# Proving IID Central Limit Theorem using Lindeberg Conditions.

The goal is to prove the IID Central Limit Theorem through Lindeberg's Condition.

Suppose that $$X_1,X_2,\ldots\displaystyle\sim\text{i.i.d.}$$ with $$E[X_i]=\mu$$ and $$Var[X_i]=\sigma^2<\infty$$.

Let $$Y_i=X_i-\mu$$ and $$s_n^2=\sum_{i=1}^{n}Var[Y_i]=n\sigma^2$$.

Prove that $$Z_n:=\frac{\sum_{k=1}^{n}(X_k-\mu)}{s_n}\rightarrow$$N$$(0,1)$$ in distribution.

Lindeberg's condition is as follows:

If the following holds:

$$\displaystyle\lim_{n\rightarrow\infty}\frac{1}{s_n^2}\sum_{i=1}^{n}E\big[Y_i^2\cdot\mathbb{I}_{( \ |Y_i|\geq\epsilon\cdot s_n \ )}\big]=0$$ for all $$\epsilon>0$$

Then $$Z_n\rightarrow$$N$$(0,1)$$ in distribution.

So going right to lindbergs condition:

$$\displaystyle\lim_{n\rightarrow\infty}\frac{1}{n\sigma^2}\sum_{i=1}^{n}E\big[Y_i^2\cdot\mathbb{I}_{( \ |Y_i|\geq\epsilon\cdot \sigma\sqrt{n} \ )}\big]=\displaystyle\lim_{n\rightarrow\infty}\frac{1}{\sigma^2}E\big[Y_i^2\cdot\mathbb{I}_{( \ |Y_i|\geq\epsilon\cdot \sigma\sqrt{n} \ )}\big]$$

Now at this point I do know I can use Lebesgue Dominated Convergence theorem due to the following:

$$|Y_i^2\cdot\mathbb{I}_{( \ |Y_i|\geq\epsilon\cdot \sigma\sqrt{n} \ )}|=Y_i^2\cdot\mathbb{I}_{( \ |Y_i|\geq\epsilon\cdot \sigma\sqrt{n} \ )}\leq Y_i^2$$ and $$Y_i^2$$ is integrable as $$E[Y_i^2]=\sigma^2<\infty$$

This means:

$$\displaystyle\lim_{n\rightarrow\infty}\frac{1}{\sigma^2}E\big[Y_i^2\cdot\mathbb{I}_{( \ |Y_i|\geq\epsilon\cdot \sigma\sqrt{n} \ )}\big]=\frac{1}{\sigma^2}E\big[\displaystyle\lim_{n\rightarrow\infty}Y_i^2\cdot\mathbb{I}_{( \ |Y_i|\geq\epsilon\cdot \sigma\sqrt{n} \ )}\big]$$

Now I do not understand how this above is zero, and that's where I am stuck.

I do know that by Markov's inequality:

$$P(|Y_i|\geq\epsilon\sigma\sqrt{n})\leq\frac{E\big[|Y_i|\big]}{\epsilon\sigma\sqrt{n}}\rightarrow0$$ as $$n\rightarrow\infty$$ as $$E[|Y_i|]<\infty$$, and so the probability of this event becomes zero.

Any help with understand this final step would be much appreciated!

All you need is $$EY_1^{2} I_{\{|Y_1| >\epsilon \sigma \sqrt n\}} \to 0$$ as $$n \to \infty$$ and this follows from DCT. [$$Y_1^{2} I_{\{|Y_1| >\epsilon \sigma \sqrt n\}}$$ is dominated by $$Y_1^{2}$$ which is integrable. Of course, the events $$\{|Y_1| >\epsilon \sigma \sqrt n\}$$ decrease to empty set so $$Y_1^{2} I_{\{|Y_1| >\epsilon \sigma \sqrt n\}} \to 0$$ almost surely. ].
• I think MCT will also work. Because by MCT we have, $\mathbb E \left [Y_1^2 I_{\left \{\left \lvert Y_1 \right \rvert \leq \varepsilon \sigma \sqrt {n} \right \}} \right ] \bigg\uparrow\ \mathbb E \left [Y_1^2 \right ]$ and hence the desired result follows from the decomposition $$\mathbb E \left [Y_1^2 \right ] = \mathbb E \left [Y_1^2 I_{\left \{\left \lvert Y_1 \right \rvert \leq \varepsilon \sigma \sqrt {n} \right \}} \right ] + \mathbb E \left [Y_1^2 I_{\left \{\left \lvert Y_1 \right \rvert \gt \varepsilon \sigma \sqrt {n} \right \}} \right ].$$ Commented Apr 24 at 16:30