Problem 167 - An Introduction to Mathematical Statistics and Its Applications

Question

The four random variables \(W, X, Y\), and \(Z\) have the multivariate pdf $$ f_{W, X, Y, Z}(w, x, y, z)=16 w x y z $$ for \(0 \leq w \leq 1,0 \leq x \leq 1,0 \leq y \leq 1\), and \(0 \leq z \leq 1\). Find the marginal pdf, \(\bar{f}_{W, X}(w, x)\), and use it to compute \(P\left(0 \leq W \leq \frac{1}{2}, \frac{1}{2} \leq X \leq 1\right) .\)

Step-by-Step Solution

Answer

The marginal pdf is \(\bar{f}_{W, X}(w, x) = 4wx\), and the required probability is 1/6.

Step 1: Define the Given Joint pdf

The multivariate probability density function \(f_{W, X, Y, Z}(w, x, y, z) = 16 w x y z \) is given for \(0 \leq w \leq 1\), \(0 \leq x \leq 1\), \(0 \leq y \leq 1\), and \(0 \leq z \leq 1\).

Step 2: Find the Marginal pdf

To find the marginal pdf, \(\bar{f}_{W, X}(w, x)\), we have to integrate the given function over the variables Y and Z. This gives us: \[\bar{f}_{W, X}(w, x) = \int_0^1 \int_0^1 f_{W, X, Y, Z}(w, x, y, z) dy dz = \int_0^1 \int_0^1 (16wxyz) dy dz.\] Simplifying it further using the fact that \(\int_0^1 y dy = 1/2\) and \(\int_0^1 z dz = 1/2\), we get for the final result: \[\bar{f}_{W, X}(w, x) = 4wx.\] This is valid for \(0 \leq w \leq 1\) and \(0 \leq x \leq 1\).

Step 3: Compute the Probability

The probability \(P(0 \leq W \leq \frac{1}{2}, \frac{1}{2} \leq X \leq 1)\) can then be found by integrating the marginal pdf over the stated intervals. This is the definition of the probability for a continuous random variable, i.e., \[P(a \leq W \leq b, c \leq X \leq d) = \int_b^a \int_d^c \bar{f}_{W, X}(w, x) dw dx .\] So we can write: \[ P(0 \leq W \leq \frac{1}{2}, \frac{1}{2} \leq X \leq 1) = \int_{0}^{\frac{1}{2}} \int_{\frac{1}{2}}^{1} \bar{f}_{W, X}(w, x) dw dx = \int_{0}^{\frac{1}{2}} \int_{\frac{1}{2}}^{1} (4wx) dw dx. \] Evaluating the above integrals gives us the final probability which is 1/6.

Key Concepts

Multivariate Probability Density Function

When dealing with more than one random variable at a time, we often use a Multivariate Probability Density Function (pdf). In the given exercise, we encounter four random variables: \( W, X, Y, \) and \( Z \), each ranging from 0 to 1. The multivariate pdf is defined as \( f_{W, X, Y, Z}(w, x, y, z) = 16wxyz \). This function describes the joint probability distribution of our random variables over their entire range.
To manage such functions, we must understand the structure of joints: in this case, the combination of the probabilities via their product over the given intervals. When you have multiple random variables, the joint pdf helps in describing how their values, when considered together, behave within a given space. The integration of these variables is necessary to simplify or find marginal pdfs or specific derived probabilities.

Integration in Probability

Integration is a fundamental operation in probability, especially when dealing with continuous random variables. It's used to find the marginal probability density functions and specific probabilities computed over certain intervals. In the exercise, we found the marginal pdf, \( \bar{f}_{W, X}(w, x) \), by integrating the multivariate pdf over the unwanted variables \( Y \) and \( Z \). This step is crucial in breaking down the joint dependency of multiple variables into simpler forms.
The integration steps involved simplifying the term \( 16wxyz \). First, integrate over \( y \) and \( z \) individually from 0 to 1, with each of these integrals resulting in \( 1/2 \). As a result, combining these outcomes gives us the marginal pdf \( \bar{f}_{W, X}(w, x) = 4wx \).
In solving probabilities like \( P(0 \leq W \leq \frac{1}{2}, \frac{1}{2} \leq X \leq 1) \), integration again plays a role. Here, the marginal pdf is further integrated within specific bounds to find the required probability, showing how integration helps translate densities into actual probability measures.

Continuous Random Variables

Continuous random variables can take any value within a given range. In contrast to discrete variables, we cannot list all the possible values, but we can describe them using functions. The pdf of a continuous variable essentially provides the likelihood of the variable falling within a particular small interval.
For \( W, X, Y, \) and \( Z \) involved here, each is a continuous variable spanning from 0 to 1. The challenge with continuous random variables is effectively computing probabilities, as some might hope to find specific values, but continuous variables focus on ranges. This is where probability distribution functions come into play; through integration, these functions help derive meaningful statistics such as probabilities over specified intervals.
It's essential to understand that for continuous variables, the probability of them taking an exact value is effectively zero; instead, their distributions are channeled through functions and calculated over intervals, just as accomplished in the exercise's solution.