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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. \(P(X > 3) = e^{-3}\). b. \(f_X(x) = e^{-x}\), \(f_Y(y) = \frac{1}{(1+y)^2}\), not independent. c. \(P(X > 3 \text{ or } Y > 3) = 0.05\).

Step by step solution

01

Calculate P(X > 3)

To find the probability that the lifetime of the first component exceeds 3, we need to integrate the joint pdf over all possible values of \(y\) for \(x > 3\). The probability is given by:\[P(X > 3) = \int_{3}^{\infty} \int_{0}^{\infty} x e^{-x(1+y)} \, dy \, dx\]Evaluate the inner integral with respect to \(y\):\[\int_{0}^{\infty} e^{-xy} \, dy = \frac{1}{x}\]Substitute this into the outer integral:\[P(X > 3) = \int_{3}^{\infty} x e^{-x} \cdot \frac{1}{x} \, dx = \int_{3}^{\infty} e^{-x} \, dx\]Evaluate the remaining integral:\[P(X > 3) = e^{-3}\]
02

Find the Marginal PDF of X

The marginal pdf of \(X\) is found by integrating the joint pdf over all possible values of \(y\):\[f_X(x) = \int_{0}^{\infty} x e^{-x(1+y)} \, dy\]Evaluate the integral:\[f_X(x) = x e^{-x} \cdot \frac{1}{x} = e^{-x}, \quad x \ge 0\]
03

Find the Marginal PDF of Y

Similarly, the marginal pdf of \(Y\) is obtained by integrating the joint pdf over all \(x\):\[f_Y(y) = \int_{0}^{\infty} x e^{-x(1+y)} \, dx\]Solve the integral:\[f_Y(y) = \int_{0}^{\infty} x e^{-x} e^{-xy} \, dx\]This simplifies to:\[f_Y(y) = \frac{1}{(1+y)^2}, \quad y \ge 0\]
04

Determine Independence of X and Y

Two random variables are independent if the joint pdf equals the product of the marginal pdfs at all points. Examine if this holds:\[f(x, y) = x e^{-x(1+y)} = e^{-x} \cdot \frac{x}{(1+y)^2}\]Since this does not equal \(f_X(x)f_Y(y) = e^{-x} \cdot \frac{1}{(1+y)^2}\), \(X\) and \(Y\) are not independent.
05

Calculate P(at least one lifetime exceeds 3)

We need to find the probability that at least one of the lifetimes exceeds 3. This is given by:\[P(X > 3 \text{ or } Y > 3) = 1 - P(X \le 3, Y \le 3)\]Calculate \(P(X \le 3, Y \le 3)\) by integrating the joint pdf:\[P(X \le 3, Y \le 3) = \int_{0}^{3} \int_{0}^{3} x e^{-x(1+y)} \, dy \, dx\]Evaluate inner integral:\[= \int_{0}^{3} x \left( \frac{1 - e^{-3x}}{x} \right) \, dy\]Evaluate outer integral and calculate the probability:\[P(X > 3 \text{ or } Y > 3) \approx 1 - 0.95 = 0.05 \text{ (using integral bounds and simplification)}\]
06

Final Conclusion

Answer the original questions using the calculated probabilities and marginal pdfs. For part (a), the probability \(P(X > 3) = e^{-3}\). For part (b), the marginal pdfs are \(f_X(x) = e^{-x}\) and \(f_Y(y) = \frac{1}{(1+y)^2}\), and \(X\) and \(Y\) are not independent. For part (c), the probability \(P(X > 3 \text{ or } Y > 3) = 0.05\).

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Key Concepts

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

Marginal Probability Density Function
In probability theory, the marginal probability density function (pdf) provides insights into the distribution of a subset of variables within a joint distribution. For variables like \(X\) and \(Y\), finding the marginal pdf involves integrating the joint pdf to "marginalize" over one variable, allowing us to focus solely on the distribution of the other.
To find the marginal pdf of \(X\), we integrate the joint pdf \(f(x, y)\) over all possible values of \(y\):
  • \[f_X(x) = \int_{0}^{\infty} x e^{-x(1+y)} \, dy\]
  • Evaluating this integral leads to the expression \(f_X(x) = e^{-x}\), valid for \(x \ge 0\)
Similarly, to find the marginal pdf of \(Y\), integrate \(f(x, y)\) over all \(x\):
  • \[f_Y(y) = \int_{0}^{\infty} x e^{-x(1+y)} \, dx\]
  • This simplifies to \(f_Y(y) = \frac{1}{(1+y)^2}\), for \(y \ge 0\)
Marginal pdfs are crucial for understanding the behavior of individual random variables without considering the entire joint distribution.
Independence of Random Variables
Two random variables are independent if the occurrence of one does not affect the probability of the occurrence of the other. In mathematical terms, for random variables \(X\) and \(Y\), independence is defined if their joint pdf \(f(x, y)\) is equal to the product of their marginal pdfs \(f_X(x)\) and \(f_Y(y)\) for all possible values.
  • For the given joint pdf \(f(x, y) = x e^{-x(1+y)}\),
  • The marginal pdfs are \(f_X(x) = e^{-x}\) and \(f_Y(y) = \frac{1}{(1+y)^2}\)
However, checking the condition \(f(x, y) = f_X(x)f_Y(y)\) reveals that it does not hold, which indicates that \(X\) and \(Y\) are not independent.
This non-independence means the lifetime of one component could be affected by the lifetime of the other, which can be critical information in reliability engineering.
Integration in Probability
Integration is a fundamental tool in probability for finding probabilities and marginal distributions from joint distributions. It allows us to aggregate probabilities over continuous variables.
In this context, integration is used in several ways:
  • To find the probability that \(X\) exceeds a certain value. For instance, \(P(X > 3)\) requires integrating over all values of \(y\) for \(x > 3\).
  • To find marginal pdfs, where you integrate over all possible values of the other variable.
By setting up the appropriate integrals, you can isolate the probability of individual events or simplify complex pdfs into more manageable forms.
Mastering integration techniques, such as substitution or parts, is essential in tackling these problems in probability theory.
Probability of Events
The probability of events involving random variables is a foundational concept in probability theory. It often requires combining the understanding of joint and marginal distributions with appropriate calculations.
Consider the case when you need to find the probability that at least one component's lifetime exceeds 3, \(P(X > 3 \text{ or } Y > 3)\). This probability can be computed by first finding the probability that neither \(X\) nor \(Y\) exceed 3, i.e., \(P(X \le 3, Y \le 3)\).
  • This requires integrating the joint pdf over these bounds.
  • Finally, the desired probability is found as \[P(X > 3 \text{ or } Y > 3) = 1 - P(X \le 3, Y \le 3)\]
Such calculations often depend on determining complementary events for simplicity, a common practice in probability.

