# Convergence in Central Limit Thorem

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

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

One can have a simpler example for "convergence in distribution does not imply convergence in probability".

Let $X ∼ N(0, 1)$. Let $X_n=-X$ for all $n$. Then $X_n ∼ N(0, 1)$ for all $n$ (which can be shown by using characteristic functions). Trivially we have $X_n\to X$ in distribution. But

$$P(|X_n − X| > \epsilon) = P(|2X| > \epsilon) = P(|X| >\epsilon /2) \not= 0.$$ So $X_n$ does not converge to $X$ in probability.

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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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