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Chapter 5: Problem 12

Two components of a minicomputer have the following joint pdf for their useful lifetimes \(X\) and \(Y\) : $$ f(x, y)=\left\\{\begin{array}{cc} x e^{-x(1+y)} & x \geq 0 \text { and } y \geq 0 \\ 0 & \text { otherwise } \end{array}\right. $$ a. What is the probability that the lifetime \(X\) of the first component exceeds 3 ? b. What are the marginal pdf's of \(X\) and \(Y\) ? Are the two lifetimes independent? Explain. c. What is the probability that the lifetime of at least one component exceeds 3 ?

Short Answer

Expert verified
a. \( e^{-3} \); b. \( f_X(x) = e^{-x}, f_Y(y) = \frac{1}{(1+y)^2} \); c. \( 2e^{-3} - e^{-6} \). X and Y are not independent.

Step by step solution

01

Understanding the problem

We have a joint probability density function (pdf) \( f(x, y) = x e^{-x(1+y)} \) for the lifetimes \(X\) and \(Y\) of two components. We need to determine probabilities related to these lifetimes and their marginal pdfs.
02

Step 2a: Find P(X > 3) [Part a]

To find the probability that the lifetime \(X\) exceeds 3, we compute \( P(X > 3) \) by integrating the marginal pdf of \(X\) from 3 to infinity. First, find the marginal pdf of \(X\):\[ f_X(x) = \int_{0}^{\infty} f(x, y) \, dy = \int_{0}^{\infty} x e^{-x(1+y)} \, dy\]This integral evaluates to:\[ f_X(x) = x e^{-x} \int_{0}^{\infty} e^{-xy} \, dy = x e^{-x} \left[\frac{1}{x} e^{-xy} \right]_{0}^{\infty} = e^{-x}\]Thus, \( f_X(x) = e^{-x} \) for \(x \geq 0\).Now compute \( P(X > 3) \):\[ P(X > 3) = \int_{3}^{\infty} e^{-x} \, dx = \left[-e^{-x} \right]_{3}^{\infty} = 0 - (-e^{-3}) = e^{-3}\]
03

Step 2b: Find marginal pdf's [Part b]

We've already found \( f_X(x) = e^{-x} \) for \( x \geq 0 \).Now, find \( f_Y(y) \):\[ f_Y(y) = \int_{0}^{\infty} f(x, y) \, dx = \int_{0}^{\infty} x e^{-x(1+y)} \, dx\]This integrates to:\[ f_Y(y) = \int_{0}^{\infty} x e^{-x} e^{-xy} \, dx = \int_{0}^{\infty} x e^{-x(1+y)} \, dx\]By substituting \(u = x(1+y)\) and finding derivatives, the integral simplifies to:\[ f_Y(y) = \frac{1}{(1+y)^2}, \text{ for } y \geq 0\]The marginal pdfs are \( f_X(x) = e^{-x} \) and \( f_Y(y) = \frac{1}{(1+y)^2} \). Since \( f(x, y) eq f_X(x)f_Y(y) \), \(X\) and \(Y\) are not independent.
04

Probability of either exceeding 3 [Part c]

To find the probability that the lifetime of at least one component exceeds 3, use the complement rule:\[ P(X > 3 \text{ or } Y > 3) = 1 - P(X \leq 3 \text{ and } Y \leq 3)\]First, compute \( P(X \leq 3 \text{ and } Y \leq 3) \) by integrating over both \( x \) and \( y \):\[ \int_{0}^{3} \int_{0}^{3} x e^{-x(1+y)} \, dy \, dx\]Evaluating the integral:\[ \int_{0}^{3} \left( x e^{-x} \left[ \frac{e^{-xy}}{x} \right]_{0}^{3} \right) dx = \int_{0}^{3} e^{-x}(1 - e^{-3x}) dx\]Solving this gives:\[ P(X \leq 3, Y \leq 3) = 1 - (2e^{-3} - e^{-6})\]Therefore, \( P(X > 3 \text{ or } Y > 3) = 1 - (1 - (2e^{-3} - e^{-6})) = 2e^{-3} - e^{-6} \).

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Key Concepts

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

Marginal PDF
Understanding marginal probability density functions (PDFs) is crucial when dealing with multiple random variables. In essence, a marginal PDF represents the probability distribution of one random variable in a joint setting, with all others integrated out.

To compute the marginal PDF of a single variable, we need to integrate the joint PDF over the other variables. For instance, given a joint PDF of two variables, \(f(x, y)\), the marginal PDF for \(X\), denoted as \(f_X(x)\), is calculated by integrating over all possible values of \(Y\):
\[ f_X(x) = \int_{-\infty}^{\infty} f(x, y) \, dy \]
For our specific problem, this computation leads us to the marginal PDF \(f_X(x) = e^{-x}\), indicating an exponential distribution for \(X\).

Similarly, the marginal PDF for \(Y\), \(f_Y(y)\), involves integrating over all \(X\):
\[ f_Y(y) = \int_{-\infty}^{\infty} f(x, y) \, dx \]
In this problem, it produces \(f_Y(y) = \frac{1}{(1+y)^2}\), showcasing a different distribution for \(Y\). Each marginal PDF reveals vital insights about the behavior of individual variables, independent of each other.
Probability Integration
Integration in probability is essential to find probabilities over continuous random variables. It involves integrating the probability density function over the desired range. This process captures the total probability mass, or likelihood, of the variable's occurrence within specified intervals.
In the problem, to find \(P(X > 3)\), we compute:
\[ P(X > 3) = \int_{3}^{\infty} f_X(x) \, dx \]
Here, \(f_X(x)\) is the marginal PDF of \(X\). We integrate from 3 to infinity, covering all values that exceed 3, obtaining \(e^{-3}\), which is the likelihood of \(X\) surpassing 3 units.

