To solve this problem, we need to find the maximum likelihood estimators (MLE) for the parameters \(\theta\) and \(\lambda\) of the shifted exponential distribution given the sample data. ### Part a: Obtain the Maximum Likelihood Estimators 1. **Likelihood Function**: The likelihood function for the shifted exponential distribution is: \[ L(\lambda, \theta) = \prod_{i=1}^{n} \lambda e^{-\lambda (x_i - \theta)} \] 2. **Log-Likelihood Function**: Taking the natural logarithm, the log-likelihood function is: \[ \log L(\lambda, \theta) = n \log \lambda - \lambda \sum_{i=1}^{n} (x_i - \theta) \] 3. **Partial Derivatives**: - Differentiate with respect to \(\lambda\): \[ \frac{\partial \log L}{\partial \lambda} = \frac{n}{\lambda} - \sum_{i=1}^{n} (x_i - \theta) \] - Differentiate with respect to \(\theta\): \[ \frac{\partial \log L}{\partial \theta} = \lambda n \] 4. **Setting Derivatives to Zero**: - For \(\lambda\): \[ \frac{n}{\lambda} = \sum_{i=1}^{n} (x_i - \theta) \] Solving for \(\lambda\): \[ \hat{\lambda} = \frac{n}{\sum_{i=1}^{n} (x_i - \theta)} \] - For \(\theta\), since \(\theta\) is the smallest value in the sample: \[ \hat{\theta} = \min(x_1, x_2, \ldots, x_n) \] ### Part b: Calculate MLEs for Given Data Given data: \(13.11, 10.64, 12.55, 12.20, 15.44, 13.42, 20.39, 18.93, 27.82, 11.30\) 1. **Estimate \(\theta\)**: - \(\hat{\theta} = \min(13.11, 10.64, 12.55, 12.20, 15.44, 13.42, 20.39, 18.93, 27.82, 11.30) = 10.64\) 2. **Estimate \(\lambda\)**: - Calculate \(\sum_{i=1}^{n} (x_i - \hat{\theta})\): \[ \sum_{i=1}^{n} (x_i - 10.64) = (13.11 - 10.64) + (10.64 - 10.64) + \ldots + (11.30 - 10.64) \] \[ = 2.47 + + 1.91 + 1.56 + 4.80 + 2.78 + 9.75 + 8.29 + 17.18 + .66 = 49.4 \] - Solve for \(\hat{\lambda}\): \[ \hat{\lambda} = \frac{10}{49.4} \approx .202 \] ### Final Answer - \(\hat{\theta} = 10.64\) - \(\hat{\lambda} \approx .202\)