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Chapter 2: Problem 57

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

Expert verified
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

01

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).
02

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 \(r = 0, 1, \dots, n+m\).
03

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 \(k + l = r\).
04

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 \[ P(Z = r) = \sum_{k = 0}^{n} \binom{n}{k} \binom{m}{r - k} p^r (1-p)^{n+m-r} \]
05

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 \((n + m, p)\).

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