Question

If X is a random variable such that,
\(P(X \le x) = \begin{cases} 0 & ; x<0 \\ 1-e^{-x\theta} & ; x \ge 0 \end{cases}\)
based on 'n' independent observations on X, the Maximum Likelihood Estimator (MLE) of E(X) is

• A. \( \sum x_i \)
• B. \( \bar{x} \)
• C. \( \frac{1}{\bar{x}} \)
• D. \( \frac{1}{\sum x_i} \)

Hint:

The invariance property of MLEs is a powerful shortcut. Once you find the MLE for a parameter, you can find the MLE for any function of that parameter by simply plugging in the parameter's MLE into the function.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The given cumulative distribution function (CDF) is that of an exponential distribution. The problem asks for the Maximum Likelihood Estimator (MLE) of the expected value, E(X). We will use the invariance property of MLEs, which states that if \(\hat{\theta}\) is the MLE of \(\theta\), then the MLE of a function \(g(\theta)\) is \(g(\hat{\theta})\).

Step 2: Key Formula or Approach:
1. Identify the distribution and its parameter(s) from the CDF. The PDF is \(f(x) = F'(x)\). 2. Find the expected value, E(X), in terms of the parameter \(\theta\). 3. Find the MLE of the parameter \(\theta\), denoted \(\hat{\theta}\). 4. Apply the invariance property to find the MLE of E(X).

Step 3: Detailed Explanation:
1. The CDF is \(F(x) = 1 - e^{-x\theta}\) for \(x \ge 0\). The corresponding PDF is \(f(x) = \frac{d}{dx}F(x) = \theta e^{-x\theta}\). This is an exponential distribution with rate parameter \(\theta\). 2. The expected value (mean) of an exponential distribution with rate \(\theta\) is \(E(X) = \frac{1}{\theta}\). 3. To find the MLE of \(\theta\), we first write 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: \[ l(\theta) = \ln(L(\theta)) = n \ln(\theta) - \theta \sum_{i=1}^n x_i \] Differentiate with respect to \(\theta\) and set to zero: \[ \frac{dl}{d\theta} = \frac{n}{\theta} - \sum_{i=1}^n x_i = 0 \] Solving for \(\theta\), we get the MLE of \(\theta\): \[ \hat{\theta} = \frac{n}{\sum x_i} = \frac{1}{\bar{x}} \] 4. Using the invariance property of MLEs, the MLE of \(E(X) = 1/\theta\) is: \[ \widehat{E(X)} = \frac{1}{\hat{\theta}} = \frac{1}{1/\bar{x}} = \bar{x} \]
Step 4: Final Answer:
The MLE of E(X) is the sample mean, \(\bar{x}\).