# Proof of Convergence in Distribution with unbounded moment

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.

• I see now. If you truncate the random variables at $n$ then the variance of each is going to be $2 \log n$, so the variance of the sum of these truncated random variables would be $2 n \log n$, the square of the denominator in your expresssion. So I think you need to apply a theorem for sums of truncated random variables $Y = X 1(|X| < n)$. – Xiaomi Nov 13 '18 at 23:03
• let $T$ be the sum of the truncated random variables. Then if you can show the truncated sum converges to normal, and also show that $S -T$ converges in probability to 0, then the result would follow by Slutskys theorem – Xiaomi Nov 13 '18 at 23:06
• If I did it right, the characteristic function of the density would boil down to $2\int_1^\infty \cos(tx)/x^3 dx$, and then one could use the result ${\displaystyle \int {\frac {\cos ax}{x^{n}}}\,dx=-{\frac {\cos ax}{(n-1)x^{n-1}}}-{\frac {a}{n-1}}\int {\frac {\sin ax}{x^{n-1}}}\,dx\: {\mbox{(for }}n\neq 1),}$. With a tractable c.f. and the fact that the cf of the sum of independent rvs is the product of the cfs, you may then be able to expand that in a series and show that higher order terms go to 0. – Glen_b -Reinstate Monica Nov 13 '18 at 23:38
• @Glen_b Thanks for the comment. I tried this route, but how do you actually show the higher order terms go to 0 since it involves integration from 1 to $\infty$? – davidolohowski Nov 14 '18 at 0:01
• The integral is already done above; there's no integration left for you to do (other than to confirm my calculation). The characteristic function of the normalized sum will be function of t (and n), which I envisioned expanding in a Taylor series in a similar manner to the classical CLT. I haven't checked that it comes out but I expect you can just follow analogous steps -- it shouldn't be any harder than the integral that's already done above. – Glen_b -Reinstate Monica Nov 14 '18 at 0:05

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

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

• Thanks for your answer. We have not learned anything about Slutsky's theorem or Lindeberg condition. So I guess this route is not what I am looking for. I want to know how to approach it by using characteristic function? – davidolohowski Nov 14 '18 at 0:03
• I would then take glens advice. He is very smart and knows what he is talking about :) – Xiaomi Nov 14 '18 at 0:10

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.

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{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}

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

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

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:

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

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 $$n \rightarrow \infty$$.

• Thanks for your suggestion. This is the exact steps I have come up with. But I got stuck at proving the last line you wrote for 3 days. How do you come to the conclusion that the higher order terms in the expansion can be ignored? All the higher order terms diverge because of the improper integral that goes unbounded. – davidolohowski Nov 14 '18 at 2:25
• In my answer here I haven't actually attempted to expand the integrand out into a power series, so there are no higher-order terms at the moment. If you were to pursue this line of attack, I guess it is going to require you to find an expansion of the integrand where the higher-order terms are using inverse powers of $x$, so that they are vanishing instead of diverging. A standard Taylor expansion doesn't do this, so you're going to have to look at other ways of expanding it. Like I said, I'm not sure myself how to do this. – Reinstate Monica Nov 14 '18 at 2:29
• @David Leigh: I have made a further change in the form of the characteristic function that might make it easier to expand into a form with vanishing higher-order terms. Please have a look at the updated answer and see if this helps. – Reinstate Monica Nov 14 '18 at 3:05
• Thanks so much your effort. But I think after the transformation, it still diverges because the integration starts from 0 for the inverse power of $y$, assuming a power series expansion of $\cos(x)$. I have no idea what other form of expansion exists for $\cos(x)$. – davidolohowski Nov 14 '18 at 3:24