# What is the connection between binomial and poisson distribution? [duplicate]

Binomial is counting success in $n$ trails. Poisson is also for counting events happening. What is the connection between them?

I know is that when sample size is large both can be approximated with normal.

But how are they similar or different? I learned in class their PMF are different but not really understand where does such PMF comes from and intuitive way of explaining them.

• There are a number of potential connections between the two (or indeed between other pairs of distributions); for example one is sometimes used as an approximation to the other. Can you clarify what sort of thing you're seeking? Feb 8, 2017 at 22:55
• @Glen_b I think I do not know what I do not know.. Just feel they are very similar, but no intuitive feeling about why they are similar or different. What I learned in class is just some formulas. Feb 8, 2017 at 22:57
• This question is answered in the original post at stats.stackexchange.com/questions/167212, and its answers contain proofs of the proposition that the Poisson is a limiting version of the Binomial.
– whuber
Feb 8, 2017 at 23:26

They are strongly related to each other. For $n \rightarrow \infty,\ p \rightarrow 0$ such that $np \rightarrow \lambda\$we have

$$P_{Bin(n,p)}(k) = \binom{n}{k}p^k(1-p)^{n-k} \eqsim \frac{\lambda^k}{k!}\exp^{-\lambda} = P_{Poiss(\lambda)}(k)$$

So Poisson distribution is a limiting binomial distribution with $\lambda$ being the average rate (that is $np$) of certain event occuring

• Thanks! For example, can I say binomial with n=1000 p=0.1 is almost equal to poission lambda=100 ? Feb 8, 2017 at 22:56
• I don't know precise rate of convergence w.r.t. $n$, but it will be a pretty good approximation Feb 8, 2017 at 23:31
• @hxd It depends on how close you regard as close, but for those numbers take a look at the cdfs here (binomial cdf is black, Poisson is red) Feb 9, 2017 at 0:14

One can get the Poisson from Binomial by taking limit, and the Binomial from Poisson by conditioning. More precisely, we have the following.

• If $X\sim\text{Pois}(\lambda_1)$, $Y\sim\text{Pois}(\lambda_2)$ are independent random variables, then the distribution of $X$ given $X+Y=n$ is $X_{\text{cond}}\sim\text{Bin}(n,\lambda_1/(\lambda_1+\lambda_2))$
• If $X\sim\text{Bin}(n,p)$, and if $n\to\infty$, $p\to 0$ such that $np\to\lambda$, then $\mathbb{P}(X=k)\to\mathrm{e}^{-\lambda}\dfrac{\lambda^k}{k!}$