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Show that the probability density function of a negative binomial random variable equals the probability density function of a geometric random variable when \(r=1\). Show that the formulas for the mean and variance of a negative binomial random variable equal the corresponding results for a geometric random variable when \(r=1\).

Short Answer

Expert verified
The PDFs and the mean and variance formulas are identical for both distributions when \( r=1 \).

Step by step solution

01

Identify the PDF of the Negative Binomial Distribution

The probability density function (PDF) of a negative binomial random variable is given by \( P(X=k) = \binom{k+r-1}{k}(1-p)^r p^k \), where \( r \) is the number of successes, \( p \) is the probability of success in a single trial, and \( k \) is the number of failures. We must show this is equal to the geometric distribution when \( r=1 \).
02

Substitute and Simplify for r=1

Substituting \( r=1 \) into the negative binomial PDF, we have: \( P(X=k) = \binom{k}{k}(1-p)p^k \). This simplifies to \( P(X=k) = (1-p)p^k \).
03

Identify the PDF of a Geometric Distribution

The probability density function for a geometric distribution is \( P(Y=k) = (1-p)p^k \). This distribution describes the number of Bernoulli trials needed to get one success with success probability \( p \).
04

Comparison of PDFs

By comparing \( P(X=k) = (1-p)p^k \) for the negative binomial with \( P(Y=k) = (1-p)p^k \) for the geometric distribution, we observe that they are identical when \( r=1 \).
05

Formulas for Mean and Variance of Negative Binomial

The mean of a negative binomial distribution is given by \( \frac{r(1-p)}{p} \) and the variance is \( \frac{r(1-p)}{p^2} \). Substituting \( r=1 \), the mean becomes \( \frac{1(1-p)}{p} = \frac{1-p}{p} \) and the variance becomes \( \frac{1(1-p)}{p^2} = \frac{1-p}{p^2} \).
06

Formulas for Mean and Variance of Geometric Distribution

For a geometric distribution, the mean is \( \frac{1-p}{p} \), and the variance is \( \frac{1-p}{p^2} \). These match exactly with the mean and variance derived for the negative binomial distribution when \( r=1 \).
07

Conclusion

When \( r=1 \), the probability density functions and the formulas for the mean and variance of negative binomial and geometric distributions are identical, confirming the required conditions.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Density Function
The probability density function, or PDF, is a crucial component in understanding probability distributions. For the negative binomial distribution, the PDF describes the probability of achieving a certain number of failures before a specified number of successes occurs. In the formula \( P(X=k) = \binom{k+r-1}{k}(1-p)^r p^k \), each term has a specific role:
  • \( k \) is the number of failures desired.
  • \( r \) is the number of successes needed.
  • \( p \) represents the probability of success on each trial.
When \( r=1 \), this PDF transforms to the geometric distribution's PDF, \( P(Y=k) = (1-p)p^k \). This indicates the number of trials required to get a single success. Here, the negative binomial and geometric distributions align perfectly, illustrating their beautiful mathematical crossover at \( r=1 \).
Geometric Distribution
The geometric distribution is a specific case of the negative binomial distribution. It is employed for scenarios where one wants to count the number of Bernoulli trials needed to achieve the first success.
  • The geometric PDF is expressed as \( P(Y=k) = (1-p)p^k \).
  • It requires success on the first trial after \( k \) failures.
  • The parameter \( p \) remains the probability of success per trial.
This distribution is memoryless, meaning the probability of success remains the same after any number of failures. When \( r = 1 \), the geometric distribution exactly mirrors the negative binomial distribution, reaffirming the fundamental relationship between them.
Mean and Variance
The concepts of mean and variance are pivotal when exploring probability distributions, like the negative binomial and geometric distributions. The mean indicates the expected number of trials, while the variance measures the spread or variability of the trials.
  • For the negative binomial distribution, with mean \( \frac{r(1-p)}{p} \) and variance \( \frac{r(1-p)}{p^2} \), substituting \( r=1 \) simplifies these to geometric distribution’s mean \( \frac{1-p}{p} \) and variance \( \frac{1-p}{p^2} \).
  • The calculations show that when \( r = 1 \), both distributions share the same statistical properties.
This equivalence not only underlines their theoretical link but also aids in understanding how these statistical measures predict the outcomes of random experiments within these distributions.
Negative Binomial Random Variable
A negative binomial random variable describes the number of trials required to achieve a specific number of successes. This variable becomes particularly interesting when \( r=1 \), effectively transforming it into a geometric random variable.
  • The negative binomial distribution accommodates different numbers of successes, \( r \), offering flexibility.
  • When \( r=1 \), the focus narrows to just achieving the first success (geometric distribution).
  • This variable thus captures a comprehensive range of scenarios—from just getting one success to a desired number of successes—with the probability of each scenario neatly described through its distribution.
Understanding the negative binomial as a generalization of the geometric distribution helps in visualizing its adaptability and the fundamental probability concepts it encapsulates.

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Most popular questions from this chapter

An article in Information Security Technical Report ["Malicious Software-Past, Present and Future" (2004, Vol. 9, pp. \(6-18\) ) ] provided the following data on the top 10 malicious software instances for \(2002 .\) The clear leader in the number of registered incidences for the year 2002 was the Internet worm "Klez," and it is still one of the most widespread threats. This virus was first detected on 26 October 2001 , and it has held the top spot among malicious software for the longest period in the history of virology. The 10 most widespread malicious programs for 2002 $$ \begin{array}{clc} \text { Place } & \text { Name } & \text { \% Instances } \\ \hline 1 & \text { I-Worm.Klez } & 61.22 \% \\ 2 & \text { I-Worm.Lentin } & 20.52 \% \\ 3 & \text { I-Worm.Tanatos } & 2.09 \% \\ 4 & \text { I-Worm.BadtransII } & 1.31 \% \\ 5 & \text { Macro.Word97.Thus } & 1.19 \% \\ 6 & \text { I-Worm.Hybris } & 0.60 \% \\ 7 & \text { I-Worm.Bridex } & 0.32 \% \\ 8 & \text { I-Worm.Magistr } & 0.30 \% \\ 9 & \text { Win95.CIH } & 0.27 \% \\ 10 & \text { I-Worm.Sircam } & 0.24 \% \end{array} $$ Suppose that 20 malicious software instances are reported. Assume that the malicious sources can be assumed to be independent. (a) What is the probability that at least one instance is "Klez?" (b) What is the probability that three or more instances are "Klez?" (c) What are the mean and standard deviation of the number of "Klez" instances among the 20 reported?

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Derive the expression for the variance of a geometric random variable with parameter \(p\).

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