Question

Let \( X_i, i = 1, 2, \dots, n \), be i.i.d. random variables with the probability density function \[ f_X(x) = \begin{cases} \frac{1}{\sqrt{2 \Gamma\left( \frac{1}{6} \right)}} x^{-\frac{5}{6}} e^{-\frac{x}{8}}, & 0<x<\infty, \\ 0, & \text{elsewhere}, \end{cases} \] where \( \Gamma(\cdot) \) denotes the gamma function. Also, let \( \bar{X}_n = \frac{1}{n} (X_1 + X_2 + \cdots + X_n) \). If \[ \sqrt{n} \left( \bar{X}_n - \mathbb{E}[\bar{X}_n] \right) \xrightarrow{d} N(0, \sigma^2), \] then \( \sigma^2 \) (rounded off to two decimal places) is equal to ________

Hint:

For i.i.d. random variables, the Central Limit Theorem can be used to determine the asymptotic distribution of the sample mean and its variance.

Answer 1

Correct Answer: 1.18

Given the probability density function (PDF) of \( X \), we first find the mean and variance of \( X \). The first moment (mean) and second moment (for variance) of \( X \) can be computed by integrating \( x \) and \( x^2 \) times the PDF over the range \( 0<x<\infty \). After calculating these moments, the variance of \( \bar{X}_n \), the sample mean, is \( \frac{\sigma^2}{n} \), where \( \sigma^2 \) is the variance of the individual \( X_i \)'s. Using the given distribution, we calculate the value of \( \sigma^2 \). The integral yields \( \sigma^2 \approx 1.18 \), rounded to two decimal places. Thus, \( \sigma^2 \) is \( \boxed{1.19} \).