Question

Let \( X_1, X_2, \dots, X_n \) be a random sample from a continuous distribution with the probability density function \[ f(x) = \frac{1}{2\sqrt{2\pi}} \left[ e^{-\frac{1}{2}(x - 2)^2} + e^{-\frac{1}{2}(x - 4)^2} \right], \quad -\infty<x<\infty \] If \( T_n = X_1 + X_2 + \dots + X_n \), then which one of the following is an unbiased estimator of \( \mu \)?

• A. \( \frac{T_n}{n} \)
• B. \( \frac{T_n}{2n} \)
• C. \( \frac{T_n}{3n} \)
• D. \( \frac{T_n}{4n} \)

Hint:

For unbiased estimation, use the sample mean \( \frac{T_n}{n} \), as it provides an unbiased estimate of the population mean.

Answer 1

The Correct Option is C

Step 1: Identifying the mean \( \mu \).
The mean \( \mu \) of a mixture of two normal distributions can be calculated by taking the weighted average of the means of each component. For the given distribution, the mean is \( \mu = 3 \).
Step 2: Find the unbiased estimator.
The sum \( T_n = X_1 + X_2 + \dots + X_n \) is the total of \( n \) i.i.d. samples. The expected value of \( T_n \) is \( n \mu \), and therefore \( \frac{T_n}{n} \) is an unbiased estimator for \( \mu \).
Step 3: Conclusion.
The correct answer is (A) \( \frac{T_n}{n} \).