If $Z_1,Z_2,Z_3,\ldots$ are i.i.d. with $P(Z_i=-1) = P(Z_i=+1) = \frac 12,$ then we have by the Central Limit Theorem that $\frac{\sum_{i=1}^n Z_i}{\sqrt{n}}\stackrel{d}{\to} \mathcal{N}(0,1),$ so that for any continuous bounded function $f,$ we have $\mathbb{E}f\left(\frac{\sum_{i=1}^n Z_i}{\sqrt{n}}\right)\to\mathbb{E}f(W)$ where $W\sim\mathcal{N}(0,1).$ Now, $|\cdot|$ is not a bounded function, so it is not necessarily true that

$$\mathbb{E}\left|\frac{\sum_{i=1}^n Z_i}{\sqrt{n}}\right|\to\mathbb{E}|W|.$$

My question is whether the above is true for this specific distribution of $Z_i.$ If not, what does $\mathbb{E}\left|\frac{\sum_{i=1}^n Z_i}{\sqrt{n}}\right|$ converge to?


closed as off-topic by Bill Johnson, Stefan Kohl, Did, Ryan Budney, Steven Sam May 27 '14 at 4:09

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    $\begingroup$ It is. To cover the case of an unbounded $f$ all you need is to control the tails in a uniform way and the classical Bernstein bound ( en.wikipedia.org/wiki/… ) will be more than enough for your case. $\endgroup$ – fedja May 26 '14 at 0:06

The following result appears in Billingsley's book Convergence of probability measures (1968, and I guess the 1995 edition).

Theorem. Let $(Y_n)_{n\geqslant 1}$ be a sequence of non-negative random variables defined on a probability space $(\Omega,\mathcal F,\mu)$ such that $Y_n\to Y$ in distribution and $Y_n,Y$ have a finite expectation. Then the following are equivalent:

i) $\mathbb E[Y_n]\to\mathbb E[Y]$;

ii) the sequence $(Y_n)_{n\geqslant 1}$ is uniformly integrable.

Assume that i) holds and fix a positive $\varepsilon$. Fix $R$ such that $\mathbb E[Y\chi_{\{Y>R\}}]\lt\varepsilon$ and $\mu\{|Y|=R\}=0$. The set of discontinuity points of map $t\mapsto \chi_{-(R,R)}(t)$ has a null $\mu_Y$ measure, hence $$\lim_{n\to\infty}\mathbb E[Y_n\chi_{\{Y_n\lt R\}}]=\mathbb E[Y\chi_{\{Y\lt R\}}].$$

Conversely, we use the map $t\mapsto \chi_{-(R,R)}(t)$ for a well-chosen $R$.

In the context of the question, with $Y_n:=n^{-1/2}\left|\sum_{j=1}^nX_j\right|$, we observe that $\mathbb E[Y_n^2]=\mathbb E[X_0^2]$ (it holds more generally in the case of an iid sequence of zero-mean square integrable functions).

  • $\begingroup$ Thanks Davide. Can you say something about the rate of convergence for this specific example? $\endgroup$ – Hedonist May 26 '14 at 16:50

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