# Let $P(X_j=j)=P(X_j=-j)=1/2j^{\beta}$ and $P(X_j=0)=1-j^{-\beta}$ where $\beta\in(0,1)$, then $S_n/n^{(3-\beta)/2)}\Rightarrow c\chi$

Suppose $$P(X_j=j)=P(X_j=-j)=1/2j^{\beta}$$ and $$P(X_j=0)=1-j^{-\beta}$$, where $$\beta>0$$. Show that:

(i) If $$\beta>1$$ then $$S_n\to S_\infty$$ a.s.

(ii) If $$\beta\in(0,1)$$ then $$S_n/n^{(3-\beta)/2}\Rightarrow c\chi$$.

(iii) If $$\beta=1$$ then $$S_n/n\Rightarrow\aleph$$, where $$E\exp(it\aleph)=\exp\left(-\int_0^1 x^{-1}(1-\cos(xt)\,\mathrm{d}x\right).$$

This is problem 3.4.13 in Durrett's Probability text, part (i) was rather trivial, I feel fine about that part. I am having a difficult time on part (ii) though and would like verification for part (iii).

My ideas so far for part (ii) is to define the triangular array as $$S_{n,m}=\dfrac{X_m}{n^{(3-\beta)/2}}$$, and then use the Lindeberg-Feller theorem, but I am getting hung up on the details.

For part (iii) consider:

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, $$\phi_{S_n/n}(t) = \prod\limits_{1 \leq j \leq n} \phi_{X_j/n}(t)$$. Now, $$\phi_{X_j/n}(t) = \int_{\mathbb{R}} \mathrm{d}\lambda e^{it\lambda} \mathbb{P}\left(\frac{X_j}{n} = \lambda\right) = 1-\frac{1}{j} + \frac{1}{2j}(e^{it\frac{j}{n}} + e^{-it\frac{j}{n}}) = 1-\frac{1}{j}(1-\cos(tj/n)).$$ This is clearly real-valued and positive, so that we can write $$\log\phi_{S_n/n}(t) = \sum_{j = 1}^n \log\left(1-\frac{1}{n}\cdot \frac{n}{j}(1-\cos(tj/n)\right),$$ so, up to an $$O(1/n)$$ error term, we have $$\log \phi_{S_n/n}(t) = \frac{1}{n}\sum_{j=1}^n \frac{n}{j}(1-\cos(tj/n)) + O\left(\frac{1}{n}\right).$$ The sum on the right side is a Riemann sum for the exponential, so taking $$n \rightarrow \infty$$, we get $$\phi_{S_n/n}(t) \rightarrow E\left(e^{it\aleph}\right)$$, in our notation, the latter of which is continuous at $$0$$.

• Use the same technique for part (ii) as you are in part (iii), i.e. characteristic functions. – zoidberg Dec 9 '18 at 16:30
• @norfair I tried utilizing the same technique but I was not getting the desired result. :-/ – Dragonite Dec 9 '18 at 16:47
• Write out the same formula you had for $\beta=1$ for general $\beta$ and expand out the corresponding $e^{it j/n}$ and $e^{-it j/n}$ terms in Taylor series. You should see some nice cancellation from the constant and linear terms. You're left with a quadratic and higher order terms...from the form of the characteristic function of normal, it should be clear what to do. – zoidberg Dec 9 '18 at 17:19
• @Dragonite You might want to explain what $c_{\chi}$ actually means... you didn't introduce the notation. – saz Dec 9 '18 at 18:00

