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The convergence in the Central Limit Theorem is weak convergence, which is weaker than convergence in probability. I set it as an exercise to find an example that convergence in distribution does not imply convergence in probability:

Let $(X_j)_{j\geq 1}$ be i.i.d. with $E[X_1]=0$ and $\sigma_{X_1}^2=\sigma^2<\infty$. Let $S_n=\sum_{i=1}^nX_i$. Then $$ \frac{S_n}{\sigma\sqrt{n}}\to Z\sim N(0,1) $$ which is from the CLT.

Here is my question: does $\frac{S_n}{\sigma\sqrt{n}}$ converge in probability?


I think the point is to give a non-zero lower bound of $$ P(\frac{S_n}{\sigma\sqrt{n}}>\epsilon) $$ for some $\epsilon>0$. But I'm not sure if this can lead to the conclusion that $\frac{S_n}{\sigma\sqrt{n}}$ does not converge in probability.

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Actually one needs a lower bound for $P(\frac{S_n}{\sqrt{n}}-\frac{S_m}{\sqrt{m}}>\epsilon)$. –  Jack Nov 20 '12 at 20:44
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1 Answer

up vote 2 down vote accepted

Hint. For convenience of notation, assume that $\sigma =1$. Assume that $S_n/\sqrt{n} \to Z$ in distribution (weakly converges). Choose $n$ sufficiently large. Consider $A = S_n/\sqrt{n}$ and $B = S_{2n}/\sqrt{2n}$. Both of them are “very close” to $Z$. Thus $C=(\sqrt{2}B - A)/(\sqrt{2}-1)$ is “close” to $Z$.

We have,

  • $A$ is very close to $Z$,
  • $C$ is very close to $Z$,
  • $A$ and $C$ are independent (write explicitly what $A$ and $C$ are, to check that).

This is not possible.

Specifically, for every $\varepsilon > 0$ and sufficiently large $n$, we have (this follows from the definition of convergence in probability), \begin{align} \Pr[|A-Z| > \varepsilon] &< \varepsilon,\\ \Pr[|C-Z| > \varepsilon] &< \varepsilon,\\ \Pr[|A-C| > \varepsilon] &< \varepsilon. \end{align}

Since $A$ and $C$ are independent, $$\Pr[A > 0, C > 0] = \Pr[A>0]\cdot \Pr[C>0] \leq (\Pr[Z > - \varepsilon] + \varepsilon)^2 = (1/2 + O(\varepsilon))^2 = 1/4 + O(\varepsilon).$$

On the other hand, $\Pr[A > 0 | Z > \varepsilon] > 1 - \varepsilon$ and $\Pr[C > 0 | Z > \varepsilon] > 1 - \varepsilon$. Thus $\Pr[A > 0, Z >0 | Z > \varepsilon] > 1 - 2\varepsilon$. We get, $$ \Pr[A > 0, Z >0]\geq \Pr[A > 0, Z >0 | Z > \varepsilon]\cdot \Pr[Z > \varepsilon] \geq (1-2\varepsilon) \Pr[Z > \varepsilon] = 1 - O(\varepsilon).$$ We get a contradiction.

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I don't quite follow your hint. –  Jack Nov 20 '12 at 22:05
    
What part of the argument don't you follow? –  Yury Nov 20 '12 at 22:14
    
How do you come up with the last sentence? –  Jack Nov 20 '12 at 23:40
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