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Let $X_1, X_2, \dots$ be independent random variables and $$X_n = Y_n + Z_n$$ where $Y_n$ takes values $1$ and $-1$ with chance $1/2$ each, and $$P(Z_n = \pm n) = 1/(2n^2) = (1 - P(Z_n = 0))/2$$ and $S_n:=X_1+ \dots X_n$. Show that Lindeberg condition does not hold, yet $$S_n/\sqrt{n} \rightarrow N(0,1).$$

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    $\begingroup$ Welcome to Math.SE! Please, consider updating your question to include what you have tried and where you are getting stuck. That way, people on this site will know exactly what help you need. $\endgroup$ – Davide Giraudo Mar 18 '14 at 19:48
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    $\begingroup$ @DavideGiraudo Thank you! I think that the key of this problem is to find Var($S_n/\sqrt{n} $). Lindeberg condition does not hold, if it turns out that Var($S_n/\sqrt{n})\neq 1$. $\endgroup$ – Gabriel Mar 18 '14 at 19:57
  • $\begingroup$ @DavideGiraudo Could you please take a look at the answer below and the comments that we added? What do you think? Thanks in advance! $\endgroup$ – Gabriel Mar 19 '14 at 16:45
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Partial answer: Calculation that $$Var\left(\frac{S_n}{\sqrt{n}}\right)\neq1$$ Firstly we have that $$E[Y_n]=-1\cdot \frac12 + 1\cdot \frac12=0$$ and $$E[Y_n^2]=(-1)^2\frac12+(1)^2\frac12=1$$ which gives $$Var(Y_n)=1-0^2=1$$ Similarly we have that $$E[Z_n]=-n\cdot\frac{1}{2n^2}+n\cdot\frac{1}{2n^2}+0=0$$ and $$E[Z_n^2]=(-n)^2\frac{1}{2n^2}+(n)^2\frac{1}{2n^2}+0=1$$ which gives $$Var(Z_n)=1-0^2=1$$ Therefore, assuming independence of $Y_n$ and $Z_n$ we have that $$Var(X_n)=Var(Y_n)+Var(Z_n)=1+1=2$$ which gives that $$Var\left(\frac{S_n}{\sqrt{n}}\right)=\frac1n \sum_{i=1}^{n} Var(X_i)=\frac1n \cdot n2=2 \neq1$$ as you mentioned in the comment.


Not an answer (just calculations of characteristics functions). We have that $$φ_{\frac{S_n}{\sqrt{n}}}(t)=φ_{X_1}\left(\frac{1}{\sqrt{n}}t\right)\cdotφ_{X_2}\left(\frac{1}{\sqrt{n}}t\right)\cdot \ldots\cdotφ_{X_n}\left(\frac{1}{\sqrt{n}}t\right)$$ where $$φ_{X_k}\left(\frac{1}{\sqrt{n}}t\right)=φ_{Y_k+Z_k}\left(\frac{1}{\sqrt{n}}t\right)=φ_{Y_k}\left(\frac{1}{\sqrt{n}}t\right)\cdotφ_{Z_k}\left(\frac{1}{\sqrt{n}}t\right)$$ with $$φ_{Y_k}\left(\frac{1}{\sqrt{n}}t\right)=E[e^{\frac{it}{\sqrt{n}}Y_k}]=\frac12 e^{-\frac{it}{\sqrt{n}}}+\frac12 e^{\frac{it}{\sqrt{n}}}$$ for all $k \in \mathbb N$ and $$φ_{Z_k}\left(\frac{1}{\sqrt{n}}t\right)=E[e^{\frac{it}{\sqrt{n}}Z_k}]=\frac1{2k^2} e^{-\frac{itk}{\sqrt{n}}}+\frac1{2k^2} e^{\frac{itk}{\sqrt{n}}}+\left(1-\frac1{k^2}\right)\cdot1$$ Thus $$φ_{X_k}\left(\frac{1}{\sqrt{n}}t\right)=\left(\frac12 e^{-\frac{it}{\sqrt{n}}}+\frac12 e^{\frac{it}{\sqrt{n}}} \right)\cdot \left(\frac1{2k^2} e^{-\frac{itk}{\sqrt{n}}}+\frac1{2k^2} e^{\frac{itk}{\sqrt{n}}}+\left(1-\frac1{k^2}\right) \right)=\\=\frac{1}{4k^2}\left(e^{\frac{-it}{\sqrt{n}}(1+k)}+e^{\frac{-it}{\sqrt{n}}(1-k)}+e^{\frac{-it}{\sqrt{n}}(k-1)}+e^{\frac{-it}{\sqrt{n}}(-1-k)}\right)+\frac12\left(1-\frac1{k^2}\right)\left(e^{\frac{-it}{\sqrt{n}}}+e^{\frac{it}{\sqrt{n}}}\right)$$ and $$φ_{\frac{S_n}{\sqrt{n}}}(t)=\prod_{k=1}^{n}\left[\frac{1}{4k^2}\left(e^{\frac{-it}{\sqrt{n}}(1+k)}+e^{\frac{-it}{\sqrt{n}}(1-k)}+e^{\frac{-it}{\sqrt{n}}(k-1)}+e^{\frac{-it}{\sqrt{n}}(-1-k)}\right)+\frac12\left(1-\frac1{k^2}\right)\left(e^{\frac{-it}{\sqrt{n}}}+e^{\frac{it}{\sqrt{n}}}\right) \right]$$ which shows that calculating the characteristic function of $S_n/\sqrt{n}$ is not easy.


