Showing that Lindeberg condition does not hold

Question

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 $$S_n/\sqrt{n} \rightarrow N(0,1).$$

Answer 1

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.

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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.

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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 $$\frac{S_n}{\sqrt{n}} \to N(0,2)$$