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Let $\{ \xi_n \}^{\infty}_{n=1}$ be iid nondegenerate random variables with finite second moment. Let $\mathbb{E} \xi_i = a$, $S_n = \xi_1 + \cdots + \xi_n$. Prove that $\sqrt{n} \left( \dfrac{S_n}{n} - a\right)$ converges in distribution and has no limit for the convergence in probability.

Сonvergence in distribution is proved, using the central limit theorem. How to prove that there is no convergence in probability?

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    http://math.stackexchange.com/questions/2029863/central-limit-theorem-and-convergence-in-probability-from-durrett http://math.stackexchange.com/questions/241394/convergence-in-central-limit-thorem – 2017-02-16

2 Answers 2

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There may be a more elementary solution, but one way to show this is to note that convergence in probability implies convergence almost surely along a subsequence. Define $A_n:=\sqrt n \big(\frac{S_n}{n} -a\big)$. If $A_n \to A$ is probability, then there is a subsequence $A_{n_k} \to A$ almost surely.

There are now two methods which can be used to derive a contradiction here:

Method 1: On one hand, $A$ must be normally distributed since convergence in probability implies convergence in distribution. On the other hand, it is easily shown that $A$ is measurable with respect to the tail $\sigma$-algebra of the increments of $S_n$. This is indeed the case because for any fixed $N$, clearly $\big(\frac{S_n }{\sqrt n} - a \sqrt n\big)$ and $\big(\frac{S_n-S_N}{\sqrt n}-a\sqrt n\big)$ will have the same subsequential limit as $n \to \infty$. Consequently $A$ must be constant, by Kolmogorov's $0$-$1$ law. This contradits the fact that $A$ was normally distributed with positive variance.

Method 2: To obtain a contradiction, we will show that the random set $\{A_{n_k}: k \in \Bbb N\}$ must be dense in $\Bbb R$. To prove this, we mimic the argument given in the answers here: Normalized partial sums of normal random variables are dense in $\mathbb{R}$

Specifically, for real numbers $a0$$ where $Z$ is normally distributed and the final equality follows from the central limit theorem. So we see that $P(E_{a,b})=1$. Consequently we get that $P(\{A_{n_k}\}$ is dense$)=P\big( \bigcap_{a

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    Actually, what we "know" is that the random set $\{A_n:n\in\mathbb N\}$ is dense in $\mathbb R$. Where is the argument that the random set $\{A_{n_k}:k\in\mathbb N\}$ is dense in $\mathbb R$? – 2017-02-15
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    @Did I just assumed that the same argument given for the density of $(A_n)$ would work if we replaced the index $n$ with any deterministic subsequence $n_k$ because there are still only tail events involved. Perhaps I missed some detail here? – 2017-02-15
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    If the argument carries on, please show how. What I am saying is that your link does not suffice (and at present, your mention that "The full argument for this statement is given..." is misleading). – 2017-02-15
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    @Did Okay done. Though there could still be some error. And I would still be interested in a more elementary solution. – 2017-02-15
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A possible way is the following.

Assume that $\left(Z_n\right)$ and $(Z'_n)$ are two sequences of random variables such that

  1. For any $n\geqslant 1$, $Z_n$ is independent of $Z'_n$.
  2. The sequence $\left(Z_n+Z'_n\right)_{n\geqslant 1}$ converges to $0$ in probability.
  3. For any $R>0$, there exists $ \delta\gt 0$ and $n_0(R)$ such that for any $n\geqslant n_0(R)$, $\mathbb P\left(\left|Z'_n\right|\leqslant R\right)\gt \delta$.

Then the sequence $(Z_n)$ converges to $0$ in probability.

To see this, let $\varepsilon\gt 0$. Then the following inclusion holds: $$\left\{\left|Z_n+Z'_n\right|\geqslant \varepsilon\right\}\supset\left\{\left|Z_n\right|\geqslant 2\varepsilon\right\}\cap \left\{\left| Z'_n\right|\leqslant \varepsilon\right\}.$$ Taking the probabilities on both sides and using independence, we derive that for $n\geqslant n_0(\varepsilon)$, $$\mathbb P\left\{\left|Z_n+Z'_n\right|\geqslant \varepsilon\right\}\geqslant\mathbb P\left\{\left|Z_n\right|\geqslant 2\varepsilon\right\}\delta.$$ To conclude, use this with $$Z_n=\frac{S_{2n}-S_n }{2\sqrt n}\mbox{ and }Z'_n:=\left(\frac 1{\sqrt 2}-1\right)\frac{S_n}{\sqrt n}. $$