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

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?

• Observe that $Z_{2n}=\frac1{\sqrt2}(Z_n+Z'_n)$ where $Z'_n$ is an independent copy of $Z_n$. Commented Jun 13 at 7:37
• A perhaps quicker argument than in @geetha290krm's link: If $Z_n$ converges in probability, then so does $Z'_n$. Let $Z,Z'$ denote the respective limits in probability. Passing to the limit in $Z_{2n}=\frac1{\sqrt 2}(Z_n+Z'_n)$ gives $Z=\frac1{\sqrt 2}(Z+Z')$ a.s., that is $(\sqrt2-1)Z=Z'$. But this contradicts $Z\stackrel d=Z'\sim\mathcal N(0,\sigma^2)$. Commented Jun 13 at 8:35

1 Answer

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.