Does a Binomial converge to Poisson or Normal? I have read the answer here. Here the distinction is that

*

*If $n\to\infty$ and $p\to0$ while $np$ approaches some positive number $\lambda,$ then the binomial distribution approaches a Poisson distribution with expected value $\lambda.$


*If $n\to\infty$ as $p$ stays fixed, and $X\sim\operatorname{Binomial}(n,p)$ then the distribution of $(X-np)/\sqrt{np(1-p)}$ approaches the standard normal distribution, i.e. the normal distribution with expected value $0$ and standard deviation $1.$
I am finding it hard to wrap my head around this. In the derivation of the central limit theorem nowhere is $p$ taken into consideration. So even if $p$ is very small, according to CLT the standardized Binomial should limit to a standard normal. And the two limiting behaviors are both for $n \to \infty$
Please help me understand this concept a bit more. How can CLT not be valid when $p$ is really small?
 A: The difficulty disappears when you are careful in formulating the limits. In the first case, $p$ is not constant, so it would be more precise to write it as $p_n$, as $p$ varies with $n$. We can write $n \cdot p_n \to \lambda>0$ another way as $p_n \sim \lambda/n$, where $\sim$ means that the quotient between the two sides converges to unity with $n \to\infty$.
For the second case, $p$ is constant, and however small, when $n$ is large enough, $np$ is no longer small. The CLT is still valid when $p>0$ is small and constant.
To understand this better, you could try to use the CLT for IID variables in the first case. Write the binomial out as a sum of $n$ IID Bernoulli variables, as $X_n= B_1 + \dotsm + B_n$. Now check the assumptions of the CLT. You will find that the $B_1, B_2, \dotso, B_n$ all must have the same distribution, and that distribution should not depend on $n$. Is that the case?
A: One can write the CLT as:
$$
\frac{\sum_{i=1}^{n}X_i - n\mu}{\sigma\sqrt{n}} \stackrel{d}{\to} N(0,1)
$$
If we consider each $X_i$ here as a Bernoulli random variable that are independent and identically distributed, recalling that the mean and variance of a Bernoulli random variable are $p$ and $p(1-p)$ respectively, we can rewrite this as
$$
\frac{\sum_{i=1}^{n}X_i - np}{\sqrt{np(1-p)}} \stackrel{d}{\to} N(0,1)
$$
And then note that a binomial is simply a sum of n Bernoulli random variable. That is that for  $X = X_1 + \dots + X_n$ where each $X_i \sim Bernoulli(p)$ we have  $X \sim Binomial(n,p)$. So we can finally get the form of the CLT given in your question.
$$
\frac{X - np}{\sqrt{np(1-p)}} \stackrel{d}{\to} N(0,1)
$$
This is true even when p is very small, so long as the bernoulli random variables are identically distributed. In the first case where the binomial converges to a poisson distribution, p is growing smaller as n goes to infinity, and so the requirement that the random variables be identically distributed is not satisfied.
