) = \begin{cases} 2(1-z)z, & 0 \leq z \leq 1 \ 0, & \text{otherwise} \end{cases}\]
The pdfs enable us to understand how probability is distributed over time durations, providing deeper insights into task completion and the variability of event outcomes.
Cumulative Distribution Function
The Cumulative Distribution Function (CDF) of a continuous random variable is the probability that the variable takes on a value less than or equal to \(x\). For a random variable \(Y\), the CDF \(F_Y(y)\) is the integral of its pdf from \(-\infty\) to \(y\).
In practical terms, the CDF offers a way to model cumulative probabilities and is especially useful when deriving pdfs for transformed variables. In the current exercise:
  • The CDF for \(Y = X_1 + X_2\) is expressed as: \[F_Y(y) = \int_{0}^{\min(1, y/2)} \int_{x_1}^{\min(1, y-x_1)} 2(x_1+x_2)\, dx_2\, dx_1\]
  • It is differentiated to obtain the pdf \(f_Y(y)\).
Similarly:
  • For the difference \(Z = X_2 - X_1\), integration to find \(F_Z(z)\) and differentiating yields the pdf \(f_Z(z)\).
CDFs act as building blocks in probability theory, allowing us to transition from cumulative probabilities to specific probability densities, thus formulating the required distributions effectively.

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Most popular questions from this chapter

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