# Does a Binomial converge to Poisson or Normal?

• 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. – microhaus Apr 10 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 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 at 12:35

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?