Question

Let \( X_1, X_2, \dots, X_n \) be a random sample from the distribution with the probability density function \[ f(x) = \frac{1}{4} e^{-|x-4|} + \frac{1}{4} e^{-|x-6|}, x \in \mathbb{R}. \] Then \( \frac{1}{n} \sum_{i=1}^{n} X_i \) converges in probability to ................ 
 

Hint:

For mixtures of distributions, the expected value is the weighted average of the means of the components.

Answer 1

Correct Answer: 4.75 - 5.25

Step 1: Understand the distribution.
The given probability density function \( f(x) \) represents a mixture of two exponential distributions, one centered at \( 4 \) and the other at \( 6 \). This is a bimodal distribution where each mode has equal weight \( \frac{1}{4} \).

Step 2: Identify the expected value of the distribution.
Since the distribution is symmetric with modes at \( 4 \) and \( 6 \), the expected value \( E(X) \) will be the average of the two modes: \[ E(X) = \frac{4 + 6}{2} = 5. \]

Step 3: Apply the Law of Large Numbers.
By the Weak Law of Large Numbers, as \( n \to \infty \), the sample mean \( \frac{1}{n} \sum_{i=1}^{n} X_i \) will converge in probability to the expected value of the distribution, which is 5.

Step 4: Conclusion.
Thus, \( \frac{1}{n} \sum_{i=1}^{n} X_i \) converges in probability to \( 5 \).