So, you should proceed as in the Proof of CLT presented here. For any random variable, $X$, with zero mean and a unit variance ($\mathrm{Var}(X) = 1$), the characteristic function of $X$ is, by Taylor's theorem, $$\varphi_X(t) = 1 - {t^2 \over 2} + o(t^2), \quad t \rightarrow 0$$ Now, the random variables $\frac{X_k}{\sqrt{2}}$ have the desired property. Thus, although they are not i.i.d. the random variable $$Z_n=\frac{S_n}{\sqrt{2n}}$$ converges to the standard nomral distribution from which you have that $$\frac{S_n}{\sqrt{n}} \to N(0,2)$$

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  • $\begingroup$ Thank you very much! So it remains to be proven that $P(S_n/\sqrt{n}\leq x) \rightarrow \frac{1}{\sqrt(2\pi)}\int_{-\infty}^x e^{-t^2/2} dt$. Or it is better to compute the characteristic function and apply Lévy's theorem? $\endgroup$ – Gabriel Mar 18 '14 at 21:46
  • $\begingroup$ @Gabriel Since $E[X_n]=0$ and $Var(X_n)=2$ the random variable $S_n/\sqrt{n}$ converges to $N(0,2)$ (according to the CLT) and not $N(0,1)$. Am I wrong? Look here $\endgroup$ – Jimmy R. Mar 18 '14 at 21:51
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    $\begingroup$ These are not i.i.d. random variables. The Lindeberg condition should have been verified to apply CLT. But this is an example where Lindeberg condition does not hold. $\endgroup$ – Gabriel Mar 18 '14 at 22:02
  • $\begingroup$ @Gabriel Yes, you are right about that. Sorry, I did not see it. Yes, calculate the characteristic function, it sounds ok. Give an update when you have calculated it $\endgroup$ – Jimmy R. Mar 18 '14 at 22:07
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    $\begingroup$ @Gabriel. Ok, If I find the time today or tomorrow, I will calculate the characteristic function. $\endgroup$ – Jimmy R. Mar 22 '14 at 11:34
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It seems that the goal of the example is to give an example where the classical central limit theorem holds but the Lindeberg condition is not satisfied. Hence the later is sufficient for the central limit theorem but not necessary.

I will assume that the collection of random variables $\left\{Y_i,Z_j,i,j\geqslant 1\right\}$ is independent.

By the Borel-Cantelli lemma, there exists $\Omega'$ of probability one such that for all $\omega\in \Omega'$, there exists an integer $N(\omega)$ for which $X_n(\omega)=0$ for all $n\geqslant N(\omega)$ hence $n^{-1/2}\sum_{i=1}^nZ_i\to 0$ almost surely. From the classical central limit theorem applied to the i.i.d. sequence $\left(Y_i\right)_{i\geqslant 1}$ and Slutsky's theorem, it follows that $n^{-1/2}\sum_{i=1}^nX_i\to N(0,1)$.

Observe that $\left(X_i\right)_{i\geqslant 1}$ is independent and that $\mathbb E\left[X_i^2\right]= \mathbb E\left[Y_i^2\right]+\mathbb E\left[Z_i^2\right]=2$ hence Lindeberg's condition reads $$ \forall \varepsilon\gt 0,\frac 1n\sum_{i=1}^n \mathbb E\left[X_i^2\mathbb 1\left\{\left\lvert X_i\right\rvert \gt \varepsilon \sqrt n\right\}\right]\to 0. $$ Observe that for all $i$, $$X_i^2\mathbb 1\left\{\left\lvert X_i\right\rvert \gt \varepsilon \sqrt n\right\}\geqslant \left(i+1\right)^2\left[i+1>\varepsilon\sqrt n\right] \mathbf 1\{Y_i=1\}\mathbf 1\{Z_i=i\}$$ where $[P]$ equal $i$ if assertion $P$ holds and $0$ otherwise. Therefore, $$\mathbb E\left[X_i^2\mathbb 1\left\{\left\lvert X_i\right\rvert \gt \varepsilon \sqrt n\right\}\right]\geqslant \left(i+1\right)^2\left[i+1>\varepsilon\sqrt n\right] \Pr\{Y_i=1\}\Pr 1\{Z_i=i\}\geqslant \frac 14\left[i+1>\varepsilon\sqrt n\right].$$ It follows that $$ \frac 1n\sum_{i=1}^n \mathbb E\left[X_i^2\mathbb 1\left\{\left\lvert X_i\right\rvert \gt \varepsilon \sqrt n\right\}\right]\geqslant \frac 1n\sum_{\varepsilon\sqrt n\leqslant i\leqslant n}\frac 14=\frac{n-\varepsilon\sqrt n}{4n}. $$

