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In john rice Mathematical Statistics and Data Analysis we find a proof about the central limit theorem.

Let $X_1, X_2,\ldots$ be a sequence of independent random variables having variance $\sigma^2$ and the commmon distribution function $F$ and moment generating function $M$ defined in a neighborhood of zero. Let

$$S_n = \sum_{i = 1}^n X_i$$

Then:

$$\lim_{n \longrightarrow \infty}P \left( \frac{S_n}{\sigma \sqrt n} \le x\right) = \Phi(x)$$

Proof:

Let $Z_n = \frac{S_n}{\sigma \sqrt{n}}$

Since $S_n$ is a sum of independent random variables:

$$M_{S_n}(t) = [M(t)]^n$$

$$M_{Z_n}(t) = \left[M\left(\frac t {\sigma \sqrt n}\right)\right]^n$$

$M(s)$ has a Taylor series expansion about zero.

$$M(s) = M(0) + s M'(0) + \frac{1}{2}s^2 M''(0) + \varepsilon_s$$

where $\frac{\varepsilon_s}{s^2} \longrightarrow 0$ as $s \longrightarrow 0$.

Why is this? And why is it useful? I thought that if the denominator of a fraction converges to zero you would essentially have an expression that have a zero value in the denominator which would produce an undefined value.

Since $E(X) = 0$, $M'(0) = 0$ and $M''(0) = \sigma^2$.

As $n \longrightarrow \infty, t/(\sigma \sqrt{n}) \longrightarrow 0$

The statement $n \longrightarrow \infty, t/(\sigma \sqrt{n}) \longrightarrow 0$ made, but why is it useful?

$$M\left(\frac{t}{\sigma \sqrt{n}}\right) = 1 + \frac{1}{2}\sigma^2 \left(\frac{t}{\sigma \sqrt{n}}\right)^2 + \varepsilon_n$$

I assume the above follows from the fact that $s = \frac{t}{\sigma \sqrt{n}}$ So you replace this into the taylor expansion, but why is $\varepsilon_{s}$ replaced with $\varepsilon_n$?

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  • $\begingroup$ consider $2^{-n}/2^{-n}$ denominator is going to zero but always equal to 1 for any integer n $\endgroup$
    – seanv507
    Oct 28, 2021 at 11:58
  • $\begingroup$ (Partial hint, hence comment) - For $\epsilon_s / s^2 \rightarrow 0$ as $s \rightarrow 0$, the reason this is the case will be more apparent if you expand the Taylor series of $M(s)$ for a few more terms and compare the latter $s$ terms in relation to $1/s^2$. $\endgroup$
    – B.Liu
    Oct 28, 2021 at 12:50
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    $\begingroup$ What do you mean by the latter $s$ terms? $\endgroup$ Oct 28, 2021 at 13:32
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    $\begingroup$ "if the denominator of a fraction converges to zero you would essentially have an expression that have a zero value in the denominator which would produce an undefined value." $${}$$ THAT IS FALSE. All derivatives are limits as the numerator and denominator approach $0.$ If that made them undefined, then all derivatives would be undefined. If the denominator approaches $0$ while the numerator approaches something else, then the limit is undefined, but if they BOTH approach $0,$ then in many cases the limit exists. $\endgroup$ Oct 29, 2021 at 0:02

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if the denominator of a fraction converges to zero you would essentially have an expression that have a zero value in the denominator which would produce an undefined value."

THAT IS FALSE. All derivatives are limits as the numerator and denominator approach $0.$ If that made them undefined, then all derivatives would be undefined. If the denominator approaches $0$ while the numerator approaches something else, then the limit is undefined, but if they BOTH approach $0,$ then in many cases the limit exists.

Note that if the numerator is $s^3$ and the denomoinator is $s^2,$ then certainly the fraction approaches $0$ as $s^2\to0.$

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  • $\begingroup$ Ok that clears up part of my confusion. Why is the fact that $\frac{\epsilon_{s}}{s^2} \longrightarrow 0$ as $s \longrightarrow 0$ even useful in the proof? $\endgroup$ Oct 30, 2021 at 8:52
  • $\begingroup$ @DanielDeWet : Recall from calculus that $$ \lim_{n\to\infty} \left( 1 + \frac a n \right)^n = e^a,$$ so in particular, $$ \lim_{n\to\infty} \left( 1+\frac{t^2}{2n} \right)^n = e^{t^2/2}.$$ But what you have is $$ \lim_{n\to\infty} \left( 1+\frac{t^2}{2n} + \cdots \right)^n.$$ The idea is to show that the terms appearing where those dots are do not alter the bottom line. $\qquad$ $\endgroup$ Oct 30, 2021 at 10:29
  • $\begingroup$ What do you mean the terms appearing where those dots are do not alter the bottom line? What is this bottom line you are talking about? $\endgroup$ Oct 31, 2021 at 7:39
  • $\begingroup$ @DanielDeWet : By "the bottom line" I meant the expression to the right of the "equals" sign, i.e. $e^{t^2/2}. \qquad$ $\endgroup$ Oct 31, 2021 at 14:30
  • $\begingroup$ Sorry for all of the questions but I'm not sure what you mean by "altering" the expression $e^{\frac{t^{2}}{2}}$ $\endgroup$ Oct 31, 2021 at 18:29

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