Problem 6 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

Let \(X\) and \(Y\) be two independent \(\operatorname{Ber}\left(\frac{1}{2}\right)\) random variables. Define random variables \(U\) and \(V\) by: $$ U=X+Y \quad \text { and } \quad V=|X-Y| $$ a. Determine the joint and marginal probability distributions of \(U\) and \(V\). b. Find out whether \(U\) and \(V\) are dependent or independent.

Step-by-Step Solution

Answer

U and V are dependent random variables.

Step 1: Define Random Variables

Consider the random variables given: \( X \) and \( Y \) are independent and both have a Bernoulli distribution with parameter \( p = \frac{1}{2} \). The random variable \( U = X + Y \). Here, \( U \) takes values \( 0, 1, \) and \( 2 \). The random variable \( V = |X - Y| \). Here, \( V \) takes values \( 0 \) and \( 1 \) since \( X \) and \( Y \) are either \( 0 \) or \( 1 \).

Step 2: Compute Probability Distribution of U

For \( U = 0 \), both \( X \) and \( Y \) need to be 0. Since \( X \) and \( Y \) are independent, \( P(U = 0) = P(X = 0) P(Y = 0) = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \). For \( U = 1 \), either \( X = 1, Y = 0 \) or \( X = 0, Y = 1 \). Thus, \( P(U = 1) = P(X = 1)P(Y = 0) + P(X = 0)P(Y = 1) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \). For \( U = 2 \), both \( X \) and \( Y \) must be 1. Hence, \( P(U = 2) = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \).

Step 3: Compute Probability Distribution of V

For \( V = 0 \), both \( X \) and \( Y \) need to be equal, i.e., both are 0 or both are 1. So, \( P(V = 0) = P(X = 0, Y = 0) + P(X = 1, Y = 1) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \). For \( V = 1 \), either \( X = 1, Y = 0 \) or \( X = 0, Y = 1 \), which already have been calculated as \( \frac{1}{4} + \frac{1}{4} = \frac{1}{2} \).

Step 4: Joint Distribution of U and V

We obtain joint probabilities by considering each combination of \( (U, V) \):- \( P(U = 0, V = 0) = P(X = 0, Y = 0) = \frac{1}{4} \) - \( P(U = 1, V = 1) = P(X = 1, Y = 0) + P(X = 0, Y = 1) = \frac{1}{2} \) - \( P(U = 2, V = 0) = P(X = 1, Y = 1) = \frac{1}{4} \)

Step 5: Determine Independence of U and V

For \( U \) and \( V \) to be independent, \( P(U = u, V = v) = P(U = u) \cdot P(V = v) \) for all \( u, v \). Checking specific probabilities: \( P(U = 0, V = 0) = \frac{1}{4} \) and \( P(U = 0) \cdot P(V = 0) = \frac{1}{4} \cdot \frac{1}{2} = \frac{1}{8} \), which are not equal, indicating dependence.

Key Concepts

Bernoulli Distribution

In probability theory, a Bernoulli distribution is one of the simplest kinds of probability distributions. It models a random experiment that has exactly two possible outcomes. These outcomes are usually called "success" and "failure." The Bernoulli distribution is defined by a single parameter, denoted as \( p \), which represents the probability of "success." Thus, the probability of "failure" is \( 1 - p \).

For example, flipping a fair coin is a scenario that can be described by a Bernoulli distribution. In such a distribution, the probability \( p \) for heads (or success) and the probability for tails (or failure) equals \( \frac{1}{2} \).

In the context of the given problem, both \( X \) and \( Y \) are independently Bernoulli distributed with parameter \( p = \frac{1}{2} \). This means each trial (in this case, each variable) can result in a 0 or 1, with equal likelihood. This setup forms the basis for calculating further joint or marginal probabilities involving these random variables.

Random Variables

Random variables are a fundamental concept in probability theory. They are variables that take on numerical values based on the outcomes of a random phenomenon. These variables are not deterministic, meaning their values are not fixed and instead have associated probabilities.

In the problem, we have two initial random variables \( X \) and \( Y \), each following a Bernoulli distribution. We also introduce two derived random variables, \( U \) and \( V \).

• \( U = X + Y \), which sums the outcomes of \( X \) and \( Y \).
• \( V = |X - Y| \), which measures the absolute difference between \( X \) and \( Y \).

Since \( X \) and \( Y \) each can only be 0 or 1, the possible outcomes for \( U \) are 0, 1, or 2. Similarly, \( V \) can either be 0 or 1. Understanding these random variables and how they relate helps establish the basis to find joint probability distributions.

Joint Probability Distribution

The joint probability distribution of two random variables provides the probability of each possible pair of outcomes. This distribution offers a complete description of how multiple random variables are related, which is valuable in understanding their potential correlations and dependencies.

For the given situation, we determine the joint distribution of the random variables \( U \) and \( V \). This involves calculating the probabilities of all combinations, such as \( P(U = 0, V = 0) \), \( P(U = 1, V = 1) \), and so on.

• \( P(U = 0, V = 0) \) results when both \( X \) and \( Y \) are 0.
• \( P(U = 1, V = 1) \) occurs when \( X \) and \( Y \) are different, leading to either \( X = 1, Y = 0 \) or \( X = 0, Y = 1 \).
• \( P(U = 2, V = 0) \) arises when both \( X \) and \( Y \) are 1.

These joint probabilities are critical in determining whether \( U \) and \( V \) are independent or dependent. If the joint probability equals the product of the marginal probabilities for each possible event, the variables are independent. In this exercise, evidence of inequality in probabilities confirmed that \( U \) and \( V \) are indeed dependent.