Proof of Convergence in Distribution with unbounded moment

Question

I posted the question here, but no one has provided an answer, so I am hoping I could get an answer here. Thanks very much!

Prove that given $\{X_n\}$ being a sequence of iid r.v's with density $|x|^{-3}$ outside $(-1,1)$, the following is true: $$ \frac{X_1+X_2 + \dots +X_n}{\sqrt{n\log n}} \xrightarrow{\mathcal{D}}N(0,1). $$

The original post has a 2 in the square root of the denominator. There should not be a 2.

Answer 1

Firstly define $Y_{k,n} = X_k 1\{ |X_k| \leq n \}$. Then it is easy to see that $Var(Y_{k,n}) = 2 \log n$ and that

$$Var (T_n ) = Var \left( \sum_{k=1}^n Y_{k,n} \right) = 2n \log n$$

Letting $S_n = \sum_{k=1}^n X_k$ we also see that

$$P(S_n \neq T_n) \leq P(\cup_k X_k \neq Y_{k,n}) \leq n P(X_k > n) = \frac{n}{2n^2} \to 0$$

So that it is enough to show

$$\frac{T_n}{\sqrt{2n \log n}} \to N(0,1)$$

and the result follows by Slutsky's theorem for the original sum $S_n$.

This new sum $T_n$ now has finite variance so we can apply the Lindeberg-Feller theorem (otherwise called Lindeberg condition).

Let $Z_{k,n} = \frac{Y_{k,n}}{\sqrt{2 n \log n}}$. Then we see that if the two two conditions of Lindeberg-Feller theorem hold:

1. $\sum_{k=1}^n Var(Z_{k,n}) = 1 > 0$ for all $n$ (holds trivially)

2. For all $\epsilon > 0$, $\sum_{k=1}^n E[|Z_{k,n}|^2 1\{ |Z_{k,n}| > \epsilon \}] \to 0$

then the result follows. So you only need to verify the second condition.

With the second condition you should note that you can rewrite $1\{ |Z_{k,n}| > \epsilon \}$ as

$$1\{ |Z_{k,n}| > \epsilon \} = 1\{ |Y_{k,n}| > \sqrt{n \log n }\epsilon \} = 1\{ X_k 1 \{ |X_k| \leq n \log n \} > \sqrt{n \log n }\epsilon \}$$

Answer 2

This is a proof by c.f. approach:

The c.f. of $X_i$ is $$ \phi_i(t) = \int_{R}e^{itx}|x|^{-3}\boldsymbol{1}_{x \notin (-1,1)}dx = 2\int_{1}^{\infty}\frac{\cos(tx)}{x^3}dx. $$ Hence, for $Y_n = (X_1+X_2+\dots+X_n)(\sqrt{n\log n})^{-1}$, we have \begin{align*} \phi_{Y_n}(t) =& \phi_i\left(\frac{t}{\sqrt{n\log n}}\right)^n\\ =& \left(2\int_{1}^{\infty}\cos\left(\frac{tx}{\sqrt{n\log n}}\right)\frac{1}{x^3}dx\right)^n.\\ \end{align*} We first consider the integral: \begin{align*} 2\int_{1}^{\infty}\cos\left(\frac{tx}{\sqrt{n\log n}}\right)\frac{1}{x^3}dx =& 1 + 2\int_{1}^{\infty}\cos\left(\frac{tx}{\sqrt{n\log n}}\right)\frac{1}{x^3}-\frac{1}{x^3}dx\\ =& 1 + 2\int_{1}^{\sqrt{n\log\log n}}\cos\left(\frac{tx}{\sqrt{n\log n}}\right)\frac{1}{x^3}-\frac{1}{x^3}dx \\ +& 2\int_{\sqrt{n\log\log n}}^{\infty}\cos\left(\frac{tx}{\sqrt{n\log n}}\right)\frac{1}{x^3}-\frac{1}{x^3}dx, \end{align*} since for $x \in [1, \sqrt{n\log\log n}]$, ${\displaystyle \frac{tx}{\sqrt{n\log n}}} \to 0$ as $n \to \infty$. Hence, we can apply the Taylor expansion of the cosine term in the first integral around $0$. Then we have \begin{align*} 2\int_{1}^{\infty}\cos\left(\frac{tx}{\sqrt{n\log n}}\right)\frac{1}{x^3}dx =& 1 + 2\int_{1}^{\sqrt{n\log\log n}}-\frac{t^2}{2n\log nx} + \left[\frac{t^4x}{24(n\log n)^2 }-\dots\right]dx \\ +& 2\int_{\sqrt{n\log\log n}}^{\infty}\cos\left(\frac{tx}{\sqrt{n\log n}}\right)\frac{1}{x^3}-\frac{1}{x^3}dx\\ =& 1 + 2\int_{1}^{\sqrt{n\log\log n}}-\frac{t^2}{2n\log nx}dx + o(1/n)\\ +& 2\int_{\sqrt{n\log\log n}}^{\infty}\cos\left(\frac{tx}{\sqrt{n\log n}}\right)\frac{1}{x^3}-\frac{1}{x^3}dx\\ =& 1 -\frac{t^2\log( n\log\log n)}{2n\log n} + o(1/n)\\ +& 2\int_{\sqrt{n\log\log n}}^{\infty}\cos\left(\frac{tx}{\sqrt{n\log n}}\right)\frac{1}{x^3}-\frac{1}{x^3}dx\\ \end{align*} Now \begin{align*} \int_{\sqrt{n\log\log n}}^{\infty}|\cos\left(\frac{tx}{\sqrt{n\log n}}\right)\frac{1}{x^3}-\frac{1}{x^3}|dx \leq& \int_{\sqrt{n\log\log n}}^{\infty}\frac{2}{x^3}dx\\ =& \frac{1}{n\log\log n} \in o(1/n). \end{align*} Hence, $$ 2\int_{1}^{\infty}\cos\left(\frac{tx}{\sqrt{n\log n}}\right)\frac{1}{x^3}dx = 1 -\frac{t^2\log( n\log\log n)}{2n\log n} + o(1/n). $$ Let $n \to \infty$, we have $$ \lim_{n \to \infty}\left(2\int_{1}^{\infty}\cos\left(\frac{tx}{\sqrt{n\log n}}\right)\frac{1}{x^3}dx\right)^n = \lim_{n \to \infty}\left(1 -\frac{t^2\log( n\log\log n)}{2n\log n}\right)^n = \lim_{n \to \infty}\left(1-\frac{t^2}{2n}\right)^n = e^{-t^2/2}, $$

