# Central limit theorem and convergence in probability from Durrett

I saw following exercise from Durrett's probability theory book and I managed to solve the 1st part, but couldn't get the 2nd part.

Let $X_1, X_2, \dots$ be i.i.d samples with mean $0$, and finite non-zero variance. Denote $S_n = X_1 + X_2+\dots + X_n$.

1. Use central limit theorem and Kolmogorov $0-1$ law to show $\limsup S_n/\sqrt{n} = \infty$ a.s.
2. Use contradiction to show $S_n/\sqrt{n}$ does not converge in probability. (Hint: consider $n=m!$)

For 1st part: (Please point out errors if you see any)

W.L.O.G, we may assume $Var(X_1)=1$. Then by C.L.T, we have $S_n/\sqrt{n} \sim Z=N(0,1)$. Then the event $$\left\{\limsup S_n/\sqrt{n} = \infty\right\} = \left\{\bigcap _{m=1}^{\infty} \bigcup_{n=m}^{\infty} S_n/\sqrt{n} = \infty\right\}$$ is in the tail $\sigma$-algebra formed by $X_1,X_2,\dots$. By $0-1$ law, we have $P(\limsup S_n/\sqrt{n} = \infty)$ is either $0$ or $1$. Now, the event $\{\limsup S_n/\sqrt{n} = \infty\}$ is equivalent to: for any $N$, we have $\{\limsup S_n/\sqrt{n} > N\}$. Then \begin{align} P\left(\limsup S_n/\sqrt{n} > N\right)&=\lim_{m\rightarrow \infty} P\left(\bigcup_{n=m}^\infty S_n/\sqrt{n} > N\right) \\[0.2cm] & \ge \lim_{m\to \infty} P\left( S_m/\sqrt{m} > N\right) \\[0.3cm] & = 1-\Phi(N) >0 \end{align}

For 2nd part: Suppose it converges, then it must converge to normal $Z$. Then I am not sure how to proceed. I think it may somehow connect with part $1$. The hint also reminds me of Stirling's approximation, not sure.

Thanks.

• @Jimmy R. Many thanks for your editing. Nov 25, 2016 at 11:46

For the first, you need more details at the end: we have to use the fact that for each $N$, the event $\left\{\limsup S_n/\sqrt{n} > N\right\}$ is a tail event and by the previous work, its probability is $1$. Therefore, the intersection has a probability equal to $1$.
For the second part, we can show that the sequence $\left(S_{m!} / \sqrt{m!}\right)_{m\geqslant 1}$ does not converge in probability. Otherwise this sequence converges to some $Y$ in probability, we would have $S_{(m+1)!} / \sqrt{(m+1)!}-S_{m!} / \sqrt{m!}\to 0$ in probability. Define $$U_m:=\left(\frac 1{\sqrt{(m+1)!}} -\frac 1{\sqrt{m!}} \right) S_{m!}$$ $$V_m :=\frac 1{\sqrt{(m+1)!}}\left(S_{(m+1)!}-S_{m!}\right).$$ Then the sequences $\left(U_m\right)_{m\geqslant 1}$ and $\left(V_m\right)_{m\geqslant 1}$ are independent. Since for any positive $\epsilon$, $$\mathbb P\left(\left|U_m+V_m\right | \gt \epsilon\right)\geqslant \mathbb P\left(\left\{\left|U_m\right | \gt 2\epsilon\right\}\cap \left\{\left|V_m\right|\leqslant\epsilon \right\}\right),$$ we get that $U_m\to 0$ in probability.
• @LeiHao Maybe there is something I have not seen. It works fine with $n$ and $2n$ for example (not sure with $n$ and $n+1$). Nov 26, 2016 at 11:42