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I will prove more general statement. Let's assume $(Y_n)_n$ i.i.d. such that $\mathbf{E}Y_1=0$ and $\mathbf{E}Y_1^2=1$. To show that $S_n/\sqrt{n}\to N(0,1)$ we need two things. Let $S_n^Y=Y_1+\dots+Y_n$, and $S_n^Z=Z_1+\dots+Z_n$. Firstly, by Lindeberg-Levy theorem (https://en.wikipedia.org/wiki/Central_limit_theorem#Classical_CLT) $S_n^Y/\sqrt{n}\overset{D}{\longrightarrow}N(0,1)$. Now we are going to show second thing that we need, i.e. that $S_n^Z/\sqrt{n}\overset{a.s.}{\longrightarrow} 0$. Since $$\sum\limits_{n=1}^{\infty} \mathbf{P}(Z_n\ne0)=\sum\limits_{n=1}^{\infty} \frac{1}{n^2}<\infty$$ we have by Borel-Cantelli lemma that $\mathbf{P}(Z_n\ne0\text{ i.o.})=0$. From that we have that $S_n^Z/\sqrt{n}\overset{a.s.}{\longrightarrow} 0$ because almost surely for some $n_0\in\mathbb{N}$ and all $n\ge n_0$ $Z_n=0$. From these two facts we have that $S_n/\sqrt{n}=S_n^Y/\sqrt{n}+S_n^Z/\sqrt{n}\overset{D}{\longrightarrow}N(0,1)$ (using for example Slutsky's theorem).

Now it's time to show that Lindeberg's condition does not hold. I use this version of expression inside condition: $$ \frac{1}{s_n^2}\sum\limits_{k=1}^n \mathbf{E}[(X_k-m_k)^2;|X_k-m_k|\ge \varepsilon s_k],$$ where $m_k=\mathbf{E}X_k$ and $s_n^2=Var(X_1+\dots+X_n)$. Obviously, in our case $m_k=\mathbf{E}Y_k+\mathbf E Z_k=0$, and because $Var(Z_k)=1$ we have $s_n^2=2n$, $\forall n\ge 1$. Let's calculate

$\begin{align} \mathbf{E}[&(X_k-m_k)^2;|X_k-m_k|\ge \varepsilon s_k]= \mathbf{E}[(Y_k+Z_k)^2;|Y_k+Z_k|\ge \varepsilon s_k]\\ &=\mathbf{E}[(Y_k+k)^2;|Y_k+k|\ge \varepsilon s_k,Z_k=k]+ \mathbf{E}[(Y_k+0)^2;|Y_k+0|\ge \varepsilon s_k,Z_k=0]+\mathbf{E}[(Y_k-k)^2;|Y_k-k|\ge \varepsilon s_k,Z_k=-k]\\ &\overset{independece}{\ge} \mathbf{E}[(Y_k+k)^2;|Y_k+k|\ge \varepsilon s_k]\frac{1}{2k^2}+ \mathbf{E}[(Y_k-k)^2;|Y_k-k|\ge \varepsilon s_k]\frac{1}{2k^2}\\ &\overset{iden. distr.}{\ge} \mathbf{E}[(Y_1+k)^2;|Y_1+k|\ge \varepsilon s_k]\frac{1}{2k^2}+\mathbf{E}[(Y_1-k)^2;|Y_1-k|\ge \varepsilon s_k]\frac{1}{2k^2}.\tag{1} \end{align}$ Now from Fatou's lemma $\liminf\limits_{k\to\infty}\mathbf{E}[(Y_1\pm k)^2;|Y_1\pm k|\ge \varepsilon s_k]\frac{1}{2k^2}\ge\mathbf{E}\left[\liminf\limits_{k\to\infty}\left(\frac{(Y_1\pm k)^2}{2k^2}\mathbb{1}_{\{|Y_1\pm k|\ge \varepsilon \sqrt{2k}\}}\right)\right]=\mathbf E[\frac12]=\frac12$. So, for sufficiently large $n_0\in\mathbf N$ we have that $\forall k> n_0$ $\mathbf{E}[(Y_1\pm k)^2;|Y_1\pm k|\ge \varepsilon s_k]\frac{1}{2k^2}\ge \frac13$. Finally, using $(1)$ we have $\begin{align} \liminf_{n\to\infty}\frac{1}{s_n^2}\sum\limits_{k=1}^n \mathbf{E}[(X_k-m_k)^2;|X_k-m_k|\ge \varepsilon s_k]\ge\liminf_{n\to\infty}\frac{1}{2n}\left(\sum\limits_{k=n_0+1}^n \mathbf{E}[(Y_1+k)^2;|Y_1+k|\ge \varepsilon s_k]\frac{1}{2k^2}+\sum\limits_{k=n_0+1}^n \mathbf{E}[(Y_1-k)^2;|Y_1-k|\ge \varepsilon s_k]\frac{1}{2k^2}\right)\ge \liminf_{n\to\infty}\frac{1}{2n}\cdot \left((n-n_0)\cdot\frac13+(n-n_0)\cdot\frac13\right)=\frac13. \end{align}$ So we have showed that Lindeberg's condition does not hold.

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