Question

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

Short answer

(a) \( k = \frac{1}{64} \), (b) \( P(X > 2) = \frac{3}{8} \), (c) \( E(X) = \frac{8}{3} \)

Step-by-step solution

Find the Constant k

The joint PDF must integrate to 1 over the entire support. So, we need to solve the following integral for \( k \):\[\int_0^2 \int_0^4 \int_0^y k x y \, dx \, dy \, dz = 1\]Let's first integrate with respect to \(x\):\[\int_0^y k x y \, dx = k y \left[\frac{x^2}{2}\right]_0^y = k y \left(\frac{y^2}{2}\right) = \frac{k y^3}{2}\]Next, integrate the result with respect to \(y\):\[\int_0^4 \frac{k y^3}{2} \, dy = \frac{k}{2} \left[\frac{y^4}{4}\right]_0^4 = \frac{k}{2} \left(\frac{256}{4}\right) = 32k\]Now, integrate with respect to \(z\):\[\int_0^2 32k \, dz = 64k\]Set the total integral equal to 1 to solve for \(k\):\[64k = 1 \rightarrow k = \frac{1}{64}\]

Calculate P(X > 2)

We need to find \( P(X > 2) \). This is given by the integral:\[P(X > 2) = \int_0^2 \int_0^4 \int_2^y \frac{1}{64} x y \, dx \, dy \, dz\]First, integrate with respect to \(x\):\[\int_2^y \frac{1}{64} x y \, dx = \frac{y}{64} \left[\frac{x^2}{2}\right]_2^y = \frac{y}{64} \left(\frac{y^2 - 4}{2}\right) = \frac{y(y^2 - 4)}{128}\]Next, integrate the result with respect to \(y\):\[\int_2^4 \frac{y(y^2 - 4)}{128} \, dy = \int_2^4 \frac{y^3 - 4y}{128} \, dy\]Calculate:\[= \frac{1}{128} \left[\frac{y^4}{4} - 2y^2\right]_2^4 \]\[= \frac{1}{128} \left(\left(\frac{256}{4} - 32\right) - \left(\frac{16}{4} - 8\right)\right)\]\[= \frac{1}{128} (32 - 8) = \frac{1}{128} \times 24 = \frac{3}{16}\]Finally, integrate with respect to \(z\), which leads to:\[P(X > 2) = \int_0^2 \frac{3}{16} \, dz = \frac{3}{16} \times 2 = \frac{3}{8}\]

Calculate E(X)

The expected value \(E(X)\) is calculated using the integral:\[E(X) = \int_0^2 \int_0^4 \int_0^y x \cdot \frac{1}{64} x y \, dx \, dy \, dz\]First, integrate with respect to \(x\):\[\int_0^y \frac{1}{64} x^2 y \, dx = \frac{y}{64} \left[\frac{x^3}{3}\right]_0^y = \frac{y^4}{192}\]Next, integrate the result with respect to \(y\):\[\int_0^4 \frac{y^4}{192} \, dy = \frac{1}{192} \left[\frac{y^5}{5}\right]_0^4\]Calculate:\[= \frac{1}{192} \times \frac{1024}{5} = \frac{256}{192} = \frac{32}{24} = \frac{4}{3}\]Lastly, integrate with respect to \(z\):\[E(X) = \int_0^2 \frac{4}{3} \, dz = \frac{4}{3} \times 2 = \frac{8}{3}\]

Key concepts

Understanding Random Variables Integration

When dealing with joint probability density functions, it's important to understand how integration is used to analyze random variables. For any joint PDF involving multiple variables, the total probability across the defined region should equal 1. To ensure this, a constant \(k\) is introduced, which needs to be determined through integration.

Here's how it works in practice:

• Imagine a three-dimensional space outlined by constraints such as \((0 \leq x \leq y ; 0 \leq y \leq 4 ; 0 \leq z \leq 2)\).
• To find \(k\), integrate the function \(kxy\) over this entire space.
• Conduct the integration gradually over each variable: begin with \(x\), then \(y\), and finally \(z\).
• By setting the resulting total integral equal to 1, you can solve for \(k\).

This process allows us to properly normalize the function, ensuring that it represents a valid probability distribution.

Calculating Expected Value

The expected value or expectation of a random variable gives a measure of the "center" of the distribution. When calculating the expected value \(E(X)\) for a joint probability density function, integrate over all combinations of other variables.

Here's the step-by-step process:

• To find \(E(X)\), modify the original function to include \(x imes f(x, y, z)\).
• Again, integrate over the limits of all defined variables: starting with \(x\), then \(y\), and lastly \(z\).
• The inclusion of \(x\) in the integration process means focusing on the contributions of each potential value of \(x\) to the overall expectation.
• The result gives a weighted average value, which is essentially the expected value \(E(X)\).

This approach provides a comprehensive understanding of where most values are centered within the space defined by the joint PDF.

Understanding Conditional Probability

Conditional probability is a measure of the probability of an event given that another event has occurred. When dealing with joint PDFs, conditional probabilities often involve integrating parts of the function.

Let's break it down:

• To find a conditional probability like \(P(X > 2)\), focus on the region where \((X > 2)\) holds within the constraint set by \((0 \leq x \leq y)\).
• Set up the integration by altering the limits of \(x\) from \((0, y)\) to \((2, y)\), reflecting the condition \(X > 2\).
• Integrate over the specified range for all variables \(y\) and \(z\) as before, only keeping ranges that satisfy \(X > 2\).
• This process allows you to find the likelihood of \(X > 2\) with respect to the entire joint distribution.

Through these steps, you accurately calculate the probability of certain outcomes given fixed conditions, essential for deeper probabilistic insights.