I was doing a problem about the converge of the sum of random variables which has two parts:

Let $X_1, X_2 ,\dots$ be independent and identically distributed random variables with $E X_i = 0$, $ 0 <\operatorname{Var}(X_i) < \infty $, and let $S_n = X_1 + \dots+ X_n$.

(a) Use the central limit theorem and Kolmogorov's zero-one law to conclude that $ \limsup S_n / \sqrt{n} = \infty$ almost surely.

(b) Use an argument by contradiction to show that $S_n / \sqrt{n}$ does not converge in probability. Hint: Consider $n = m! $.

I did part (a) but I'm not sure about my proof and people are welcome to go through it:

(a) Let $\operatorname{Var} (X_i) = \sigma ^2$, then by central limit theorem $\frac{S_n} {\sigma \sqrt{n}} \Rightarrow \chi$ where $\chi$ has the standard normal distribution. Therefore for any $x > 0$, $P( \limsup \frac{S_n}{\sigma \sqrt{n}} > x ) \ge P (\chi > x ) > 0$, thus $P ( \limsup \frac{S_n}{\sigma \sqrt{n}} > x) = 1$ for any $x>0$ by Kolmogorov's zero-one law. By monotonicity this implies $ P ( \limsup \frac {S_n}{\sqrt{n}} = \infty) =1 $, which is $\limsup \frac{S_n}{\sqrt{n}} = \infty$ a.s.

But I stuck with part (b), my approach is the following:

(b) Suppose $ \frac{ S_n}{\sqrt{n}}$ converges in probability, then the subsequence $\frac{S_{m!}}{\sqrt{m!}}$ has a further subsequence that converges almost surely.

But it doesn't seem to work out well and I don't think I can go on with it. How to prove part (b)? Any idea is appreciated.

  • $\begingroup$ For part (b): if there was convergence in probability, the limit would be independent of the $X_i$'s (adapt the arguments of this related answer) and at the time $\sigma(X_i ; i \geq 1)$-measurable. So it should be constant, which is absurd (by the CLT). $\endgroup$ – Siméon Dec 6 '13 at 17:02
  • $\begingroup$ @user112564, why is $P(\chi > x)>0$? $\endgroup$ – Sahiba Arora Nov 11 '16 at 21:03

The first part looks ok, but I would apply central limit theorem, not the law of large number. The lower bound of the probability of the $\limsup$ has to be justified (portmanteau theorem).

For part b), we can use the following idea: in the case $S_n/\sqrt n\to \chi$ in probability, we would have $$Y_n:=\frac{S_{2n}}{\sqrt{2n}}-\frac{S_n}{\sqrt n}\to 0 \mbox{ in probability}.$$ But $Y_n=\frac{S_{2n}-S_n}{\sqrt{2n}}+\frac{S_n}{\sqrt n}\left(\frac 1{\sqrt 2}-1\right)=:Y'_n+Y''_n$. Since $S_{2n}-S_n$ is independent of $S_n$, we can compute the limit in distribution of each of the two terms which compose $Y_n$. Notice that $Y'_n$ has the same distribution as $S_n/\sqrt{2n}$ which converges in distribution to a centered normal random variable of variance $\sigma^2/2$, while $Y''_n$ converges in distribution to a centered normal random variable of variance $\sigma^2(1-\sqrt 2)^2/2$. Therefore, it can be shown that $Y_n$ converges to a non-degenerated Gaussian random variable.

  • $\begingroup$ You are welcome! $\endgroup$ – Davide Giraudo Dec 8 '13 at 9:50
  • $\begingroup$ What is the convergence of $\frac{S_{2n}-S_{n}}{\sqrt{2n}}$? $\endgroup$ – kayak May 3 '17 at 13:54
  • $\begingroup$ @kayak I have added more details. $\endgroup$ – Davide Giraudo May 3 '17 at 14:26
  • $\begingroup$ Thanks for your comment. I appreciate it much. But I now know $Y_{n}\rightarrow 0$ in probability and $Y_{n}\rightarrow Y$ in distribution where $Y$ is a non-degenerated Gaussian random variable. But how can we get a contradiction here? $\endgroup$ – kayak May 3 '17 at 14:42
  • $\begingroup$ Convergence in probability to $0$ implies convergence in distribution to $0$, and the limit in distribution is unique. $\endgroup$ – Davide Giraudo May 3 '17 at 14:51

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