Problem 44 - Calculus Early Transcendentals: Pearson New International Edition

Question

Suppose that the random variables \((X, Y, Z)\) have joint PDF $$ f(x, y, z)= \begin{cases}k x y, & \text { if } 0 \leq x \leq y ; 0 \leq y \leq 4 ; 0 \leq z \leq 2 \\ 0, & \text { otherwise }\end{cases} $$ Find each of the following: (a) \(k\) (b) \(P(X>2)\) (c) \(E(X)\)

Step-by-Step Solution

Answer

(a) \(k = \frac{1}{64}\), (b) \(P(X>2) = \frac{7}{12}\), (c) \(E(X) = \frac{32}{15}\).

Step 1: Determine the Region of Integration

The joint PDF is given for the region where \(0 \leq x \leq y\), \(0 \leq y \leq 4\), and \(0 \leq z \leq 2\). This means \(x\) must also be between 0 and 4, since it must be less than or equal to \(y\). Integrating outside these bounds should yield 0.

Step 2: Find Normalizing Constant \(k\)

To ensure the PDF integrates to 1 across the specified region, we integrate: \[\int_0^2 \int_0^4 \int_0^y kxy \, dx \, dy \, dz = 1.\] Solving this, first integrate with respect to \(x\):\[\int_0^y x \, dx = \frac{1}{2}y^2.\] Then integrate with the result for \(y\):\[ k \int_0^4 \frac{1}{2}y^3 \, dy = k \cdot \frac{1}{8}y^4\Big|_0^4 = k \cdot 32.\]Finally, integrate with respect to \(z\):\[ k \cdot 32 \int_0^2 \, dz = 64k.\] Set this equal to 1: \(64k = 1\), giving \(k = \frac{1}{64}\).

Step 3: Probability \(P(X>2)\)

The event \(X > 2\) happens when \(2 < x \leq y \leq 4\) and for all \(z\). Calculate\[\int_0^2 \int_2^4 \int_x^4 \frac{1}{64}xy \, dy \, dx \, dz.\] Integrating with respect to \(y\),\[\int_x^4 y \, dy = \frac{1}{2}(16 - x^2).\] Then with respect to \(x\),\[\int_2^4 \left(\frac{16x - x^3}{2}\right) \, dx = \left[8x^2 - \frac{x^4}{8}\right]_2^4 = \frac{56}{3}.\] Integrate with respect to \(z\):\[\frac{1}{64} \cdot \frac{56}{3} \cdot 2 = \frac{7}{12}.\]

Step 4: Expectation \(E(X)\)

Calculate the expectation:\[ E(X) = \int_0^2 \int_0^4 \int_0^y x \left(\frac{1}{64}xy\right) \, dx \, dy \, dz. \] First integrate with respect to \(x\):\[ \int_0^y x^2 \, dx = \frac{1}{3}y^3.\] Integrate with respect to \(y\):\[ \int_0^4 \frac{1}{3}y^4 \, dy = \frac{1}{15}y^5\Bigg|_0^4 = \frac{1024}{15}.\] Integrate with respect to \(z\):\[ \frac{1}{64} \cdot \frac{1024}{15} \cdot 2 = \frac{32}{15}. \]

Key Concepts

Joint Probability Density Function

A Joint Probability Density Function (Joint PDF) describes the probability distribution of two or more random variables together. It helps in understanding how multiple random variables interact with each other within a specified range. The concept becomes crucial when you want to calculate probabilities that involve more than one variable. In this exercise, the joint PDF is given for the random variables \(X, Y,\text{ and } Z\), represented as \(f(x, y, z)\). This function is defined with conditions specifying the ranges where the probability density is non-zero:

• \(0 \leq x \leq y\)
• \(0 \leq y \leq 4\)
• \(0 \leq z \leq 2\)

Outside these specified intervals, the joint PDF is zero. This means all the probability mass is concentrated within the stated bounds; beyond these bounds, no probability exists. The articualtion of these conditions is critical for determining which sections of the space contribute to the actual calculations, such as normalization and expected values.

Expectation in Probability

Expectation in probability measures the expected value or average of a random variable. It provides an idea of what a random variable will look like on average if the experiment were repeated many times. In multivariable calculus, this involves integrating over all variables within their limits. For our function \(f(x, y, z)\), calculating \(E(X)\) involves:

• Integrating the function \(x \cdot f(x, y, z)\) over the entire region.
• This involves computing multiple integrals, starting from \(x\) and moving through \(y\) and \(z\).

The process requires first calculating the integral of \(x^2\) with respect to \(x\), followed by \(y\) over its range, and lastly \(z\). The solution concludes with an expectation value of \(\frac{32}{15}\). This number provides a weighted average value of \(X\), taking into account how \(X\) varies with \(Y\) and \(Z\).

Region of Integration

The region of integration in a multivariable calculus context defines the space over which the variables exist, particularly for the expressions involved. In this problem, specifying the region explicitly is crucial to ensure correct computations.This specific region is characterized by the inequalities:

• \(0 \leq x \leq y\): \(X\) must be less than or equal to \(Y\).
• \(0 \leq y \leq 4\): \(Y\) has an upper limit of 4, dictating the maximum value \(X\) can take.
• \(0 \leq z \leq 2\): \(Z\) ranges between 0 to 2.

The region forms a defined and precise bounded space where all calculations — normalization, probabilities, and expectations, occur. Integrating outside these bounds results in zero, meaning they do not contribute to the overall integral value or calculations. Mapping out this region thoroughly ensures we focus on only pertinent sections when performing integrations, ensuring accuracy in results.

Probability Calculations

When calculating probabilities involving random variables, we focus on finding the measure of a part of the total possible outcomes. Our main task, in this case, is not just integrating over any region but over specified bounds that fulfill a condition.For instance, in this exercise, the computation of \(P(X>2)\) is carried out over a subset of the predefined region where \(X > 2\). The solution involves:

• Setting the bounds for \(x\) as between 2 and \(y\), due to the condition \(2 < x \leq y\).
• Integrating for values of \(y\) and \(z\) within their boundaries.
• Executing separate integrations for \(x\), \(y\), then \(z\), finally yielding \(\frac{7}{12}\).

These steps illustrate how we can pinpoint specific probabilities by focusing calculations only within desired parameters. Such calculations are common in statistics and help us derive meaningful insights from random variables' behavior.