Prove : $\frac{X_1+…+X_{n-1}-\log n }{\sqrt{\log n} }\rightarrow N(0,1)$
where $X_n\sim \operatorname{Ber}({\lambda}_n)$ and $\lambda _n = \log \frac{n+1}{n}$
I tried using a similar proof to CLT with characteristic function but didnt manage to get there. Here is my attempt:
$$ S_n = X_1 +…+X_n$$
$$\varphi_{\frac{S_{n-1}-\log n }{\sqrt{\log n} }}(t)=\varphi_{S_{n-1}-\log n}(\frac{t}{\sqrt{\log n}})$$
$$\varphi_{X_i-\lambda _i}=(\lambda _ie^{it}+1-\lambda _i)e^{-it\lambda _i}$$
So I get a multiplacation of $\varphi_{X_i-\lambda _i}(t/\sqrt(\log n))$ and I don't see why it should go to $e^{\frac{-t^2}{2}}$
I might add that I don't know any representation of e as a form of an infinite changing multiplication, so I don't see how this way would lead me to the solution.
Thank you,
Best Answer
Let $Y_{n}=X_{n}-\log(\frac{n+1}{n})$ . Then $\operatorname{Var}(Y_{n})=\operatorname{Var}(X_{n})=(1-\log(\frac{n+1}{n}))\log(\frac{n+1}{n})$.
Then $\displaystyle \operatorname{Var}(\sum_{k=1}^{n}Y_{k})=\sum_{k=1}^{n}\operatorname{Var}(Y_{k})=\sum_{k=1}^{n}(1-\log(\frac{k+1}{k}))\log(\frac{k+1}{k})$
Now it can be proven by applying Stolz-Cesaro's Theorem that $$\frac{\sum_{k=1}^{n}(1-\log(\frac{k+1}{k}))\log(\frac{k+1}{k})}{\sum_{k=1}^{n}\log(\frac{k+1}{k})}\to 1$$
Or just observe that $$\frac{\sum_{k=1}^{n}\log^{2}(1+1/k)}{\log(n+1)}\to 0$$ as $\sum_{k=1}^{n}\log^{2}(1+1/k)$ is a convergent series by comparison with $\sum_n\frac{1}{n^2}$ to show that the above limit tends to 1.
This gives that $\operatorname{Var}(\sum_{k=1}^{n}Y_{k})\sim\sum_{k=1}^{n}\log(\frac{k+1}{k})= \log(n+1)$
Thus, if we can verify Lindeberg's condition that $\displaystyle\frac{1}{\sqrt{\operatorname{Var}(\sum_{k=1}^{n}Y_{n})}^{2}}\sum_{k=1}^{n}E(Y_{k}^{2}\mathbf{1}_{|Y_{k}|>t\sqrt{\operatorname{Var}(\sum_{k=1}^{n}Y_{k})}})\to 0$ for all $t>0$ , then you will have by the Lindeberg-Feller CLT that $\frac{\sum_{k=1}^{n}Y_{n}}{\sqrt{\log(n)}}\xrightarrow{d} N(0,1)$ by an application of Slutsky's theorem.
So we try and verify Liapunov's condition i.e. show that there exists $\delta>0$ such that $\displaystyle\dfrac{\sum_{k=1}^{n}E(|X_{k}|^{2+\delta})}{\sqrt{\operatorname{Var}(\sum_{k=1}^{n}Y_{k})}^{2+\delta}}\to 0$
Now, $(\operatorname{Var}(\sum_{k=1}^{n}Y_{k}))^{2}\sim\log^{2}(n+1)$.
So we show that for $\delta=2$, we have
$$\frac{\sum_{k=1}^{n}E(|Y_{k}|^{4})}{\sqrt{\operatorname{Var}(\sum_{k=1}^{n}Y_{k})}^{4}}\to 0$$
Thus, Liapunov's Condition will be satisfied. Thus, by Lindeberg-Feller CLT, you have that $\frac{\sum_{k=1}^{n}Y_{n}}{\sqrt{\operatorname{Var}(\sum_{k=1}^{n}Y_{k})}}\xrightarrow{d}N(0,1)$ which is what we wanted to show. Now as $\operatorname{Var}(\sum_{k=1}^{n}Y_{k})\sim\log(n+1)$ , we have by Slutsky's theorem that $\frac{\sum_{k=1}^{n}Y_{n}}{\sqrt{\log(n)}}\xrightarrow{d}N(0,1) $
$ \blacksquare$
Now, to show that $$\frac{\sum_{k=1}^{n}E(|Y_{k}|^{4})}{\sqrt{\operatorname{Var}(\sum_{k=1}^{n}Y_{k})}^{4}}\sim\frac{\sum_{k=1}^{n}E(|Y_{k}|^{4})}{\log^{2}(n+1)}\to 0$$.
We first see that $$E|Y_{k}|^{4}=(1-\log(\frac{k+1}{k}))^{4}\log(\frac{k+1}{k})+\log^{4}(\frac{k+1}{k})(1-\log(\frac{k+1}{k}))$$.
But again we have that $$\dfrac{\sum_{k=1}^{n}\log^{m}(1+\frac{1}{k})}{\log^{2}(n+1)}\to 0$$
for all $m> 1$ as the series $\sum_{k}\log^{m}(1+\frac{1}{k})$ is convergent by comparison with $\sum_{k}\frac{1}{k^{m}}$.
Thus by expanding the numerator, we will only be left with $$\displaystyle\frac{2\cdot\sum_{k=1}^{n}\log(\frac{k+1}{k})}{\log^{2}(n+1)}=\frac{2\cdot\log(n+1)}{\log^{2}(n+1)}\to 0$$ Thus, we have shown Liapunov's condition.
$ \blacksquare$
For references on Lindegerg-Feller CLT,Liapunov's condition etc see Sidney Resnick, A Probability Path section 9.8 page 314.