Suppose X₁, X2, ..., Xn is a random sample from an exponential distribution with parameter A. Because of independence, the likelihood function is a [选择] ,[sum],[devision] of the individual pdf's:
Suppose X₁, X2, ..., Xn is a random sample from an exponential distribution with parameter A. Because of independence, the likelihood function is a [选择] ,[sum],[devision] of the individual pdf's:
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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option: product/sum/division
![Certainly! Below is the transcription of the provided text with a detailed explanation suitable for an educational website:
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### 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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffb989860-4d1c-4fb5-9e6b-42a4528dce9c%2Fecb009fe-bcc2-45fe-b23b-248eecd53c31%2F7bvzvy_processed.jpeg&w=3840&q=75)
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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