Discussion
Let \(X_{n}\) be asymptotically normally distributed with mean \(\mu\) and variance \(\sigma_{n}^{2}\), such that \(\sigma_{n}^{2} \to 0\) as \(n \to \infty\). Which of the following statements is not true ?
Answer: A
The statement that is not true is option A, \(1/X_{n} \sim AN(1/\mu, \sigma_{n}^{2}/\mu^{2})\).
By the delta method, \(g(X_{n}) \sim AN(g(\mu), [g'(\mu)]^{2}\sigma_{n}^{2})\) when \(g'(\mu) \neq 0\). For \(g(x) = 1/x\), \(g'(\mu) = -1/\mu^{2}\), so the correct variance is \(\sigma_{n}^{2}/\mu^{4}\), not \(\sigma_{n}^{2}/\mu^{2}\).
The others are right: \(g = e^{x}\) gives variance \(e^{2\mu}\sigma_{n}^{2}\) (B), and \(g = \log|x|\) gives \(\sigma_{n}^{2}/\mu^{2}\) (D). Option C, as printed 'log|Xn|σn|', evidently means \(\log|X_{n}/\sigma_{n}|\): when \(\mu = 0\), \(X_{n}/\sigma_{n} \to N(0, 1)\) and the continuous map \(\log|x|\) carries this over.
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