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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).2012-11-20

3 Answers 3

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Hint. For convenience of notation, assume that $\sigma =1$. Assume that $S_n/\sqrt{n} \to Z$ in probability. Choose a sufficiently large $n$. 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 $\Pr[A > 0, C >0 | Z > \varepsilon] > 1 - 2\varepsilon$. We get, $ \Pr[A > 0, C >0]\geq \Pr[A > 0, C >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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    very interesting argument! just a some mistake that i think Pr[A>0,Z>0] should be Pr[A>0,C>0] in the last four lines.2016-11-12
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Assume $n=2m$

$P(|\frac{S_{n}}{\sqrt{n}}-\frac{S_{m}}{\sqrt{m}}|>\varepsilon)=$

$1-\int_{-\varepsilon \sqrt{nm}}^{\varepsilon \sqrt{nm}}\int\cdots \int f(z-\sqrt{m}\sum x_{i}-(\sqrt{m}-\sqrt{n})\sum x_{i} ) f((\sqrt{m}-\sqrt{n})x_{2})\cdots f((\sqrt{m}-\sqrt{n})x_{m})f(\sqrt{m}x_{m+1})\cdots f(\sqrt{m}x_{n})dz dx_{2}\cdots dx_{n}= $

$1-\frac{1}{\sqrt{m}^{2m-1}2^{m-\frac{1}{2}}(1-\sqrt{2})^{m}}\int_{-\varepsilon \sqrt{nm}}^{\varepsilon \sqrt{nm}}\int\cdots \int f(z-\sum x_{i}-\sum x_{i} ) f(x_{2})\cdots f(x_{m})f(x_{m+1})\cdots f(x_{n})dz dx_{2}\cdots dx_{n}\to 1-0=1.$

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Assume that $\sigma=1$. Suppose $S_n/\sqrt{n}\overset{P}\to Z$, then $|S_n/\sqrt{n}-S_{2n}/\sqrt{2n}|\overset{P}\to 0\quad(*)$.

Next we want to deduce a contradiction with equation (*). Notice that

(S_{2n}-S_n)/[\sqrt{n}\cdot(\sqrt{2}-1)]=(\sqrt{2}\cdot S_{2n}/\sqrt{2n}-S_n/\sqrt{n})/(\sqrt{2}-1)\overset{d}\to Z Therefore,$(S_{2n}-S_n)/\sqrt{n}\overset{d}\to (\sqrt{2}-1)Z$. By the independence of $S_n$ and $S_{2n}-S_n$, we have

P(10$ Hence, $\liminf_{n\to\infty}P(S_n/\sqrt{n}>1,S_{2n}/\sqrt{2n}<-1/\sqrt{2})>0 $$ By which we can get a contradiction with equation (*).