Discussion
Suppose \(X_{1}, X_{2}, \ldots, X_{n}\) are independent and identically distributed random variables each having exponential distribution with mean \(\theta\). Then the asymptotic distribution of \(\sqrt{n}\left(\dfrac{\overline{X}}{S} - 1\right)\) where \(S^{2} = \dfrac{1}{n}\sum_{1}^{n} X_{i}^{2} - (\overline{X})^{2}\) is :
Answer: A
The asymptotic distribution is N(0, 1).
For the exponential with mean \(\theta\): \(\sigma^{2} = \theta^{2}\), \(\mu_{3} = 2\theta^{3}\), \(\mu_{4} = 9\theta^{4}\). By the delta method for \(g(m, v) = m/\sqrt{v}\) at \((\theta, \theta^{2})\), with gradient \((1/\theta, -1/(2\theta^{2}))\) and asymptotic covariance of \((\bar{X}, S^{2})\) equal to \(1/n\) times the matrix with rows \((\theta^{2}, 2\theta^{3})\) and \((2\theta^{3}, 8\theta^{4})\):
\[ \begin{aligned} n\,\mathrm{Var} &\approx \frac{\theta^{2}}{\theta^{2}} - 2\cdot\frac{2\theta^{3}}{2\theta^{3}} + \frac{8\theta^{4}}{4\theta^{4}} \\ &= 1 - 2 + 2 = 1 \end{aligned} \]
Since \(\bar{X}/S \to 1\), \(\sqrt{n}(\bar{X}/S - 1) \to N(0, 1)\). The ratio is scale-free, so \(\theta\) cannot appear in the limit, which rules out \(N(0, \theta)\) and \(N(0, \theta^{2})\); and the limit is centred at 0, not 1.
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