Step 1: Concept Overview:
We are given the CDF of an exponential distribution and need to find the Maximum Likelihood Estimator (MLE) of its expected value, E(X). We will leverage the invariance property of MLEs: if \(\hat{\theta}\) is the MLE of \(\theta\), then the MLE of \(g(\theta)\) is \(g(\hat{\theta})\).
Step 2: Methodology:
1. Determine the distribution and its parameter(s) from the provided CDF. Obtain the PDF using \(f(x) = F'(x)\).2. Express the expected value, E(X), as a function of the parameter \(\theta\).3. Calculate the MLE of the parameter \(\theta\), denoted as \(\hat{\theta}\).4. Apply the invariance property to determine the MLE of E(X).
Step 3: Detailed Solution:
1. The CDF is given as \(F(x) = 1 - e^{-x\theta}\) for \(x \ge 0\). Differentiating, the PDF is \(f(x) = \frac{d}{dx}F(x) = \theta e^{-x\theta}\). This corresponds to an exponential distribution with rate parameter \(\theta\).2. For an exponential distribution with rate \(\theta\), the expected value is \(E(X) = \frac{1}{\theta}\).3. To find the MLE of \(\theta\), we construct the likelihood function for a sample \(x_1, \dots, x_n\): \[ L(\theta) = \prod_{i=1}^n f(x_i; \theta) = \prod_{i=1}^n \theta e^{-x_i\theta} = \theta^n e^{-\theta \sum x_i} \] The log-likelihood function is: \[ l(\theta) = \ln(L(\theta)) = n \ln(\theta) - \theta \sum_{i=1}^n x_i \] Taking the derivative with respect to \(\theta\) and setting it to zero: \[ \frac{dl}{d\theta} = \frac{n}{\theta} - \sum_{i=1}^n x_i = 0 \] Solving for \(\theta\) yields the MLE of \(\theta\): \[ \hat{\theta} = \frac{n}{\sum x_i} = \frac{1}{\bar{x}} \]4. Using the invariance property, the MLE of \(E(X) = 1/\theta\) is: \[ \widehat{E(X)} = \frac{1}{\hat{\theta}} = \frac{1}{1/\bar{x}} = \bar{x} \]
Step 4: Conclusion:
The MLE of E(X) is the sample mean, \(\bar{x}\).