# Steps in CLT proof unclear

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 $$\varepsilon_n$$?

• consider $2^{-n}/2^{-n}$ denominator is going to zero but always equal to 1 for any integer n Oct 28, 2021 at 11:58
• (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 $1/s^2$. Oct 28, 2021 at 12:50
• What do you mean by the latter $s$ terms? Oct 28, 2021 at 13:32
• "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. Oct 29, 2021 at 0:02

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 $$s^2\to0.$$
• 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? Oct 30, 2021 at 8:52
• @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. $\qquad$ Oct 30, 2021 at 10:29
• @DanielDeWet : By "the bottom line" I meant the expression to the right of the "equals" sign, i.e. $e^{t^2/2}. \qquad$ Oct 31, 2021 at 14:30
• Sorry for all of the questions but I'm not sure what you mean by "altering" the expression $e^{\frac{t^{2}}{2}}$ Oct 31, 2021 at 18:29