Question about the proof of Central Limit Theorem

Question

My instructor proved the central limit theorem using the characteristic function. I think the proof is a standard one because I found basically the same proof in wikipedia.

So for i.i.d. ${X_1, X_2,\cdots, X_n}$ with $E[X_i]=\mu$ and $\text{Var}(X_i)=\sigma^2<+\infty$, he then defined $$Y_i=\frac{X_i-\mu}{\sqrt{n}}$$

Then $E[Y_i]=0$ and $\text{Var}(Y_i)=\sigma^2/n$.

So the characteristic function of $Y_i$ is $$\phi(k)=1-\frac{k^2\sigma^2}{2n}+o\left(\frac{1}{n}\right)$$

Then he further define $$Z=Y_1+Y_2+\cdots+Y_n$$ and so the characteristic function of $Z$ is $$\Phi(k)=\phi(k)^n=\left(1-\frac{k^2\sigma^2}{2n}+o\left(\frac{1}{n}\right)\right)^n$$

Then he claimed that when $n\rightarrow\infty$ $$\Phi(k)=\lim_{n\rightarrow\infty}\left(1-\frac{k^2\sigma^2}{2n}+o\left(\frac{1}{n}\right)\right)^n=\exp\left(-\frac{k^2\sigma^2}{2}\right)$$

Performing the inverse Fourier transform we get $$P(Z)=\frac{1}{\sqrt{2\pi}\sigma}\exp\left(-\frac{Z^2}{2\sigma^2}\right)$$

Then finally defining $$\tilde{X}=\frac{Z}{\sqrt{n}}+\mu=\frac{X_1+X_2+\cdots+X_n}{n}$$ it can be shown that $$P(\tilde{X})=\frac{1}{\sqrt{2\pi}(\sigma/\sqrt{n})}\exp\left(-\frac{(\tilde{X}-\mu)^2}{2(\sigma/\sqrt{n})^2}\right)$$

I found slightly different proof in wiki but the basic idea is the same.

Now where I have problem is the step $$\lim_{n\rightarrow\infty}\left(1-\frac{k^2\sigma^2}{2n}+o\left(\frac{1}{n}\right)\right)^n=\exp\left(-\frac{k^2\sigma^2}{2}\right)$$

I know that I cannot stupidly put the limit inside the bracket for otherwise I'd get the answer $1^n=1$. But why do I know I cannot do it to the $O(1/n)$ term, but can do it to the $o(1/n)$ terms?

Are there something I am missing?

Answer 1

When we take the logarithm, we have to be careful, since the little "o" is complex (at least we have to choose a good determination). Alternatively, we can use and show the fact that if $\left(z_n\right)_{n\geqslant 1}$ is a sequence of complex numbers which converges to $z$, then $$\lim_{n\to +\infty}\left(1+\frac{z_n}n\right)^n=e^z.$$

To see this, notice that if $(x_n)_{n\geqslant 1}$ and $(y_n)_{n\geqslant 1}$ are two sequences of complex numbers such that $\max\left\{|x_n|,|y_n|\right\}\leqslant R$, then $$\tag{*}\left|\prod_{j=1}^nx_j-\prod_{j=1}^ny_j\right|\leqslant R^{n-1}\sum_{j=1}^n\left|x_j-y_j\right|$$ (this can be done by induction).

Now, define $x_n:=\left(1+\frac{z_n}n\right)^n$ and $y_n:=\left(1+\frac{z}n\right)^n$. By (*) with $R:=1+\sup_{j\geqslant 1}|z_j|/n$, we get that $$\left|\left(1+\frac{z_n}n\right)^n-\left(1+\frac{z}n\right)^n\right| \leqslant \left(1+\sup_{j\geqslant 1}|z_j|/n\right)^{n-1}\left|z_n-z\right|,$$ which gives what we wanted, as $\left(1+\sup_{j\geqslant 1}|z_j|/n\right)^{n}$ converges to $\exp\left(\sup_{j\geqslant 1}|z_j|\right)$.

Answer 2

To prove that if a complex sequence $$\lim_{n\rightarrow+\infty}z_n=z,$$ then $$\lim_{n\rightarrow+\infty}\left(1+\frac{z_n}{n}\right)^n=e^z$$

Proof:

Let $(x_j)_{n\ge j \ge 1}$ and $(y_j)_{n\ge j \ge 1}$ be two sequences of complex numbers.

Let $|x_j| \le R$ and $|y_j| \le R$ for all $1 \le j \le n$.

Then $$\bigg|\prod^n_{j-1}x_j-\prod^n_{j=1}y_j\bigg|\le R^{n-1}\sum_{j=1}^n|x_j-y_j|.$$

Let's prove the above by induction.

For $n=1$, it is obviously true.

Suppose $$\bigg|\prod^n_{j-1}x_j-\prod^n_{j=1}y_j\bigg|\le R^{n-1}\sum_{j=1}^n|x_j-y_j|$$

Then $$\bigg|\prod^{n+1}_{j-1}x_j-\prod^{n+1}_{j=1}y_j\bigg|=\frac{1}{2}\bigg|\left(x_{n+1}-y_{n+1}\right)\left(\prod_{j=1}^nx_j+\prod_{j=1}^ny_j\right)+\left(x_{n+1}+y_{n+1}\right)\left(\prod_{j=1}^nx_j-\prod_{j=1}^ny_j\right)\bigg|$$ $$\le \frac{1}{2}|x_{n+1}-y_{n+1}|\bigg|\prod_{j=1}^nx_j+\prod_{j=1}^ny_j\bigg|+\frac{1}{2}|x_{n+1}+y_{n+1}|R^{n-1}\sum_{j=1}^n|x_j-y_j|$$ $$\le \frac{1}{2}|x_{n+1}-y_{n+1}|2R^n+\frac{1}{2}2RR^{n-1}\sum_{j=1}^n|x_j-y_j|$$ $$= R^n\sum_{j=1}^{n+1}|x_j-y_j|$$

Now for fixed $n$, let $$x_j=1+\frac{z_n}{n}$$ $$y_j=1+\frac{z}{n}$$ $$R=1+\frac{\sup_{i\ge 1}|z_i|}{n}$$

Then we have $$\bigg|\left(1+\frac{z_n}{n}\right)^n-\left(1+\frac{z}{n}\right)^n\bigg|\le \left(1+\frac{\sup_{i\ge 1}|z_i|}{n}\right)^{n-1}|z_n-z|$$

Take limit $n \rightarrow +\infty$, we have $$\lim_{n\rightarrow+\infty}\bigg|\left(1+\frac{z_n}{n}\right)^n-\left(1+\frac{z}{n}\right)^n\bigg| \le \lim_{n\rightarrow+\infty}\left(1+\frac{\sup_{i\ge 1}|z_i|}{n}\right)^{n-1}|z_n-z|$$ $$=\lim_{n\rightarrow+\infty}\left(1+\frac{\sup_{i\ge 1}|z_i|}{n}\right)^n \left(1+\frac{\sup_{i\ge 1}|z_i|}{n}\right)^{-1}|z_n-z|$$ $$=\exp\left(\sup_{i\ge 1}|z_i|\right) \times 1 \times 0$$ $$=0$$

Hence $$\lim_{n\rightarrow+\infty}\left(1+\frac{z_n}{n}\right)^n=\lim_{n\rightarrow+\infty}\left(1+\frac{z}{n}\right)^n$$ $$=e^z$$