Chapter 5: Problem 4
Find the generating function of the negative binomial mass function $$ f(k)=\left(\begin{array}{l} k-1 \\ r-1 \end{array}\right) p^{r}(1-p)^{k-r}, \quad k=r, r+1, \ldots $$ where \(0
Short Answer
Expert verified
Generating function: \( G(t) = \left(\frac{pt}{1-(1-p)t}\right)^r \); Mean: \( \frac{r(1-p)}{p} \); Variance: \( \frac{r(1-p)}{p^2} \).
Step by step solution
01
Understand the Negative Binomial Mass Function
The negative binomial mass function is given by \( f(k)=\binom{k-1}{r-1} p^{r}(1-p)^{k-r} \), which reduces to the probability of \( k \) trials being needed to achieve \( r \) successes, where \( p \) is the probability of success in each trial.
02
Define the Generating Function
The generating function of a sequence \( a_k \) is given by \( G(t) = \sum_{k=0}^{\infty} a_k t^k \). For the negative binomial distribution, this becomes \( G(t) = \sum_{k=r}^{\infty} \binom{k-1}{r-1} p^r (1-p)^{k-r} t^k \).
03
Rewrite the Summation
Rewrite the sum by factoring out constants that don't depend on \( k \): \( G(t) = p^r t^r \sum_{k=r}^{\infty} \binom{k-1}{r-1}((1-p)t)^{k-r} \). Let \( m = k - r \), then \( k = m + r \), and the sum becomes \( \sum_{m=0}^{\infty}\binom{m+r-1}{r-1}((1-p)t)^{m} \).
04
Utilize Known Series Identity
The series identity \( \sum_{m=0}^{\infty} \binom{m+r-1}{r-1} x^{m} = \frac{1}{(1-x)^r} \) is used. Hence, substitute \( x = (1-p)t \) which yields \( \sum_{m=0}^{\infty} \binom{m+r-1}{r-1} ((1-p)t)^m = \frac{1}{(1-(1-p)t)^r} \).
05
Calculate the Generating Function
Taking the result from Step 4, the generating function becomes: \( G(t) = p^r t^r \frac{1}{(1-(1-p)t)^r} \). This simplifies to: \( G(t) = \left(\frac{pt}{1-(1-p)t}\right)^r \).
06
Deduce the Mean and Variance
For the negative binomial distribution, if the generating function is \( \left(\frac{pt}{1-(1-p)t}\right)^r \), then the mean is \( \frac{r(1-p)}{p} \) and the variance is \( \frac{r(1-p)}{p^2} \). These are standard results derived from the generating function using differentiation techniques.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Generating Function
The generating function is a powerful tool in probability theory and combinatorics. It provides a way to encapsulate an entire sequence of numbers into a single algebraic expression. For the negative binomial distribution, its generating function is crucial to understanding its properties. We start with the probability mass function: \[ f(k) = \binom{k-1}{r-1} p^r (1-p)^{k-r}, \] for \( k = r, r+1, \ldots \). The generating function, denoted by \( G(t) \), is given by the infinite sum: \[ G(t) = \sum_{k=r}^{\infty} f(k) t^k. \] By substituting the mass function into this sum, the generating function becomes: \[ G(t) = \sum_{k=r}^{\infty} \binom{k-1}{r-1} p^r (1-p)^{k-r} t^k. \] To simplify, we factor out constants: \[ G(t) = p^r t^r \sum_{k=r}^{\infty} \binom{k-1}{r-1}((1-p)t)^{k-r}. \] Introducing \( m = k-r \) allows us to use a known series identity, giving the final generating function form:
Mean and Variance
The mean and variance of a probability distribution are fundamental measures that describe its central tendency and spread. With the generating function \( G(t) = \left(\frac{pt}{1-(1-p)t}\right)^r \), we can derive these statistics for the negative binomial distribution. The mean (expected number of trials needed to achieve \( r \) successes) is calculated as: \[ \text{Mean} = \frac{r(1-p)}{p}. \] This result tells us that the higher the probability of failure \((1-p)\), or the more successes \( r \) desired, the longer it will take. Variance measures how much the number of trials can vary. For this distribution, the variance is: \[ \text{Variance} = \frac{r(1-p)}{p^2}. \] The variance is larger than the mean, indicating that the distribution is more spread out, particularly when the probability \( p \) of success in each trial is low.
Probability Mass Function
The probability mass function (PMF) is key to understanding how probabilities are assigned across possible outcomes in a discrete distribution. For the negative binomial distribution, the PMF is given by: \[ f(k) = \binom{k-1}{r-1} p^r (1-p)^{k-r}, \] where \( k = r, r+1, \ldots \), \( p \) is the probability of success in each trial, and \( r \) is the number of successes you want. The term \( \binom{k-1}{r-1} \) accounts for the different combinations of how the \( r \) successes can be achieved within the trials. The remaining terms \( p^r (1-p)^{k-r} \) provide the probability of those specific successes and failures. Together, these elements reflect the likelihood of requiring exactly \( k \) trials to achieve \( r \) successes when trials are identically and independently carried out.
Series Identity
A series identity is a mathematical expression that allows complex series to be expressed in a simpler form. In this case, it plays a pivotal role in simplifying the generating function for the negative binomial distribution. We use the identity: \[ \sum_{m=0}^{\infty} \binom{m+r-1}{r-1} x^m = \frac{1}{(1-x)^r}, \] by substituting \( x = (1-p)t \). This lets us rewrite part of our sum from the generating function development process. Essentially, it provides a way to switch from a complex summation to a simple fraction, making calculations more manageable. This identity is central to the step where from the complex sum \( \sum_{m=0}^{\infty} \binom{m+r-1}{r-1} ((1-p)t)^m \), we can use the series identity to neatly arrive at the expression: \[ \frac{1}{(1-(1-p)t)^r}. \] It's a crucial step that makes further mathematical deductions, such as those for mean and variance, much easier.