Discussion
Suppose \(\{X_{1}, X_{2}, \ldots, X_{n}\}\) are independent and identically distributed random variables each having the probability density function :
\(f(x) = \dfrac{\theta}{x^{2}},\ x \ge \theta\)
Then :
\[ \begin{aligned} &\sqrt{n}\left(X\left(\left[\tfrac{n}{4}\right] + 1\right) - \tfrac{4\theta}{3}\right) \\ &\xrightarrow{L} z, \end{aligned} \]
Answer: C
The limit is \(z \sim N(0, 16\theta^{2}/27)\).
\(X([n/4] + 1)\) is the sample lower quartile. Its population counterpart \(\xi\) solves \(F(\xi) = 1 - \theta/\xi = 1/4\), so \(\xi = 4\theta/3\). A sample \(p\)-quantile is asymptotically normal with variance \(p(1 - p)/f(\xi)^{2}\):
\[ \begin{aligned} f(\xi) &= \frac{\theta}{\xi^{2}} = \frac{9}{16\theta} \\ \frac{(1/4)(3/4)}{f(\xi)^{2}} &= \frac{3}{16}\cdot\frac{256\theta^{2}}{81} = \frac{16\theta^{2}}{27} \end{aligned} \]
The variance must scale with \(\theta^{2}\), since \(\theta\) is a scale parameter, which rules out the forms in \(\theta\) (two of which are printed identically).
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