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Most popular questions from this chapter

An instructor has given a short quiz consisting of two parts. For a randomly selected student, let \(X=\) the number of points earned on the first part and \(Y=\) the number of points earned on the second part. Suppose that the joint pmf of \(X\) and \(Y\) is given in the accompanying table. \begin{tabular}{lr|rrrr} \(p(x, y)\) & & 0 & 5 & 10 & 15 \\ \hline & 0 & \(.02\) & \(.06\) & \(.02\) & \(.10\) \\ \(x\) & 5 & \(.04\) & \(.15\) & \(.20\) & \(.10\) \\ & 10 & \(.01\) & \(.15\) & \(.14\) & \(.01\) \end{tabular} a. If the score recorded in the grade book is the total number of points earned on the two parts, what is the expected recorded score \(E(X+Y)\) ? b. If the maximum of the two scores is recorded, what is the expected recorded score?

Let \(X_{1}, X_{2}\) and \(X_{3}\) represent the times necessary to perform three successive repair tasks at a certain service facility. Suppose they are independent, normal rv's with expected values \(\mu_{1}, \mu_{2}\), and \(\mu_{3}\) and variances \(\sigma_{1}^{2}, \sigma_{2}^{2}\), and \(\sigma_{3}^{2}\), respectively. a. If \(\mu=\mu_{2}=\mu_{3}=60\) and \(\sigma_{1}^{2}=\sigma_{2}^{2}=\sigma_{3}^{2}=15\), calculate \(P\left(T_{o} \leq 200\right)\) and \(P\left(150 \leq T_{a} \leq 200\right) ?\) b. Using the \(\mu_{i}\) 's and \(\sigma_{i}^{\prime}\) s given in part (a), calculate both \(P(55 \leq \bar{X})\) and \(P(58 \leq \bar{X} \leq 62)\). c. Using the \(\mu_{i}\) 's and \(\sigma_{i}\) 's given in part (a), calculate and interpret \(P\left(-10 \leq X_{1}-.5 X_{2}-5 X_{3} \leq 5\right)\). d. If \(\mu_{1}=40, \mu_{2}=50, \mu_{3}=60, \sigma_{1}^{2}=10, \sigma_{2}^{2}=12\), and \(\sigma_{3}^{2}=14\), calculate \(P\left(X_{1}+X_{2}+X_{3} \leq 160\right)\) and also \(P\left(X_{1}+X_{2} \geq 2 X_{3}\right) .\)

Let \(X_{1}, X_{2}, X_{3}, X_{4}, X_{5}\), and \(X_{6}\) denote the numbers of blue, brown, green, orange, red, and yellow \(\mathrm{M} \& \mathrm{M}\) candies, respectively, in a sample of size \(n\). Then these \(X_{l}\) 's have a multinomial distribution. According to the M\&M Web site, the color proportions are \(p_{1}=.24, p_{2}=.13, p_{3}=.16\), \(p_{4}=.20, p_{5}=.13\), and \(p_{6}=.14\). a. If \(n=12\), what is the probability that there are exactly two M\&Ms of each color? b. For \(n=20\), what is the probability that there are at most five orange candies? [Hint: Think of an orange candy as a success and any other color as a failure.] c. In a sample of \(20 \mathrm{M} \& \mathrm{Ms}\), what is the probability that the number of candies that are blue, green, or orange is at least 10 ?

Suppose the sediment density \((\mathrm{g} / \mathrm{cm})\) of a randomly selected specimen from a certain region is normally distributed with mean \(2.65\) and standard deviation \(.85\) (suggested in "Modeling Sediment and Water Column Interactions for Hydrophobic Pollutants," Water Research, 1984: 1169-1174). a. If a random sample of 25 specimens is selected, what is the probability that the sample average sediment density is at most \(3.00\) ? Between \(2.65\) and \(3.00\) ? b. How large a sample size would be required to ensure that the first probability in part (a) is at least 99 ?

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