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I have a solution to the following exercise and I want to know if there is another way to solve it:

Let $X_1,X_2,\dots$ be i.i.d. real r.v. with $\operatorname{Var}(X_1)=1,\ E(X_1)=0$ and let $S_n=X_1+\cdots + X_n,\quad n\in\mathbb{N}$

Deduce $\lim\limits_{n\to \infty}E(|S_n|)=\infty$


Solution: if $N\sim\mathcal{N}(0,1)$ is a r.v. then $p:=P(N\geq 1)>0$ and the CLT gives $$P\left(\left|\frac{S_n}{\sqrt{n}}\right|\geq 1\right)\xrightarrow{n\to\infty}P(|N|\geq 1)=2p$$

With such $p>0$ and $n_0\in\mathbb{N}$ by Markov's inequality $$\forall n\geq n_0,\quad E(|S_n|)\geq \sqrt{n}P(|S_n|\geq \sqrt{n})\geq p\sqrt{n}$$ hence $\lim\limits_{n\to \infty}E(|S_n|)=\infty$

For me this $p$ just looks like something arbitrary and not really intuitively. It can be with Markov chains or martingales or something like that.

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  • $\begingroup$ Are you sure that the sum diverges? As a random walk it is symmetrical, so it would stay centered. $\endgroup$ – Cehhΐro Jan 16 '17 at 16:20
  • $\begingroup$ @O.VonSeckendorff Unless the solution is wrong. But I'm pretty confident that it is not...? $\endgroup$ – MarcE Jan 16 '17 at 16:37
  • $\begingroup$ Are the $X_n$ always normally distributed? $\endgroup$ – Cehhΐro Jan 16 '17 at 16:37
  • $\begingroup$ @O.VonSeckendorff We don't know the distribution of the $X_n$ besides of the expected value and the variance $\endgroup$ – MarcE Jan 16 '17 at 16:39
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With a martingale argument, you can drop the assumption that the variances exist.

The process $(S_n)_{n\in\mathbb{N}}$ is a martingale and so $(|S_n|)_{n\in\mathbb{N}}$ is a submartingale. This means that the sequence $\mathbb{E}(|S_n|)$ is non-decreasing, and so it has a limit (possibly $+\infty$). If the limit is finite, then $\sup_n \mathbb{E}(|S_n|)<\infty$ so that $(S_n)_{n\in\mathbb{N}}$ is an $L^1$-bounded martingale.

By the martingale convergence theorem, $S_n$ converges almost surely to some random variable $Y$. On the other hand, the process $S_n-X_1=X_2+\cdots + X_n$ is an identically distributed martingale which converges to a limit $Z$, where $Z$ is independent of $X_1$ and has the same distribution as $Y$.

This gives $Z+X_1\stackrel{D}{=}Z$ where $Z$ is independent of $X_1$. Taking characteristic functions gives $\varphi_Z(t)\varphi_{X_1}(t)=\varphi_Z(t)$ for all $t$. Since $\varphi_Z(t)$ is continuous and takes the value 1 at $t=0$, we can divide out to obtain $\varphi_{X_1}(t)=1$ for all sufficiently small $t$. You can now show that $\mathbb{P}(X_1=0)=1$, and so $S_n\equiv 0$ for all $n$.

This is the only one of these random walks that is $L^1$-bounded.

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  • $\begingroup$ Thank you very much for your answer! $\endgroup$ – MarcE Jan 22 '17 at 12:57
  • $\begingroup$ Glad to help out. $\endgroup$ – user940 Jan 22 '17 at 14:12

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