# 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?

• On first pass, my instinct is that ambiguity arises here due to insufficient precision in how the semantics of "approaches" is interpreted mathematically. Apr 10, 2021 at 12:17
• The answer you reference clearly articulates the fundamental point, beginning with "It is sloppy to say something approaches something depending on n as n→∞, unless it is precisely defined and not meant literally...."
– whuber
Apr 10, 2021 at 12:25
• You appear to be making a common mistake about the central limit theorem: stats.stackexchange.com/questions/473455/…. The central limit theorem is about a sampling distribution, not about the original population.
– Dave
Apr 10, 2021 at 12:35

## 2 Answers

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?

• If np is small but not infinitesimal (smaller than 4), the binomial distribution is asymmetrical. Even if n very very large I don't understand how it'll look like bell curve. So my question is why CLT fails here? Apr 4, 2022 at 2:02
• @AhmedAbdullah: If $n$ is large and $np$ is, say 4 or 2 or 1 ... then the Poisson fit better than the normal. The CLT does not fail, but willl maybe need a very large $n$ Apr 4, 2022 at 2:17

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.