MATLAB: An Introduction with Applications
MATLAB: An Introduction with Applications
6th Edition
ISBN: 9781119256830
Author: Amos Gilat
Publisher: John Wiley & Sons Inc
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Certainly! Below is the transcription of the provided text with a detailed explanation suitable for an educational website: --- ### Understanding Likelihood Functions in Exponential Distributions **Problem Statement:** Suppose \( X_1, X_2, \cdots, X_n \) is a random sample from an exponential distribution with parameter \( \lambda \). Because of independence, the likelihood function is a [choice],[sum],[division] of the individual pdf's: \[ f(x_1, \ldots, x_n; \lambda) = (\lambda e^{-\lambda x_1}) \times \cdots \times (\lambda e^{-\lambda x_n}) = \lambda^n e^{\lambda \sum x_i} \] ### Detailed Explanation: 1. **Likelihood Function:** - The likelihood function \( L(\lambda; x_1, \ldots, x_n) \) measures how likely it is that the observed data \( x_1, \ldots, x_n \) came from a distribution with parameter \( \lambda \). 2. **Exponential Distribution:** - The probability density function (pdf) of an exponential distribution with parameter \( \lambda \) is given by \( f(x; \lambda) = \lambda e^{-\lambda x} \) for \( x \geq 0 \). 3. **Independence:** - Given that \( X_1, X_2, \cdots, X_n \) are independent, the likelihood function for the entire sample is the product of the individual pdf's. 4. **Mathematical Derivation:** - The joint likelihood function is derived as follows: \[ L(\lambda; x_1, \ldots, x_n) = f(x_1; \lambda) \times f(x_2; \lambda) \times \cdots \times f(x_n; \lambda) \] \[ = (\lambda e^{-\lambda x_1}) \times (\lambda e^{-\lambda x_2}) \times \cdots \times (\lambda e^{-\lambda x_n}) \] - This simplifies to: \[ = \lambda^n e^{-\lambda (x_1 + x_2 + \cdots + x_n)} \] \[ = \lambda^n e^{-\lambda
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Transcribed Image Text:Certainly! Below is the transcription of the provided text with a detailed explanation suitable for an educational website: --- ### Understanding Likelihood Functions in Exponential Distributions **Problem Statement:** Suppose \( X_1, X_2, \cdots, X_n \) is a random sample from an exponential distribution with parameter \( \lambda \). Because of independence, the likelihood function is a [choice],[sum],[division] of the individual pdf's: \[ f(x_1, \ldots, x_n; \lambda) = (\lambda e^{-\lambda x_1}) \times \cdots \times (\lambda e^{-\lambda x_n}) = \lambda^n e^{\lambda \sum x_i} \] ### Detailed Explanation: 1. **Likelihood Function:** - The likelihood function \( L(\lambda; x_1, \ldots, x_n) \) measures how likely it is that the observed data \( x_1, \ldots, x_n \) came from a distribution with parameter \( \lambda \). 2. **Exponential Distribution:** - The probability density function (pdf) of an exponential distribution with parameter \( \lambda \) is given by \( f(x; \lambda) = \lambda e^{-\lambda x} \) for \( x \geq 0 \). 3. **Independence:** - Given that \( X_1, X_2, \cdots, X_n \) are independent, the likelihood function for the entire sample is the product of the individual pdf's. 4. **Mathematical Derivation:** - The joint likelihood function is derived as follows: \[ L(\lambda; x_1, \ldots, x_n) = f(x_1; \lambda) \times f(x_2; \lambda) \times \cdots \times f(x_n; \lambda) \] \[ = (\lambda e^{-\lambda x_1}) \times (\lambda e^{-\lambda x_2}) \times \cdots \times (\lambda e^{-\lambda x_n}) \] - This simplifies to: \[ = \lambda^n e^{-\lambda (x_1 + x_2 + \cdots + x_n)} \] \[ = \lambda^n e^{-\lambda
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