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This is from and exercise of probability theory:

Let $(X_j)_{j\geq 1}$ be independent and let $X_j$ have the uniform distribution on $(-j,j)$. Show that $$ \lim_{n\to\infty}\frac{S_n}{n^{3/2}}=Z\sim N(0,\frac{1}{9}) $$ in distribution.

In terms of characteristic functions, it suffices to show that $$ \lim_{n\to\infty}\prod_{j=1}^n\frac{\sin(jn^{-3/2}u)}{jn^{-3/2}u}=e^{-u^2/18}. $$ Take $\log$ on both side one gets: $$ \lim_{n\to\infty}\sum_{j=1}^n\log\frac{\sin(jn^{-3/2}u)}{jn^{-3/2}u}=-\frac{u^2}{18}. $$ How can this limit be proved?

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I think it should be an n^{-3/2} instead of n^{3/2} –  blabler Nov 27 '12 at 2:55
    
@blabler: Corrected. Thanks! –  Jack Nov 27 '12 at 12:41

1 Answer 1

up vote 4 down vote accepted

For each $u \in \Bbb{R}$, we have $n^{-1/2}|u| < 1$ for every sufficiently large $n$. Then for any $1 \leq j \leq n$ we have $jn^{-3/2}|u| < 1$. Now, from the Taylor expansion

\begin{equation} \log\left(\frac{\sin x}{x}\right) = -\frac{x^2}{6} + O\left(x^4\right), \end{equation}

we have

\begin{align} \sum_{j=1}^{n} \log\left( \frac{\sin (jn^{-3/2} u)}{jn^{-3/2} u} \right) &= \sum_{j=1}^{n} \left\{ -\frac{1}{6}\frac{j^2u^2}{n^3} + O\left(\frac{(ju)^4}{n^6}\right)\right\} \\ &= \sum_{j=1}^{n} \left\{ -\frac{1}{6}\frac{j^2u^2}{n^3} + O\left(\frac{u^4}{n^2}\right)\right\} \\ &= -\frac{u^2}{6}\left( \sum_{j=1}^{n} \frac{j^2}{n^3} \right) + O\left(\frac{u^4}{n}\right). \end{align}

Taking $n \to \infty$ we obtain the desired result.

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