For the joint event where neither \(X\) nor \(Y\) exceeds 3, a double integration over their joint PDF is necessary:
\[ \int_{0}^{3} \int_{0}^{3} f(x, y) \, dy \, dx \]
This computation gives the probability that both \(X\) and \(Y\) are less than or equal to 3, helping us eventually find the probability that at least one exceeds 3 through:
\[ P(X > 3 \text{ or } Y > 3) = 1 - P(X \leq 3, Y \leq 3) \]
Integration is thus a fundamental tool, helping us extract meaningful probabilities from density functions, and is particularly potent in multidimensional probability scenarios.
Independent Random Variables
Determining whether random variables are independent can greatly simplify complex problems and enhance understanding of their interactions. Two variables, \(X\) and \(Y\), are independent if the joint PDF \(f(x, y)\) equals the product of their marginal PDFs:
\[ f(x, y) = f_X(x) \times f_Y(y) \]
In our context, after computing the marginals, we found that \(f(x, y) eq f_X(x) \cdot f_Y(y)\). This indicates a dependency between \(X\) and \(Y\) in this particular joint scenario.

Independence implies that knowing the outcome of one variable provides no information about the other. However, when joint and marginal PDFs don't match this condition, variable interactions exist, affecting their combined probabilities.

Understanding independence in probability helps in simplifying the analysis and interpretation of pearred outcomes across variable sets. If two variables are independent, it means their behaviors are distinct and don't influence each other's probabilities.

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Most popular questions from this chapter

A company maintains three offices in a certain region, each staffed by two employees. Information concerning yearly salaries ( 1000 s of dollars) is as follows: \(\begin{array}{lcccccc}\text { Office } & 1 & 1 & 2 & 2 & 3 & 3 \\ \text { Employee } & 1 & 2 & 3 & 4 & 5 & 6 \\ \text { Salary } & 29.7 & 33.6 & 30.2 & 33.6 & 25.8 & 29.7\end{array}\) a. Suppose two of these employees are randomly selected from among the six (without replacement). Determine the sampling distribution of the sample mean salary \(\bar{X}\). b. Suppose one of the three offices is randomly selected. Let \(X_{1}\) and \(X_{2}\) denote the salaries of the two employees. Determine the sampling distribution of \(\bar{X}\). c. How does \(E(\bar{X})\) from parts (a) and (b) compare to the population mean salary \(\mu\) ?

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In cost estimation, the total cost of a project is the sum of component task costs. Each of these costs is a random variable with a probability distribution. It is customary to obtain information about the total cost distribution by adding together characteristics of the individual component cost distributions-this is called the "roll-up" procedure. For example, \(E\left(X_{1}+\ldots+\right.\) \(\left.X_{n}\right)=E\left(X_{1}\right)+\cdots+E\left(X_{n}\right)\), so the roll-up procedure is valid for mean cost. Suppose that there are two component tasks and that \(X_{1}\) and \(X_{2}\) are independent, normally distributed random variables. Is the roll-up procedure valid for the 75 th percentile? That is, is the 75 th percentile of the distribution of \(X_{1}+X_{2}\) the same as the sum of the 75 th percentiles of the two individual distributions? If not, what is the relationship between the percentile of the sum and the sum of percentiles? For what percentiles is the roll-up procedure valid in this case?

Each front tire on a particular type of vehicle is supposed to be filled to a pressure of 26 psi. Suppose the actual air pressure in each tire is a random variable- \(X\) for the right tire and \(Y\) for the left tire, with joint pdf $$ f(x, y)=\left\\{\begin{array}{cc} K\left(x^{2}+y^{2}\right) & 20 \leq x \leq 30,20 \leq y \leq 30 \\ 0 & \text { otherwise } \end{array}\right. $$ a. What is the value of \(K\) ? b. What is the probability that both tires are underfilled? c. What is the probability that the difference in air pressure between the two tires is at most 2 psi? d. Determine the (marginal) distribution of air pressure in the right tire alone. e. Are \(X\) and \(Y\) independent rv's? d. Determine the (marginal) distribution of air pressure in the right tire alone. e. Are \(X\) and \(Y\) independent rv's?

A health-food store stocks two different brands of a certain type of grain. Let \(X=\) the amount (lb) of brand \(A\) on hand and \(Y=\) the amount of brand \(B\) on hand. Suppose the joint pdf of \(X\) and \(Y\) is $$ f(x, y)= \begin{cases}k x y & x \geq 0, y \geq 0,20 \leq x+y \leq 30 \\ 0 & \text { otherwise }\end{cases} $$ a. Draw the region of positive density and determine the value of \(k\). b. Are \(X\) and \(Y\) independent? Answer by first deriving the marginal puf of each variable. c. Compute \(P(X+Y \leq 25)\). d. What is the expected total amount of this grain on hand? e. \(\operatorname{Compute} \operatorname{Cov}(X, Y)\) and \(\operatorname{Corr}(X, Y)\). f. What is the variance of the total amount of grain on hand?
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