# Show the Binomial distribution approaches a Normal distribution (using characteristic function)

Let $$X_n = Bin(n,p)$$. We fix $$p$$, and we want to show that as $$n \to \infty$$, $$\frac{X_n-np}{\sqrt{np(1-p)}}$$ converges to $$N(0,1)$$ in distribution. And I want to show this using characteristic functions, not using the CLT.

The characteristic function of $$X_n$$ is $$(pe^{it} + (1-p))^n$$, so the CF of $$\frac{X_n-np}{\sqrt{np(1-p)}}$$ is $$\left(pe^{it \frac{1}{\sqrt{np(1-p)}}}+(1-p)\right)^n\ e^{\frac{-it np}{\sqrt{np(1-p}}}.$$

I'm having trouble finding the limit as $$n \to \infty$$, as I don't think $$e^{\frac{-it np}{\sqrt{np(1-p}}}$$ has a limit as $$n \to \infty$$, and I've tried putting $$(pe^{it \frac{1}{\sqrt{np(1-p)}}}+(1-p))^n$$ into the form $$\left(1+\frac{a_n}{n}\right)^n$$ with $$a_n \to a$$, so we can find the limit to be $$e^a$$, but that has been to no avail.

Thanks.

Your idea is good. But it's easier to take logarithms and just look at what results. You will eventually want to expand the log characteristic function (aka the cumulant generating function) as a Taylor series in $$t$$ and keep track of the powers of $$n.$$ You will see this technique again and again in asymptotic applications, enough so that eventually you will be able to do it mentally.

In the following, all you need to know are two basic Taylor series:

$$e^z = 1 + z + \frac{z^2}{2!} + O(z^3),$$

for any complex number $$z,$$ which we will use repeatedly in the form

$$e^z - 1 = z + \frac{z^2}{2!} + O(z^3);$$

and

$$\log(1 + z) = z - \frac{z^2}{2} + O(z^3)$$

for any complex number $$z$$ with $$|z|\lt 1.$$

The simple algebraic manipulation in the first line below is the heart of the matter. It corresponds to the approach you intuitively wanted to take.

Because $$0\le p \lt 1$$ implies $$|p\left(e^{it}-1\right)|\lt 1$$ for sufficiently small real values of $$t,$$

\begin{aligned} \log\left(pe^{it} + 1-p)\right)^n &= n \log \left(1 + p\left(e^{it}-1\right)\right)\\ &= n\left(p\left(e^{it}-1\right) - \frac{p^2}{2} \left(e^{it}-1\right)^2 + O(t^3)\right) \end{aligned}

Plugging in $$e^{it}-1 = it + (it)^2/2! + O(t^3)$$ and collecting powers of $$t$$ (which is a purely mechanical process) gives

\begin{aligned} \log\left(pe^{it} + 1-p)\right)^n &=n\left(pit + p\frac{(it)^2}{2!} - \frac{p^2}{2}\left(it\right)^2 + O(t^3)\right)\\ &=(np)\frac{it}{1!} + np(1-p)\frac{(it)^2}{2!} + nO(t^3). \end{aligned}

Recentering (by the expectation of the sum, $$np$$) will zero out the first term, while rescaling (by the variance of the sum, $$np(1-p)$$) divides $$t^2$$ by $$np(1-p)$$ everywhere. The latter means you may replace any power $$t^k$$ by $$t^k (np(1-p))^{-k/2}.$$ Thus, bearing in mind $$p$$ is a constant, you may absorb functions of it in the $$O(t^3)$$ term to obtain

$$\frac{(it)^2}{2!} + \left(np(1-p)\right)^{-3/2} nO(t^3) = \frac{(it)^2}{2!} + O(n^{-1/2})O(t^3).$$

This tells you all the terms past $$t^2$$ are growing vanishingly small as $$n$$ increases. (It says more than that: the $$t^3$$ term, proportional to the skewness, behaves like $$n^{-1/2};$$ the $$t^4$$ term, proportional to the kurtosis, shrinks even faster like $$n^{-1},$$ and so on.)

In the limit you're left with $$(it)^2/2,$$ which is the logarithm of the standard Normal characteristic function. Because the log is continuous for $$t\ne 0,$$ the c.f. converges to the Normal c.f. and the Levy Continuity Theorem does the rest.

Apply the expension of exponential and logrithm. The CF of $$(X_n - np) / \sqrt{np(1-p)}$$ is \begin{align} f(t) &= \left(pe^{it\frac{1}{\sqrt{np(1-p)}}} -p + 1 \right)^ne^{-it\frac{np}{\sqrt{np(1-p)}}} \\ &= \exp\{n\left(\log\left(pe^{it\frac{1}{\sqrt{np(1-p)}}} -p + 1\right)\right) -it\frac{np}{\sqrt{np(1-p)}}\} \\ &\sim\exp\{n\left(\log\left(p{it\frac{1}{\sqrt{np(1-p)}}} - \frac12p{t^2\frac{1}{{np(1-p)}}}+o(n^{-3/2})+ 1\right)\right) -it\frac{np}{\sqrt{np(1-p)}}\} \\ &\sim \exp\{ n\left( pit\frac1{\sqrt{np(1-p)}}- \frac12p{t^2\frac{1}{{np(1-p)}}}+\frac12\frac{p^2t^2}{np(1-p)} + o(n^{-3/2})\right) -it\frac{np}{\sqrt{np(1-p)}}\} \\ &\to \exp\{-\frac12t^2\} \end{align}