Question

Let \( X_1, X_2, \dots, X_8 \) be a random sample taken from a distribution with the probability density function \[ f_X(x) = \begin{cases} \frac{x}{8}, & 0<x<4, \\ 0, & \text{elsewhere}. \end{cases} \] Let \( F_8(x) \) be the empirical distribution function of the sample. If \( \alpha \) is the variance of \( F_8(2) \), then 128\( \alpha \) (in integer) is equal to ________

Hint:

For empirical distribution functions, the variance can be computed using the formula \( \frac{p(1 - p)}{n} \), where \( p \) is the cumulative probability and \( n \) is the sample size.

Answer 1

Correct Answer: 3

The probability density function is given by \( f_X(x) = \frac{x}{8} \) for \( 0<x<4 \), and 0 elsewhere. The cumulative distribution function \( F_X(x) \) is the integral of \( f_X(x) \) from 0 to \( x \): \[ F_X(x) = \int_0^x \frac{t}{8} \, dt = \frac{x^2}{16}, \quad 0<x<4. \] Now, the empirical distribution function \( F_8(x) \) is the proportion of sample points less than or equal to \( x \). We are interested in \( F_8(2) \), the empirical distribution function at \( x = 2 \). Since the sample size is 8, \( F_8(2) \) is the proportion of sample points less than or equal to 2. The variance of the empirical distribution function is given by: \[ \text{Var}(F_8(2)) = \frac{F_8(2) (1 - F_8(2))}{8}. \] From the distribution function, we know \( F_X(2) = \frac{2^2}{16} = \frac{4}{16} = 0.25 \). Thus, \( F_8(2) \approx 0.25 \) and: \[ \text{Var}(F_8(2)) = \frac{0.25(1 - 0.25)}{8} = \frac{0.25 \times 0.75}{8} = \frac{0.1875}{8} = 0.0234375. \] Finally, we compute \( 128 \times \text{Var}(F_8(2)) \): \[ 128 \times 0.0234375 = 3. \] Thus, \( 128 \times \alpha \) is \( \boxed{3} \).