Consider randomly selecting n segments of pipe and determining the corrosion loss (mm) in the wall thickness for each one. Denote these corrosion losses by Y₁ Yn. The article "A Probabilistic Model for a Gas Explosion Due to Leakages in the Grey Cast Iron Gas Mains"+ proposes a linear corrosion model: Y₁ = t;R, where t; is the age of the pipe and R, the corrosion rate, is exponentially distributed with parameter . Obtain the maximum likelihood estimator of the exponential parameter (the resulting mle appears in the cited article). [Hint: If c > 0 and X has an exponential distribution, so does cX.] OÂ= »yt O λ = i=1 ବିକା ମିତ ବସିବା ପରେ i=1 i = 1 i=1 (Y/t) OX = = 1 """
Consider randomly selecting n segments of pipe and determining the corrosion loss (mm) in the wall thickness for each one. Denote these corrosion losses by Y₁ Yn. The article "A Probabilistic Model for a Gas Explosion Due to Leakages in the Grey Cast Iron Gas Mains"+ proposes a linear corrosion model: Y₁ = t;R, where t; is the age of the pipe and R, the corrosion rate, is exponentially distributed with parameter . Obtain the maximum likelihood estimator of the exponential parameter (the resulting mle appears in the cited article). [Hint: If c > 0 and X has an exponential distribution, so does cX.] OÂ= »yt O λ = i=1 ବିକା ମିତ ବସିବା ପରେ i=1 i = 1 i=1 (Y/t) OX = = 1 """
MATLAB: An Introduction with Applications
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ISBN:9781119256830
Author:Amos Gilat
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Chapter1: Starting With Matlab
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Problem 1P
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![### Corrosion Loss and Maximum Likelihood Estimation
Consider the scenario where you randomly select \( n \) segments of pipe and determine the corrosion loss (in millimeters) in the wall thickness for each segment. Denote these corrosion losses by \( Y_1, Y_2, \ldots, Y_n \).
The article "A Probabilistic Model for a Gas Explosion Due to Leakages in the Grey Cast Iron Gas Mains" proposes a linear corrosion model:
\[ Y_i = t_i R \]
where:
- \( t_i \) is the age of the pipe segment,
- \( R \), the corrosion rate, is exponentially distributed with parameter \( \lambda \).
Given this model, we aim to obtain the maximum likelihood estimator (MLE) of the exponential parameter \( \lambda \).
**Hint:** If \( c > 0 \) and \( X \) has an exponential distribution, so does \( cX \).
The possible estimators provided are:
1. \( \hat{\lambda} = \frac{n}{\sum_{i=1}^n (Y_i / t_i)} \)
2. \( \hat{\lambda} = \frac{n}{\sum_{i=1}^n \frac{Y_i}{t_i}} \)
3. \( \hat{\lambda} = \frac{n\sum_{i=1}^n t_j}{\sum_{i=1}^n Y_i} \)
4. \( \hat{\lambda} = \frac{\sum_{i=1}^n (Y_i / t_i)}{n} \)
5. \( \hat{\lambda} = \frac{\sum_{i=1}^n t_i}{\sum_{i=1}^n Y_i} \)
Your task is to determine which form correctly represents the maximum likelihood estimator for the parameter \( \lambda \).
### Explanation of Each Option
1. \( \hat{\lambda} = \frac{n}{\sum_{i=1}^n (Y_i / t_i)} \):
- This formula suggests normalizing the sum of observed corrosion losses by the ages of the pipes and then using the reciprocal for the estimator.
2. \( \hat{\lambda} = \frac{n}{\sum_{i=1}^n \frac{Y_i}{t_i}} \):
- This is](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffb989860-4d1c-4fb5-9e6b-42a4528dce9c%2Fbdb8b798-89f5-4b7b-a132-b89a31e8afa7%2Fziaxmlm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Corrosion Loss and Maximum Likelihood Estimation
Consider the scenario where you randomly select \( n \) segments of pipe and determine the corrosion loss (in millimeters) in the wall thickness for each segment. Denote these corrosion losses by \( Y_1, Y_2, \ldots, Y_n \).
The article "A Probabilistic Model for a Gas Explosion Due to Leakages in the Grey Cast Iron Gas Mains" proposes a linear corrosion model:
\[ Y_i = t_i R \]
where:
- \( t_i \) is the age of the pipe segment,
- \( R \), the corrosion rate, is exponentially distributed with parameter \( \lambda \).
Given this model, we aim to obtain the maximum likelihood estimator (MLE) of the exponential parameter \( \lambda \).
**Hint:** If \( c > 0 \) and \( X \) has an exponential distribution, so does \( cX \).
The possible estimators provided are:
1. \( \hat{\lambda} = \frac{n}{\sum_{i=1}^n (Y_i / t_i)} \)
2. \( \hat{\lambda} = \frac{n}{\sum_{i=1}^n \frac{Y_i}{t_i}} \)
3. \( \hat{\lambda} = \frac{n\sum_{i=1}^n t_j}{\sum_{i=1}^n Y_i} \)
4. \( \hat{\lambda} = \frac{\sum_{i=1}^n (Y_i / t_i)}{n} \)
5. \( \hat{\lambda} = \frac{\sum_{i=1}^n t_i}{\sum_{i=1}^n Y_i} \)
Your task is to determine which form correctly represents the maximum likelihood estimator for the parameter \( \lambda \).
### Explanation of Each Option
1. \( \hat{\lambda} = \frac{n}{\sum_{i=1}^n (Y_i / t_i)} \):
- This formula suggests normalizing the sum of observed corrosion losses by the ages of the pipes and then using the reciprocal for the estimator.
2. \( \hat{\lambda} = \frac{n}{\sum_{i=1}^n \frac{Y_i}{t_i}} \):
- This is
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