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Suppose $\sqrt{n}(X_n - \mu)\overset{d}{\longrightarrow}N\left(0,\sigma^2\right)$. Prove that $X_n \overset{p}{\longrightarrow}\mu$ is true.

I see that it's not true in general and I can construct a few examples when convergence in distribution does not imply convergence with probability one. But how to approach this problem?

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    I edit my answer. Please see it.2012-12-24

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Here $X_n$~$N(\mu,\frac{\sigma ^2}{n})$.Here I assume that $\sigma$ is known. Note that $\begin{align} P[|X_n-\mu|<\epsilon] &= P[-\epsilon So,$X_n \overset{p}{\longrightarrow}\mu$ is true.

I think your conclution is wrong.

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    @Learner: You are right. So, I edit my answer.2012-12-24