Problem 3 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

Let \(X\) and \(Y\) be two independent random variables, where \(X\) has a \(\operatorname{Ber}(p)\) distribution, and \(Y\) has a \(\operatorname{Ber}(q)\) distribution. When \(p=q=r\), we know that \(X+Y\) has a \(\operatorname{Bin}(2, r)\) distribution. Suppose that \(p=1 / 2\) and \(q=1 / 4 .\) Determine \(\mathrm{P}(X+Y=k)\), for \(k=0,1,2\), and conclude that \(X+Y\) does not have a binomial distribution.

Step-by-Step Solution

Answer

P(X+Y=0)=3/8, P(X+Y=1)=1/2, P(X+Y=2)=1/8. X+Y is not binomial.

Step 1: Understanding the Problem

We're given two independent random variables, \(X\) and \(Y\), where \(X\) follows a Bernoulli distribution with parameter \(p = \frac{1}{2}\) and \(Y\) follows a Bernoulli distribution with parameter \(q = \frac{1}{4}\). Our task is to compute \(\mathrm{P}(X+Y=k)\) for \(k=0, 1, 2\) and show that \(X+Y\) does not have a binomial distribution.

Step 2: Calculating P(X=0) and P(X=1)

For a Bernoulli random variable \(X\) with parameter \(p\), \(\mathrm{P}(X=0) = 1-p\) and \(\mathrm{P}(X=1) = p\). Thus, \(\mathrm{P}(X=0) = \frac{1}{2}\) and \(\mathrm{P}(X=1) = \frac{1}{2}\).

Step 3: Calculating P(Y=0) and P(Y=1)

For a Bernoulli random variable \(Y\) with parameter \(q\), \(\mathrm{P}(Y=0) = 1-q\) and \(\mathrm{P}(Y=1) = q\). Thus, \(\mathrm{P}(Y=0) = \frac{3}{4}\) and \(\mathrm{P}(Y=1) = \frac{1}{4}\).

Step 4: Calculating P(X + Y = 0)

The event \(X + Y = 0\) occurs when both \(X=0\) and \(Y=0\). Since \(X\) and \(Y\) are independent, \(\mathrm{P}(X + Y = 0) = \mathrm{P}(X=0) \times \mathrm{P}(Y=0) = \frac{1}{2} \times \frac{3}{4} = \frac{3}{8}\).

Step 5: Calculating P(X + Y = 1)

The event \(X + Y = 1\) occurs in two scenarios: \(X = 1, Y = 0\) or \(X = 0, Y = 1\). The probability is \(\mathrm{P}(X=1)\times\mathrm{P}(Y=0) + \mathrm{P}(X=0)\times\mathrm{P}(Y=1) = \frac{1}{2} \times \frac{3}{4} + \frac{1}{2} \times \frac{1}{4} = \frac{3}{8} + \frac{1}{8} = \frac{1}{2}\).

Step 6: Calculating P(X + Y = 2)

The event \(X + Y = 2\) occurs only when \(X = 1\) and \(Y = 1\). Thus, \(\mathrm{P}(X + Y = 2) = \mathrm{P}(X=1) \times \mathrm{P}(Y=1) = \frac{1}{2} \times \frac{1}{4} = \frac{1}{8}\).

Step 7: Conclusion about Distribution

Summing the probabilities \(\mathrm{P}(X + Y = 0)\), \(\mathrm{P}(X + Y = 1)\), and \(\mathrm{P}(X + Y = 2)\), we get \(\frac{3}{8} + \frac{1}{2} + \frac{1}{8} = 1\), confirming a valid distribution. However, the distribution does not match a binomial distribution of \(\operatorname{Bin}(2, r)\) due to the differing probabilities, thus demonstrating that \(X+Y\) is not binomial.

Key Concepts

Bernoulli Distribution

A Bernoulli distribution is one of the simplest forms of probability distribution and is used to represent a single experiment that has two possible outcomes, typically labeled as success and failure. The Bernoulli distribution is characterized by a single parameter, \( p \), which represents the probability of success.

• If \( X \) is a random variable with a Bernoulli distribution and \( p \) is the probability of success, then \( \mathrm{P}(X = 1) = p \) and \( \mathrm{P}(X = 0) = 1 - p \).
• The expected value or mean of a Bernoulli distribution is \( p \), and the variance is \( p(1-p) \).

In the context of our problem, we have two independent random variables, \( X \) and \( Y \), each following a Bernoulli distribution. \( X \) has a parameter \( p = \frac{1}{2} \), meaning there is an equal chance of success and failure. Meanwhile, \( Y \) has a parameter \( q = \frac{1}{4} \), indicating a lower probability of success compared to failure.

Binomial Distribution

The binomial distribution is a more complex distribution that deals with the number of successes in a fixed number of independent Bernoulli trials. It is defined by two parameters: \( n \), the number of trials, and \( r \), the probability of success in a single trial.

• A random variable \( Z \) that follows a binomial distribution is denoted as \( \operatorname{Bin}(n, r) \).
• The probability of obtaining exactly \( k \) successes in \( n \) trials is given by the formula \[ \mathrm{P}(Z = k) = \binom{n}{k} r^k (1-r)^{n-k}, \] where \( \binom{n}{k} \) is the binomial coefficient.

In our exercise, it's proposed that \( X+Y \) should follow a binomial distribution when \( p = q = r \). However, since \( p eq q \) in our case, where \( p = \frac{1}{2} \) and \( q = \frac{1}{4} \), the resulting distribution of \( X+Y \) does not fit the binomial model as the conditions are not met. This shows a distinctive property as the sum of independent Bernoulli random variables with different parameters doesn't form a binomial distribution.

Independent Random Variables

Random variables \( X \) and \( Y \) are said to be independent if the occurrence of an event related to \( X \) does not affect the probability of an event related to \( Y \) and vice versa. This concept is crucial when considering combinations of random variables.

• For independent random variables, the probability of simultaneous events can be determined by multiplying their individual probabilities: \( \mathrm{P}(X = x \text{ and } Y = y) = \mathrm{P}(X = x) \times \mathrm{P}(Y = y) \).
• Independence is key when dealing with the sum or product of random variables, as it allows simplification in calculations due to non-interaction of events.

In the given exercise, the independence of \( X \) and \( Y \) allows us to compute the probabilities \( \mathrm{P}(X+Y = k) \) by multiplying the respective probabilities of event outcomes for \( X \) and \( Y \). This makes it possible to separately calculate and then sum up the distinct probabilities needed to analyze the distribution of \( X + Y \).