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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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Decency dictates you put in at least some effort before asking someone to do your homework for you... –  tomasz Dec 23 '12 at 16:46
    
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? –  arkadiy Dec 23 '12 at 16:52
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I edit my answer. Please see it. –  A.D Dec 24 '12 at 5:52

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up vote 2 down vote accepted

Here $X_n$~$N(\mu,\frac{\sigma ^2}{n})$.Here I assume that $\sigma$ is known. Note that $$\begin{align} P[|X_n-\mu|<\epsilon] &= P[-\epsilon<X_n-\mu<\epsilon] \\ &= P[-\frac{\epsilon\sqrt n}{\sigma} <\frac{\sqrt n(X_n-\mu)}{\sigma}<\frac{\epsilon\sqrt n}{\sigma}] \\ &= 2 \Phi(\frac{\epsilon\sqrt n}{\sigma})-1 \to 1 ~ as ~n \to \infty\end{align}$$ So,$X_n \overset{p}{\longrightarrow}\mu$ is true.

I think your conclution is wrong.

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This is amazing. Thank you! –  arkadiy Dec 23 '12 at 18:18
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$2 \Phi\left(\dfrac{\epsilon\sqrt n}{\sigma}\right)-1 \to 1$ –  Henry Dec 23 '12 at 18:33
    
Yes, of course. Thanks. –  arkadiy Dec 23 '12 at 18:54
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@Learner: You are right. So, I edit my answer. –  A.D Dec 24 '12 at 5:53

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