Show that $\frac{1}{\sqrt{n}} S_n$ does not converge in probability. [duplicate]

Question

Let $(X_n)$ be a sequence of stochastically independent and identically distributed random variables with $\mathbb{E}X_1 = \mu < \infty$ and $\text{Var} \, X_1 = \sigma^2 < \infty$, where $\sigma^2 > 0$.

Let $Y_n = X_n - \mu$ and $S_n = \sum_{k=1}^{n} Y_k$ for all $n \in \mathbb{N}$.

The central limit theorem now gives the convergence

$$ \frac{1}{\sqrt{n}} S_n \xrightarrow{d} X, \quad n \to \infty, $$

for an $X \sim \mathcal{N}(0, \sigma^2)$.

Show that $\frac{1}{\sqrt{n}} S_n$ does not converge in probability.

Hint: Consider $Z_{2n}$, where $Z_n = \frac{1}{\sqrt{n}} S_n$.

I don't understand why the claim should hold, so what is going wrong that it doesn't converge in probability?

Answer 1

Here's another method.

Assume for sake of contradiction that $\frac{S_{n}}{\sqrt{n}}$ converged in probability to $X$ such that $X\sim N(0,\sigma^{2})$.

Then there would exist a subsequence $\frac{S_{n_{k}}}{\sqrt{n_{k}}}$ which converges almost surely to $X$.

Now we show that $\lim\sup\frac{S_{n_{k}}}{\sqrt{n_{k}}}=+\infty$ almost surely which would contradict our assumption that $\frac{S_{n_{k}}}{\sqrt{n_{k}}}\xrightarrow{a.s.} X$ as then $\lim\sup\frac{S_{n_{k}}}{\sqrt{n_{k}}}<\infty$ almost surely.

Let $K>0$ be an arbitrary positive integer

Observe that the event $B_{K}=\{\lim\sup\frac{S_{n_{k}}}{\sqrt{n_{k}}}>K\}$ is "Tail Measurable".

i.e. of $\mathcal{F}_{n}=\sigma(\bigcup_{k\geq n}\sigma(X_{k}))$ then $\displaystyle B_{K}\in \bigcap_{m\geq 1}\mathcal{F}_{m}$.

So by Kolmogorov's Zero One Law, $P(B_{K})=1$ or $P(B_{K})=0$.

But, by Fatou's Lemma, you have

$$ \mathbb{P}\left(\limsup\frac{S_{n_{k}}}{\sqrt{n_{k}}}>K\right)\geq \limsup \mathbb{P}\left(\frac{S_{n_{k}}}{\sqrt{n_{k}}}>K\right)=\mathbb{P}(N(0,\sigma^{2})>K)>0, $$

Thus, $P(B_{K})=1$.

So $P(\{\lim\sup\frac{S_{n_{k}}}{\sqrt{n_{k}}}=+\infty\})=P(\bigcap_{K\in\Bbb{N}}B_{K})=1$

This shows that $\lim\sup\frac{S_{n_{k}}}{\sqrt{n_{k}}}=+\infty$ almost surely and you have the desired contradiction.