Question

Let \( X_1, X_2, \dots, X_n \) be a random sample of size \( n \geq 2 \) from a distribution having the probability density function 

where \( \theta \in (0, \infty) \). Then the method of moments estimator of \( \theta \) equals

• A. \( \frac{1}{2n} \sum_{i=1}^{n} X_i \)
• B. \( \frac{2}{n} \sum_{i=1}^{n} X_i \)
• C. \( \frac{1}{n} \sum_{i=1}^{n} X_i \)
• D. \( \frac{n}{\sum_{i=1}^{n} X_i} \)

Hint:

In the method of moments, equate the sample moments (like the sample mean) to the theoretical moments (like the population mean) and solve for the parameter.

Answer 1

The Correct Option is A

The given probability density function is: \[ f(x; \theta) = \frac{1}{\theta} e^{-\frac{x-\theta}{\theta}}, \quad x>\theta. \] Step 1: Find the first moment (mean) of the distribution.
To find the method of moments estimator for \( \theta \), we first compute the expectation of the random variable \( X \). The expectation is given by: \[ E[X] = \int_{\theta}^{\infty} x \cdot \frac{1}{\theta} e^{-\frac{x-\theta}{\theta}} \, dx. \] By substitution and simplification, we find that: \[ E[X] = \theta + \theta = 2\theta. \] Step 2: Set the sample mean equal to the population mean.
The method of moments estimator equates the sample mean to the population mean. Thus, we set: \[ \frac{1}{n} \sum_{i=1}^{n} X_i = 2\theta. \] Step 3: Solve for \( \theta \).
From the above equation, we solve for \( \theta \): \[ \theta = \frac{1}{2n} \sum_{i=1}^{n} X_i. \] Final Answer: \[ \boxed{\frac{1}{2n} \sum_{i=1}^{n} X_i} \]