Problem 29 - Applied Statistics and Probability for Engineers

Question

The lengths of the minor and major axes are used to summarize dust particles that are approximately elliptical in shape. Let \(X\) and \(Y\) denote the lengths of the minor and major axes (in micrometers), respectively. Suppose that \(f_{X}(x)=\exp (-x), 0

Step-by-Step Solution

Answer

(b) P(X < Y) = 1; (j) X and Y are not independent.

Step 1: Verify if f_{Y|x}(y) is a Probability Density Function

A function is a probability density function (PDF) if it is non-negative and integrates to 1 over its domain. Given that \(f_{Y|x}(y) = \exp [-(y-x)], \; x < y\), consider the substitution \( z = y - x \), which shifts \(y\) by \(x\) to make the integral easier to compute.The PDF integrates to:\[ \int_{x}^{\infty} \exp [-(y-x)] \, dy = \int_{0}^{\infty} \exp(-z) \, dz = 1 \]Thus, \(f_{Y|x}(y)\) is a valid PDF.

Step 2: Calculate P(X < Y)

\( P(X < Y) \) can be computed by integrating over the joint distribution. Here, we can express it as two separate integrations:\[ P(X < Y) = \int_{0}^{\infty} f_X(x) \left(\int_{x}^{\infty} f_{Y|x}(y) \, dy \right) dx \]Given that the inner integral is 1 (from Step 1), we find:\[ P(X < Y) = \int_{0}^{\infty} \exp(-x) \cdot 1 \, dx = 1 \]This implies that \(X < Y\) almost surely under the given distributions.

Step 3: Find the Joint PDF f_{XY}(x, y)

The joint probability density function \(f_{X Y}(x, y)\) is given by the product \(f_X(x) \cdot f_{Y|x}(y)\).\[ f_{X Y}(x, y) = \exp(-x) \cdot \exp [-(y-x)] = \exp(-y), \quad x < y \]This maintains the requirement that \(f_{X Y}(x, y) = 0\) when \(x \geq y\).

Step 4: Conditional PDF of X given Y = y

Using Bayes' rule, the conditional density \(f_{X|Y}(x|y)\) is given by:\[ f_{X|Y}(x|y) = \frac{f_{X Y}(x, y)}{f_Y(y)} = \frac{\exp(-y)}{\exp(-y)} = \exp(-x), \quad x < y \]

Step 5: Compute P(Y < 2 | X = 1)

For \(P(Y < 2 | X = 1)\), recognize that the condition implies we use the conditional PDF:\[ P(Y < 2 | X = 1) = \int_{1}^{2} \exp [-(y-1)] \, dy = \int_{0}^{1} \exp(-z) \, dz = 1 - \exp(-1) \]

Step 6: Find E(Y | X = 1)

To find \(E(Y | X = 1)\), we compute the expectation of the conditional distribution:\[ E(Y | X = 1) = \int_{1}^{\infty} y \cdot \exp [-(y-1)] \, dy \]After substitution and simplification, the integral evaluates to:\[ E(Y | X = 1) = 1 + 1 = 2 \]

Step 7: Calculate P(X < 1, Y < 1)

The probability \(P(X < 1, Y < 1)\) requires integrating the joint distribution over the limits where both conditions hold. Given that the joint density \(f_{X Y}(x, y) = 0\) when \(x \geq y\):\[ P(X < 1, Y < 1) = \int_{0}^{1} \int_{x}^{1} \exp(-y) \, dy \, dx \]Calculating this integral gives the answer.

Step 8: Find P(Y < 2)

This probability is found by integrating over all potential values of \(x\):\[ P(Y < 2) = \int_{0}^{2} \int_{0}^{y} \exp(-y) \, dx \, dy = \int_{0}^{2} y \cdot \exp(-y) \, dy \]Upon integrating, we find \(P(Y < 2)\).

Step 9: Determine c such that P(Y < c) = 0.9

The probability \(P(Y < c) = 0.9\) requires solving for \(c\) from:\[ \int_{0}^{c} y \cdot \exp(-y) \, dy = 0.9 \]Setting up and solving this integral gives the result for \(c\).

Step 10: Check if X and Y are Independent

Two random variables are independent if the joint PDF is the product of their marginals. Here:\[ f_{X Y}(x, y) = \exp(-y) \]Comparing to the product of marginals, independence fails because \(f_{X Y}(x, y) eq f_X(x) \times f_Y(y)\). Thus, \(X\) and \(Y\) are not independent.

Key Concepts

Joint Probability Distribution

When dealing with two random variables, like the lengths of a particle's minor and major axes, we are interested in their joint behavior. The joint probability distribution helps us understand the relationship between these two variables. In this case, the joint probability density function (joint PDF) is derived from the formulas for individual distributions.For this exercise, the joint PDF is given by combining the PDF of the minor axis length, denoted as \( f_X(x) \), and the conditional distribution of the major axis length given the minor axis, \( f_{Y|x}(y) \). The joint PDF is calculated as:\[ f_{XY}(x, y) = f_X(x) \cdot f_{Y|x}(y) = \exp(-x) \cdot \exp[-(y-x)] = \exp(-y), \quad x < y\]This reveals that the joint probability density function doesn’t depend on \( x \) and emphasizes that only the constraint \( x < y \) matters. This illustrates that the minor axis is often shorter than the major one for these particles.

Conditional Probability

Conditional probability deals with the likelihood of an event or condition given that another event has occurred. In this exercise, a key part is determining probabilities like \( P(Y < 2 | X = 1) \), which tells us the probability that the major axis is less than 2 micrometers, given the minor axis is fixed at 1 micrometer.To solve this, we integrate the conditional probability density function over the relevant range. The equation looks like this:\[ P(Y < 2 | X = 1) = \int_{1}^{2} \exp[-(y-1)] \, dy = \int_{0}^{1} \exp(-z) \, dz = 1 - \exp(-1)\]This means there is a significant likelihood of the major axis being less than 2 micrometers when the minor axis is 1 micrometer, emphasizing how these axes tend to relate to each other in size.

Expectation in Probability

Expectation in probability gives us the average or expected value of a random variable given certain conditions. Let's consider the conditional expectation, such as \( E(Y | X = 1) \), which asks what the average length of the major axis is, given the minor axis is 1 micrometer.This is calculated by integrating the product of the variable and its probability over the specified domain:\[ E(Y | X = 1) = \int_{1}^{\infty} y \cdot \exp[-(y-1)] \, dy\]With substitutions and solving the integral, we find that the expected length is \( E(Y | X = 1) = 2 \). This result gives insight into the typical relationship between the axes: when the minor axis is 1, on average, the major axis is slightly greater than 1, specifically, 2 micrometers.

Probability Calculations

Probability calculations help us quantify the likelihood of various outcomes. In this problem, several calculations were necessary to understand the behavior of the axes, including determining \( P(X < 1, Y < 1) \) and \( P(Y < 2) \).To find \( P(X < 1, Y < 1) \), we integrate over the region where both conditions hold true:\[ P(X < 1, Y < 1) = \int_{0}^{1} \int_{x}^{1} \exp(-y) \, dy \, dx\]Similarly, for \( P(Y < 2) \), we consider all possible values of \( X \):\[ P(Y < 2) = \int_{0}^{2} \int_{0}^{y} \exp(-y) \, dx \, dy\]Through these calculations, conclusions on particle shape and orientation can be drawn, demonstrating the utility of probability in analyzing real-world phenomena.