Lecture 34 — CLT Without Characteristic Functions
We prove the classical CLT using Taylor expansion and a comparison between the sum of arbitrary iid mean-zero variables and the sum of iid normal variables.
1. Setup
Let $X_1,X_2,\dots$ be iid with
\[E[X]=0,\qquad E[X^2]=1,\qquad E[\vert X\vert ^3]<\infty.\]Let \(S_n = \frac{1}{\sqrt n}\sum_{k=1}^n X_k.\)
Let $Z_1,Z_2,\dots$ be iid $N(0,1)$ and denote \(Z_n = \frac{1}{\sqrt n}\sum_{k=1}^n Z_k.\)
We want to show: \(S_n \Rightarrow Z,\qquad Z\sim N(0,1).\)
Goal: for all bounded $C^3$ functions $f$,
\[E\big[f(S_n)\big] \to E\big[f(Z)\big]. \tag{1}\]By smoothing (Lecture 33), it suffices to check (1) for $f\in C_B^3(\mathbb R)$.
2. Comparison Decomposition
Let \(T_m = \frac{1}{\sqrt n}\Big( X_1+\cdots+X_{m-1} + Z_{m+1}+\cdots+Z_n \Big). \tag{2}\)
On page 1, the notes show the telescoping decomposition:
\[E[f(S_n)] - E[f(Z_n)] = \sum_{m=1}^n \Big( E f(T_m + X_m/\sqrt n) - E f(T_m + Z_m/\sqrt n) \Big). \tag{3}\]Each term replaces one $X_m$ by one $Z_m$ while holding all others fixed.
Let
\[I_m = \Big\vert E f(T_m + X_m/\sqrt n) - E f(T_m + Z_m/\sqrt n) \Big\vert .\]Then:
\[\vert E f(S_n) - E f(Z_n)\vert \le \sum_{m=1}^n I_m. \tag{4}\]We will bound each $I_m$ using Taylor’s theorem.
3. Taylor Expansion to Third Order
(Page 1 middle.)
For $f\in C^3$,
\[f(x+h) = f(x) + f'(x)h + \frac{f''(x)}{2} h^2 + \frac{f^{(3)}(\xi)}{6}h^3, \qquad \vert \xi-x\vert <\vert h\vert . \tag{5}\]Diagram on page 1 shows the geometric idea: the difference between the Taylor polynomial and $f$ is controlled by $\vert h\vert ^3$.
Let $B=T_m$.
Let $Y = X_m/\sqrt n$ and $W=Z_m/\sqrt n$.
Then:
\[f(B+Y) = f(B) + f'(B)Y + \frac{f''(B)}{2}Y^2 + R_Y,\] \[f(B+W) = f(B) + f'(B)W + \frac{f''(B)}{2}W^2 + R_W,\]where the remainders satisfy (page 1, formula labelled (B)):
\(\vert R_Y\vert \le C \vert Y\vert ^3, \qquad \vert R_W\vert \le C \vert W\vert ^3, \tag{6}\) with \(C = \sup_x \frac{\vert f^{(3)}(x)\vert }{6} < \infty.\)
4. The Comparison Lemma
(Page 1–2, boxed “Lemma”.)
Lemma.
Let $B,Y,W$ be real random variables such that:
- $E[Y]=E[W]$,
- $E[Y^2]=E[W^2]$,
- $B$ is independent of both $Y$ and $W$.
Then for $f\in C_B^3$:
\[\Big\vert E f(B+Y) - E f(B+W)\Big\vert \le C\, E(\vert Y\vert ^3 + \vert W\vert ^3). \tag{7}\]Proof sketch.
Because of independence:
\[E[f'(B)Y]=E[f'(B)]\,E[Y]=E[f'(B)]\,E[W]=E[f'(B)W],\] \[E\Big[\frac{f''(B)}{2}Y^2\Big] = E\Big[ \frac{f''(B)}{2}W^2 \Big].\]Thus the quadratic Taylor terms cancel exactly (page 1 bottom, page 2 top).
Only the cubic remainders remain, giving (7).
5. Apply the Lemma to Each Term $I_m$
Here:
\[Y = \frac{X_m}{\sqrt n}, \qquad W = \frac{Z_m}{\sqrt n}.\]Thus (page 2):
\[I_m \le C\, E\Big( \frac{\vert X_m\vert ^3}{n^{3/2}} + \frac{\vert Z_m\vert ^3}{n^{3/2}} \Big). \tag{8}\]Summing over $m=1,\dots,n$:
\[\sum_{m=1}^n I_m \le C\Big( \frac{n E\vert X\vert ^3}{n^{3/2}} + \frac{n E\vert Z\vert ^3}{n^{3/2}} \Big) = \frac{C(E\vert X\vert ^3 + E\vert Z\vert ^3)}{\sqrt n} \to 0. \tag{9}\]Since $E\vert Z\vert ^3<\infty$, this goes to zero.
This proves (4) tends to zero.
6. Control of “Large” $X_m$: Lindeberg–Type Step
On page 2–3 your notes add the truncation argument:
Split
\[E\vert X\vert ^3 = E\big[ \vert X\vert ^3 \mathbf{1}_{\vert X\vert \le \varepsilon\sqrt n} \big] + E\big[ \vert X\vert ^3 \mathbf{1}_{\vert X\vert > \varepsilon\sqrt n} \big].\]Using the comparison (page 2 bottom):
\[\vert Y\vert ^2\wedge \vert Y\vert ^3 = \frac{1}{n^{3/2}}\big( X^2\mathbf{1}_{\vert X\vert >\varepsilon\sqrt n} + \vert X\vert ^3\mathbf{1}_{\vert X\vert \le \varepsilon\sqrt n} \big).\]The second term vanishes because $\varepsilon\sqrt n\to\infty$. For the first:
\[E\big[X^2\,\mathbf{1}_{\vert X\vert >\varepsilon\sqrt n}\big] \to 0 \quad\text{by dominated convergence},\]since $E[X^2]=1$.
The notes on page 3 show this as:
“DCT … since $E[X^2]=1$.”
Hence the whole remainder → 0.
Thus the bound in (9) is in fact arbitrarily small for large $n$.
7. Conclusion
Putting everything together:
\[\vert E f(S_n) - E f(Z_n)\vert \to 0 \quad\text{for all } f\in C_B^3(\mathbb R).\]Because $Z_n\Rightarrow Z$ (by classical CLT for Gaussians), we have:
\[E f(Z_n)\to E f(Z).\]Therefore:
\[E f(S_n)\to E f(Z), \qquad \forall f\in C_B^3.\]By the smoothing argument from Lecture 33 (approximating general bounded continuous $f$ by smooth $f_\sigma$), this implies the full weak convergence:
\[\boxed{ \frac{1}{\sqrt n}\sum_{k=1}^n X_k \Rightarrow N(0,1). }\]Cheat–Sheet Summary — Lecture 34
- Avoid characteristic functions: use Taylor expansion and comparison variables.
- Replace each $X_m$ by $Z_m$ and sum the errors.
- Linear and quadratic Taylor terms cancel thanks to
$E[X_m]=E[Z_m]=0$, $E[X_m^2]=E[Z_m^2]=1$. - Remainder is controlled by third derivatives of $f$ and by $E\vert X\vert ^3$.
- Summed error is $O(1/\sqrt n)\to 0$.
- Finally, $Z_n\Rightarrow N(0,1)$ and smoothing gives the CLT.
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