Question

Suppose that \(X\) and \(Y\) are independent binomial random variables with parameters \((n, p)\) and \((m, p) .\) Argue probabilistically (no computations necessary) that \(X+Y\) is binomial with parameters \((n+m, p)\).

Short answer

The sum of two independent binomial random variables \(X\) and \(Y\) with parameters \((n, p)\) and \((m, p)\), respectively, is also a binomial random variable with parameters \((n+m, p)\). This can be shown by calculating the probability mass function for \(Z = X + Y\) and using Vandermonde's identity to simplify the equation: \[ P(Z = r) = \sum_{k = 0}^{n} \binom{n}{k} \binom{m}{r - k} p^r (1-p)^{n+m-r} = \binom{n+m}{r} p^r (1-p)^{n+m-r} \] Thus, \(Z = X + Y\) follows a binomial distribution with parameters \((n+m, p)\).

Step-by-step solution

Definition of binomial distribution

Recall that a random variable \(X\) follows a binomial distribution with parameters \((n, p)\) if it represents the number of successes in \(n\) independent Bernoulli trials, each with probability of success \(p\). The probability mass function is given by \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where \(k\) is the number of successes, and \(\binom{n}{k}\) is the number of ways we can choose \(k\) successes out of \(n\) trials (combinations).

Probability for the sum of independent random variables

Let \(Z = X + Y\). Since \(X\) and \(Y\) are independent binomial random variables, for any \(k\) and \(l\), the probability of \(X = k\) and \(Y = l\) is \[ P(X = k \cap Y = l) = P(X = k) P(Y = l) \] For \(Z = X + Y\), its probability mass function \(P(Z = r)\) can be calculated by summing the probability of all possible combinations of \(k\) and \(l\) such that \(k + l = r\) \[ P(Z = r) = \sum_{k = 0}^{n}\sum_{l = 0}^{m} P(X = k \cap Y = l) \] where \(r = 0, 1, \dots, n+m\).

Substitute the binomial distributions' probabilities

Next, we substitute the probability mass functions of \(X\) and \(Y\) into the equation: \[ P(Z = r) = \sum_{k=0}^{n}\sum_{l=0}^{m} \binom{n}{k} p^k (1-p)^{n-k} \binom{m}{l} p^l (1-p)^{m-l} \] with the condition that \(k + l = r\).

Simplify and rearrange the equation

Let us simplify the above equation. Since we have a product of two binomial coefficients and the probabilities raised to the corresponding powers, we can rewrite the equation as \[ P(Z = r) = \sum_{k = 0}^{n} \binom{n}{k} \binom{m}{r - k} p^{k+(r-k)} (1-p)^{(n+m)-(k+(r-k))} \] Then, we observe that the exponents in the probability terms, \((k + (r - k))\) and \(((n + m) - (k + (r - k)))\), simplify to \(r\) and \(n+m-r\), respectively. We can rewrite the equation as \[ P(Z = r) = \sum_{k = 0}^{n} \binom{n}{k} \binom{m}{r - k} p^r (1-p)^{n+m-r} \]

Prove that \(Z\) follows a binomial distribution with \((n+m, p)\)

We can see that the probability mass function \(P(Z = r)\) has the same structure as the original binomial distribution. The only difference is that instead of a single binomial coefficient, we have a sum of products of two binomial coefficients. However, this sum is equivalent to another binomial coefficient: \[ \binom{n+m}{r} = \sum_{k=0}^{n} \binom{n}{k} \binom{m}{r-k} \] This is a well-known combinatorial identity called Vandermonde's identity. Therefore, we can rewrite the probability mass function of \(Z = X + Y\) as \[ P(Z = r) = \binom{n+m}{r} p^r (1-p)^{n+m-r} \] This proves that the random variable \(Z = X + Y\) follows a binomial distribution with parameters \((n + m, p)\).

Key concepts

Probability Mass Function

In the realm of statistics, the probability mass function (PMF) plays a pivotal role when we're handling discrete random variables like those in a binomial distribution. It fundamentally maps each specific outcome to its corresponding probability. For a binomial random variable, the PMF is given by the formula:

\[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \]
where \( n \) is the number of trials, \( k \) the number of successful outcomes or 'successes', \( p \) the probability of success on an individual trial, and \( \binom{n}{k} \) represents the combinatorial number of ways to choose \( k \) successes out of \( n \) trials. Understanding the PMF is crucial because it allows you to determine the likelihood of any specific count of successes happening, paving the way for insightful data analysis and decision-making in various fields.

Independent Random Variables

When delving into the behavior of random variables, independence is a concept that frequently crops up. Two random variables, \( X \) and \( Y \), are said to be independent if the occurrence of one does not influence the other's probability. This means knowing the outcome of \( X \) offers no advantage in predicting the outcome of \( Y \), and vice versa. In mathematical terms, for independent random variables:

\[ P(X = k \text{ and } Y = l) = P(X = k) \times P(Y = l) \]
This property becomes particularly useful when we combine random variables, such as summing them. It allows for the simplification of compound probabilities and is foundational when proving that the sum of two independent binomial random variables is itself binomial.

Combinatorial Identity

Combinatorics, the study of counting, offers a treasure trove of identities that simplify complex expressions. A crucial one for binomial distributions is Vandermonde's identity, which can be seen when adding two independent binomial random variables. The combinatorial identity:

\[ \binom{n+m}{r} = \textstyle\texttt{\textsum}_\text{{k=0}}^\text{{n}} \binom{n}{k} \binom{m}{r-k} \]
reveals that the sum of products of combinatorial expressions can be condensed into a single combinatorial term. This revelation is more than a mathematical curiosity; it underpins the reasoning that the sum of independent binomial random variables carries the binomial distribution's thumbprint. By combining elementary outcomes of two separate distributions, we essentially reconstruct the probability distribution of their sum, a binomial distribution with augmented parameters.

Bernoulli Trials

At the heart of the binomial distribution lie the Bernoulli trials—experiments that are binary in nature (only two outcomes: success or failure), independent, and have a fixed probability of success \( p \) in each trial. Imagine flipping a not-quite-fair coin, where each toss is independent of the last, and the chance of heads (our 'success') is consistent. When conducting multiple such tosses, or trials, the count of heads will follow a binomial distribution. To determine the probability of getting a specific number of heads across all the tosses, we lean on the binomial distribution's PMF, which encapsulates the essence of multiple, independent Bernoulli trials condensed into a single coherent framework. This distribution is essential in both theoretical and practical applications, spanning genetics, quality control, and more.