Chapter 5: Problem 42
The number of defects per yard \(Y\) for a certain fabric is known to have a Poisson distribution with parameter \lambda. However, \lambda itself is a random variable with probability density function given by $$f(\lambda)=\left\\{\begin{array}{ll} e^{-\lambda}, & \lambda \geq 0 \\ 0, & \text { elsewhere } \end{array}\right.$$ Find the unconditional probability function for \(Y\)
Short Answer
Step by step solution
Understand the Problem
Identify the Probability Function for Y
Define the Probability Density Function of \(\lambda\)
Write the Unconditional Probability Function
Simplify the Integral
Evaluate the Gamma Function
Conclusion
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Exponential distribution
- The probability density function (PDF) is given by \( f(\lambda) = e^{-\lambda} \) for \( \lambda \geq 0 \).
- The exponential distribution is memoryless, meaning the probability of an event occurring in the future is independent of the past.
- It has one parameter: \( \lambda \), which represents the rate of occurrence.
Probability mass function
- It describes the number of events occurring within a fixed interval of time or space.
- The events must happen independently of the time since the last event.
- Parameter \( \lambda \) is the rate at which events occur.
Gamma function
- The integral defining the Gamma function is \( \Gamma(a) = \int_0^{\infty} x^{a-1} e^{-x} \, dx \).
- The relation \( \Gamma(n) = (n-1)! \) holds true for natural numbers \( n \).