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. 
 A: 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, 
$$\Pr[A > 0, C > 0] = \Pr[A>0]\cdot \Pr[C>0] \leq (\Pr[Z > - \varepsilon] + \varepsilon)^2 = (1/2 + O(\varepsilon))^2 = 1/4 + O(\varepsilon).$$ 
Let $p = \Pr[Z > \varepsilon]$. We have, 
\begin{align}
\Pr[A > 0 | Z > \varepsilon] &= 1 - \Pr[A \leq 0 | Z > \varepsilon] \\ &= 1 - \Pr[A \leq 0 \text{ and } Z > \varepsilon] / \Pr[Z > \varepsilon] \\ &\geq  1 - \varepsilon / p.
\end{align}
Similarly,  $\Pr[C > 0 | Z > \varepsilon] > 1 - \varepsilon / p$. Thus, 
$$\Pr[A > 0, C >0 | Z > \varepsilon] > 1 - 2\varepsilon / p.$$ 
We have,
\begin{align*} \Pr[A > 0, C >0] &\geq  \Pr[A > 0, C >0 | Z > \varepsilon]\cdot \Pr[Z > \varepsilon] \\ &\geq (1-2\varepsilon/p) \cdot p = p - 2\varepsilon = 1/2 - O(\varepsilon).
\end{align*}
We get a contradiction.
A: 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}=  $
$1-\frac{1}{\sqrt{m}^{2m-1}2^{m-\frac{1}{2}}(1-\sqrt{2})^{m}}\int_{-\varepsilon \sqrt{nm}}^{\varepsilon \sqrt{nm}}\int\cdots \int f(z-\sum x_{i}-\sum  x_{i} ) f(x_{2})\cdots f(x_{m})f(x_{m+1})\cdots f(x_{n})dz dx_{2}\cdots dx_{n}\to 1-0=1.$
A: 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(1<S_n/\sqrt{n}<2,(S_{2n}-S_n)/\sqrt{n}<-3)\to P(1<Z<2)\cdot P((\sqrt{2}-1)Z<-3)>0$$
Hence,
$$\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 (*).
