# Answering Part of Durrett Exercise 3.4.13 4th edition

I am trying to prove part (iii) of Durrett exercise 3.4.13

Suppose $P(X_j=j)=P(X_j=-j)=\frac{1}{2j^\beta}$ and $P(X_j=0)=1-j^{-\beta}$ where $\beta>0$. Show that if $\beta=1$ then $\frac{S_n}{n}\implies \aleph$ where $E(\exp(it\aleph))=\exp\left(-\int\limits_0^1 x^{-1}(1-\cos(xt))dx\right)$ and $\implies$ denotes convergence in distribution.

The expectation part leads me to believe that the proof involves characteristic functions somehow, but I don't really know how to proceed.

Any help would be greatly appreciated.

Thanks!

## 1 Answer

It is a well-known theorem of Levy that if $\{X_n\}$ is a collection of random variables and $Y$ is another random variable then $X_n \Rightarrow Y$ iff $\phi_{X_n}(t) \rightarrow \phi_Y(t)$ as $n \rightarrow \infty$ and $\phi_Y$ is continuous at $t = 0$. Moreover, by properties of Fourier transforms, $\phi_{S_n/n}(t) = \prod_{1 \leq j \leq n} \phi_{X_j/n}(t)$. Now, \begin{equation*} \phi_{X_j/n}(t) = \int_{\mathbb{R}} d\lambda e^{it\lambda} \mathbb{P}(\frac{X_j}{n} = \lambda) = 1-\frac{1}{j} + \frac{1}{2j}(e^{it\frac{j}{n}} + e^{-it\frac{j}{n}}) = 1-\frac{1}{j}(1-\cos(tj/n)). \end{equation*} This is clearly real-valued and positive, so that we can write \begin{equation*} \log\phi_{S_n/n}(t) = \sum_{j = 1}^n \log(1-\frac{1}{n}\cdot \frac{n}{j}(1-\cos(tj/n)), \end{equation*} so, up to an $O(1/n)$ error term, we have \begin{equation*} \log \phi_{S_n/n}(t) = \frac{1}{n}\sum_{j=1}^n \frac{n}{j}(1-\cos(tj/n)) + O\left(\frac{1}{n}\right). \end{equation*} The sum on the right side is a Riemann sum for the exponential in your problem, so taking $n \rightarrow \infty$, we get $\phi_{S_n/n}(t) \rightarrow E\left(e^{it\aleph}\right)$, in your notation, the latter of which is continuous at zero.

• Could you please explain why $E(\exp(it\aleph))$ is continuous at zero? – User112358 Nov 8 '15 at 22:02
• Taylor expand $x^{-1}(1-\cos(tx))$ to get $\frac{1}{2}t^2 x$ plus higher order terms in $t$. All of these terms vanish as $t \rightarrow 0$, so the integral in the exponent is well-defined. – Dead-End Nov 10 '15 at 16:41