Chapter 3: Problem 133
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
Step by step solution
Identify the PDF of the Negative Binomial Distribution
Substitute and Simplify for r=1
Identify the PDF of a Geometric Distribution
Comparison of PDFs
Formulas for Mean and Variance of Negative Binomial
Formulas for Mean and Variance of Geometric Distribution
Conclusion
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Probability Density Function
- \( k \) is the number of failures desired.
- \( r \) is the number of successes needed.
- \( p \) represents the probability of success on each trial.
Geometric Distribution
- 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.
Mean and Variance
- 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.
Negative Binomial 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.