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Suppose $X_1, \dots, X_n$ are i.i.d. with mean $\mu$ and variance $\sigma^2 >0$. What is the distribution of $\overline{X}_n(1- \overline{X}_n)$ as $n \to \infty$?

So $\overline{X}_n \to N(0, \frac{\sigma^2}{n})$ in distribution and $(1- \overline{X}_n) \to 1-\mu$ in probability. So $\overline{X}_n(1- \overline{X}_n) \to N(0, (1-\mu) \frac{\sigma^2}{n})$ as $n \to \infty$

Is that correct?

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No. When writing $\bar X_n\to A$, the limit $A$, whatever its nature is, must not depend on $n$. Hence the assertion that $\bar X_n\to N(0,\sigma^2/n)$ is meaningless. – Did Mar 21 '12 at 19:53
@Didier : instead it would be $\overline{X}_n \to N(0, \sigma^2)$? – ted i Mar 21 '12 at 20:27
@Didier Piau: Oh it is 0. – ted i Mar 21 '12 at 22:54
No. By the LLN, $\bar X_n\to\mu$ hence... – Did Mar 21 '12 at 23:01
@DidierPiau: Hence $\overline{X}_{n}(1-\overline{X}_n) \to \mu(1-\mu)$ in distribution. – ted i Mar 21 '12 at 23:05

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