which completes the proof.

Answer 3

To see if this gets us anywhere useful, I'm going to go some of the way along the lines suggested by Glen_b in the comments. The characteristic function of the underlying random variables is:

$$\begin{equation} \begin{aligned} \varphi_X(t) = \mathbb{E}(\exp(itX)) &= \int \limits_{\mathbb{R}} \exp(itx) f_X(x) dx \\[6pt] &= \int \limits_{\mathbb{R}-(-1,1)} |x|^{-3} \exp(itx) dx \\[6pt] &= \int \limits_{\mathbb{R}-(-1,1)} |x|^{-3} \cos(tx) dx + i \int \limits_{\mathbb{R}-(-1,1)} |x|^{-3} \sin(tx) dx \\[6pt] &= \int \limits_{\mathbb{R}-(-1,1)} |x|^{-3} \cos(tx) dx \\[6pt] &= - \int \limits_{-\infty}^{-1} \frac{\cos(tx)}{x^3} dx + \int \limits_1^\infty \frac{\cos(tx)}{x^3} dx \\[6pt] &= 2 \int \limits_1^\infty \frac{\cos(|t|x)}{x^3} dx. \\[6pt] \end{aligned} \end{equation}$$

Now, using the change of variable $y = x^{-2}$ we have $dy = -2 x^{-3} dx$ which gives:

$$\begin{equation} \begin{aligned} \varphi_X(t) &= \int \limits_0^1 \cos \Big( \frac{|t|}{\sqrt{y}} \Big) dy. \\[6pt] \end{aligned} \end{equation}$$

We can see that the characteristic function is symmetric around $t=0$. Hence, without loss of information we can take $t>0$ and write it in simpler terms as:

$$\varphi_X(t) = \int \limits_0^1 \cos \Big( \frac{t}{\sqrt{y}} \Big) dy.$$

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The required limit: Now we define the partial sums:

$$S_n = \frac{X_1 + \cdots + X_n}{\sqrt{2n \log n}}.$$

Using the rules for characteristic functions we then have:

$$\varphi_{S_n}(t) = \varphi_X \Bigg( \frac{t}{\sqrt{2n \log n}} \Bigg)^n = \Bigg[ \int \limits_0^1 \cos \Bigg( \frac{t}{\sqrt{2n}} \cdot \frac{1}{\sqrt{y \log n}} \Bigg) dy \Bigg]^n.$$

To prove the convergence result we have to show that $\lim_{n \rightarrow \infty} \varphi_{S_n}(t) = \exp( - t^2/2 )$. Using Bernoulli's expansion for $e$ it would be sufficient to prove that as $n$ becomes large we have:

$$\int \limits_0^1 \cos \Bigg( \frac{t}{\sqrt{2n}} \cdot \frac{1}{\sqrt{y \log n}} \Bigg) dy \longrightarrow 1 - \frac{t^2}{2n}.$$

I will not go any further than this for now. It is not clear to me whether this result holds, or how you would prove it, but at least this gets you to a possible pathway to a solution. To prove this limit, you would need to find some useful expansion of the integrand that will ensure that higher-order terms vanish in the integral as $n \rightarrow \infty$.

Answer 4

When I started reading this question I was a bit confused. This factor $\sqrt{n\log n}$ is not intuitive to me. It is not the typical expression in the CLT.

Below I am trying to view your question in a more intuitive way without resorting to characteristic functions and looking at the limits of higher moments (which would be something mimicking the proof of the CLT). I believe that the question is already 'failing' at a simpler level.

The variance of the distribution is $2$ so the variance of the sum is $2 \sqrt{n}$. When this sum is divided by $\sqrt{n\log n}$ you get a variance of $\frac{1}{\log n}$. So this is not gonna equal a distributie of $N(0,1)$ because the variance is not equal.

Unless I am missing something you can not prove the statement that you are trying to prove and all the magic with characteristic functions seems like distraction to me.