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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!

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1 Answer 1

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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. ].

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  • $\begingroup$ I am okay there, I just don't understand how after LDCT brings the limit in that the integral becomes zero. I also understand that the probability of the event on the indicator function goes to zero. I'm stuck trying to connect how this gives the result of the expectation being zero. $\endgroup$
    – OGV
    Commented Feb 14, 2019 at 5:51
  • $\begingroup$ @OGV You are making this complicated. You don't need any inequality for this. I hope the extra line I have added to my answer makes things clear to you. I will be glad to help you if need some more clarifications. $\endgroup$
    – Kavi Rama Murthy
    Commented Feb 14, 2019 at 6:02
  • $\begingroup$ 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 $$\mathbb E \left [Y_1^2 \right ] = \mathbb E \left [Y_1^2 I_{\left \{\left \lvert Y_1 \right \rvert \leq \varepsilon \sigma \sqrt {n} \right \}} \right ] + \mathbb E \left [Y_1^2 I_{\left \{\left \lvert Y_1 \right \rvert \gt \varepsilon \sigma \sqrt {n} \right \}} \right ].$$ $\endgroup$
    – Anacardium
    Commented Apr 24 at 16:30

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