4
$\begingroup$

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 $0$.

$\endgroup$
  • $\begingroup$ Use the same technique for part (ii) as you are in part (iii), i.e. characteristic functions. $\endgroup$ – zoidberg Dec 9 '18 at 16:30
  • $\begingroup$ @norfair I tried utilizing the same technique but I was not getting the desired result. :-/ $\endgroup$ – Dragonite Dec 9 '18 at 16:47
  • $\begingroup$ 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. $\endgroup$ – zoidberg Dec 9 '18 at 17:19
  • 2
    $\begingroup$ @Dragonite You might want to explain what $c_{\chi}$ actually means... you didn't introduce the notation. $\endgroup$ – saz Dec 9 '18 at 18:00
1
+200
$\begingroup$

$\def\e{\mathrm{e}}\def\i{\mathrm{i}}\def\d{\mathrm{d}}$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 $c = \dfrac{1}{\sqrt{3 - β}}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.