$$\def\e{\mathrm{e}}\def\i{\mathrm{i}}\def\d{\mathrm{d}}$$As is written at the start of exercise section, $$X_1, X_2, \cdots$$ are independent.
Define $$X_{n, k} = \dfrac{X_k}{n^{\frac{3 - β}{2}}}$$ for $$1 \leqslant k \leqslant n$$. Since Lindeberg's condition does not apply for $$\{X_{n, k} \mid 1 \leqslant k \leqslant n\}$$, so the proposition has to be proved directly. Since$$φ_{n, k}(t) := E(\exp(\i t X_{n, k})) = \frac{1}{k^β} \cos\frac{kt}{n^{\frac{3 - β}{2}}} + \left( 1 - \frac{1}{k^β} \right), \quad \forall t \in \mathbb{R}$$ it suffices to prove that there exists a constant $$c$$ that$$\lim_{t → ∞} \prod_{k = 1}^n φ_{n, k}(t) = \exp\left( -\frac{1}{2} c^2 t^2 \right). \quad \forall t \in \mathbb{R}$$
For a fixed $$t$$, in order to apply Exercise 3.1.1., denote $$c_{n, k} = φ_{n, k}(t) - 1 = \dfrac{1}{k^β} \left( \cos\dfrac{kt}{n^{\frac{3 - β}{2}}} - 1 \right)$$, it suffices to prove that$$\lim_{n → ∞} \max_{1 \leqslant k \leqslant n} |c_{n, k}| = 0, \quad \lim_{n → ∞} \sum_{k = 1}^n c_{n, k} = -\frac{1}{2} c^2 t^2, \quad \sup_{n \geqslant 1} \sum_{k = 1}^n |c_{n, k}| < +∞.$$ Since $$|c_{n, k}| \leqslant \dfrac{1}{k^β} · \dfrac{1}{2} \left( \dfrac{kt}{n^{\frac{3 - β}{2}}} \right)^2 = \dfrac{k^{2 - β} t^2}{2n^{3 - β}} \leqslant \dfrac{t^2}{2n}$$, then $$\lim\limits_{n → ∞} \max\limits_{1 \leqslant k \leqslant n} |c_{n, k}| = 0$$ and$$\sum_{k = 1}^n |c_{n, k}| \leqslant \sum_{k = 1}^n \frac{k^{2 - β} t^2}{2n^{3 - β}} \leqslant \frac{t^2}{2n^{3 - β}} \int_1^{n + 1} x^{2 - β} \,\d x \leqslant \frac{t^2}{2(3 - β)} \left( \frac{n + 1}{n} \right)^β,$$ which implies $$\sup\limits_{n \geqslant 1} \sum\limits_{k = 1}^n |c_{n, k}| < +∞$$.
Now, since $$\cos x = 1 - \dfrac{x^2}{2} + \dfrac{x^4}{24} + o(x^5)\ (x → 0)$$, there exists $$δ > 0$$ such that$$1 - \frac{x^2}{2} < \cos x < 1 - \frac{x^2}{2} + \frac{x^4}{23}. \quad \forall |x| < δ$$ For $$n > \left( \dfrac{t}{δ} \right)^{\frac{2}{1 - β}}$$,\begin{align*} \sum_{k = 1}^n c_{n, k} &\leqslant \sum_{k = 1}^n \frac{1}{k^β} \left( -\frac{k^2 t^2}{2n^{3 - β}} + \frac{k^4 t^4}{23n^{2(3 - β)}} \right) = -\sum_{k = 1}^n \frac{k^{2 - β} t^2}{2n^{3 - β}} + \sum_{k = 1}^n \frac{k^{4 - β} t^4}{23n^{2(3 - β)}}\\ &\leqslant -\frac{t^2}{2n^{3 - β}} \int_0^n x^{2 - β} \,\d x + n · \frac{n^{4 - β} t^4}{23n^{2(3 - β)}} = -\frac{t^2}{2(3 - β)} + \frac{t^4}{23n^{1 - β}}, \end{align*}$$\sum_{k = 1}^n c_{n, k} \geqslant -\sum_{k = 1}^n \frac{1}{k^β} · \frac{k^2 t^2}{2n^{3 - β}} \geqslant -\frac{t^2}{2(3 - β)} \left( \frac{n + 1}{n} \right)^β,$$ thus $$\lim\limits_{n → ∞} \sum\limits_{k = 1}^n c_{n, k} = -\dfrac{t^2}{2(3 - β)}$$. Applying Exercise 3.1.1., $$\dfrac{S_n}{n^{\frac{3 - β}{2}}} \Rightarrow cχ$$, where $$c = \dfrac{1}{\sqrt{3 - β}}$$.