The number of connections to a wrong phone number is often modeled by a Poisson distribution. To estimate the unknown parameter of that distribution (the mean number of wrong connections) one observes a sample x₁,..., n of wrong connections on n different days. Assum- ing that σ = ₁ + ... + x₂ > 0, (i) give the likelihood function of the sample; (ii) find the m.l.e. and MLE of A. =t. (i) If ƒ (x|λ) = e¯^, then show that the likelihood function is
The number of connections to a wrong phone number is often modeled by a Poisson distribution. To estimate the unknown parameter of that distribution (the mean number of wrong connections) one observes a sample x₁,..., n of wrong connections on n different days. Assum- ing that σ = ₁ + ... + x₂ > 0, (i) give the likelihood function of the sample; (ii) find the m.l.e. and MLE of A. =t. (i) If ƒ (x|λ) = e¯^, then show that the likelihood function is
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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![The image contains a mathematical explanation related to estimating the unknown parameter \(\lambda\) of a Poisson distribution, often used to model the number of wrong phone connections.
**Content Details:**
1. **Introduction**
- The text discusses modeling the number of wrong phone connections using a Poisson distribution.
- The goal is to estimate the unknown parameter \(\lambda\), representing the mean number of wrong connections.
2. **Tasks**
- (i) Provide the likelihood function of a sample.
- (ii) Find the maximum likelihood estimator (MLE) of \(\lambda\).
3. **Hint for Step (i)**
- If \(f(x|\lambda) = e^{-\lambda}\lambda^x/x!\), then use this to derive the likelihood function.
- The likelihood function is given as:
\[
f_n(\mathbf{x}_n|\lambda) = e^{-n\lambda}\lambda^\sigma/\prod_{j=1}^{n}x_j! \quad (\sigma = x_1 + \ldots + x_n)
\]
4. **Step (ii)**
- Define the log-likelihood function:
\[
L(\lambda) = \ln f_n(\mathbf{x}_n|\lambda)
\]
- Find its derivative \(L'(\lambda)\) and show that:
\[
L'(\lambda) = 0 \quad \text{iff} \quad \lambda = \sigma/n = \overline{x}_n
\]
5. **Critical Points and Behavior**
- \( \sigma/n \) is a critical point of \( L(\lambda) \).
- Rewrite \( L'(\lambda) \) as:
\[
L'(\lambda) = \frac{1}{\lambda_n}(\overline{x}_n - \lambda)
\]
- Analysis of \( L' \):
- Positive for \( \lambda < \overline{x}_n \)
- Zero at \( \lambda = \overline{x}_n \)
- Negative for \( \lambda > \overline{x}_n \)
6. **Conclusion**
- It’s proven that \(\overline{x}_n\) is a local and global maximum point.
- Therefore, \(\hat{\lambda} = \overline{x}_n\) is](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fdb82eee3-c597-43b7-b98f-cffc18fbca72%2F0081d0a0-9e87-46f9-b6d8-81d947b4ed6e%2Fjqbr5g_processed.png&w=3840&q=75)
Transcribed Image Text:The image contains a mathematical explanation related to estimating the unknown parameter \(\lambda\) of a Poisson distribution, often used to model the number of wrong phone connections.
**Content Details:**
1. **Introduction**
- The text discusses modeling the number of wrong phone connections using a Poisson distribution.
- The goal is to estimate the unknown parameter \(\lambda\), representing the mean number of wrong connections.
2. **Tasks**
- (i) Provide the likelihood function of a sample.
- (ii) Find the maximum likelihood estimator (MLE) of \(\lambda\).
3. **Hint for Step (i)**
- If \(f(x|\lambda) = e^{-\lambda}\lambda^x/x!\), then use this to derive the likelihood function.
- The likelihood function is given as:
\[
f_n(\mathbf{x}_n|\lambda) = e^{-n\lambda}\lambda^\sigma/\prod_{j=1}^{n}x_j! \quad (\sigma = x_1 + \ldots + x_n)
\]
4. **Step (ii)**
- Define the log-likelihood function:
\[
L(\lambda) = \ln f_n(\mathbf{x}_n|\lambda)
\]
- Find its derivative \(L'(\lambda)\) and show that:
\[
L'(\lambda) = 0 \quad \text{iff} \quad \lambda = \sigma/n = \overline{x}_n
\]
5. **Critical Points and Behavior**
- \( \sigma/n \) is a critical point of \( L(\lambda) \).
- Rewrite \( L'(\lambda) \) as:
\[
L'(\lambda) = \frac{1}{\lambda_n}(\overline{x}_n - \lambda)
\]
- Analysis of \( L' \):
- Positive for \( \lambda < \overline{x}_n \)
- Zero at \( \lambda = \overline{x}_n \)
- Negative for \( \lambda > \overline{x}_n \)
6. **Conclusion**
- It’s proven that \(\overline{x}_n\) is a local and global maximum point.
- Therefore, \(\hat{\lambda} = \overline{x}_n\) is
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