Approximation of $\operatorname{Poisson}(\lambda)$ by Binomial I don't really understand the approximation of $\operatorname{Poisson}(\lambda)$ distributions by binomial distributions (law of rare events?). For example if I consider $X\sim \operatorname{Poissson}(1)$, its distribution is not symmetric as drawed in wikipedia (orange curve). However, what I understood is that if we take say $n=1000$ ("big") and $p=0.001$ (such that $np=1$) and we consider the binomial distribution $Y\sim B(n,p)$ we will have that the distribution of $X$ and the distribution of $Y$ are almost the same. But $X$ has a distribution which is not symmetric and $Y$ has a distribution that is symmetric which is confusing me.
 A: $\text{Bin}(1000;0.001)$ is not symmetric.
It is not necessary to take n=1000. This is for $n=10$ and $p=0.1$
these are the resulting pmf: the blue one is the binomial while the green one is the poisson (1)

As you can see the probabilities are very close one each other
A: The other answers give data to support the claim.  By contrast, I'll supply some intuition for why you should trust that a binomial distribution can be a good approximation of a Poisson distribution when you have enough trials.
Let's think about an online global company whose orders per day can be modeled by a Poisson distribution with $\lambda=1$.   So, more or less one order per day.  What would be wrong with modeling a day's sales with an hourly geometric distribution with $n=24$ and $p=\frac1{24}$?  It would have the right mean, but it wouldn't model the possibility that the company got two orders in the same hour.  But getting two orders in an hour when you only expect one order per day is going to be a pretty rare event, so we shouldn't worry too much.
But, if we did worry, we could switch to a geometric model of sales per minute by setting $n=1440$ and $p=\frac1{1440}$.  Now, we're only failing to model the possibility that the company gets two orders in the same minute when we still only expect one order per day.  So the differences between this distribution and the Poisson distrition with $\lambda=pn=1$ would be virtually indistinguishable.
A: A distribution like $B(1000, 0.001)$ isn't anywhere near being symmetric.  Binomial distributions are roughly symmetric if $n$ is large and $p$ isn't too close to 0 or 1.  If you're looking at the drawings in the Wikipedia article on the binomial distribution, those will be misleading.
Here are the Poisson distributions with $\lambda = 1, 5, 10$:

And here are the Binomial(1000, p) distributions with $p = 0.001, 0.005, 0.01$:

(I did actually generate separate pictures for these, but I didn't need to - I could have used the same picture for both, because it's a very good approximation.) You can see that the Poisson gets more symmetric with the larger values of $\lambda$, and the same happens for the binomial.
A: Binomial distribution with $n = 1000, p = 0.001$ is not symmetrical and is reasonably close to Poisson with mean $\lambda=1.$
From R, here are probabilities rounded to five places for values from 0 through 10. Neither distribution has significant probability for larger values, [Ignore 'line numbers' in brackets.]
x = 0:10
pdf.b = round(dbinom(x, 1000, .001), 5)
pdf.p = round(dpois(x, 1), 5)
cbind(x, pdf.b, pdf.p)
       x   pdf.b   pdf.p
 [1,]  0 0.36770 0.36788
 [2,]  1 0.36806 0.36788
 [3,]  2 0.18403 0.18394
 [4,]  3 0.06128 0.06131
 [5,]  4 0.01529 0.01533
 [6,]  5 0.00305 0.00307
 [7,]  6 0.00051 0.00051
 [8,]  7 0.00007 0.00007
 [9,]  8 0.00001 0.00001
[10,]  9 0.00000 0.00000
[11,] 10 0.00000 0.00000

The following plot has resolution about 0.01, so
the two distributions seem identical. (Poisson probabilities are represented by vertical bars; binomial probabilities by dots.) Theoretically,
distribution $\mathsf{Pois}(\lambda = 1)$ has
positive probabilities extending to infinity,
hdr = "PDFs of POIS(1) [bars], and BINOM(1000,.001)"
plot(x, pdf.p, type="h", lwd=3, col="blue", main=hdr)
points(x, pdf.b, col="red")
abline(h=0, col="green2")
abline(v=0, col="green2")


A: This is a manifestation of a well known result similar to the central limit theorem.

Theorem: Let $X_{n,m}$, $m=1,\ldots,m_n$, be independent random random
variables with values in $\mathbb{Z}_+$. Suppose that

*

*$\sum^{m_n}_{m=1}\Pr[X_{n,m}=1]\rightarrow\lambda$ as
$n\rightarrow\infty$.

*$\max_{1\leq m\leq m_n}\Pr[X_{n,m}=1]\rightarrow0$
as  $n\rightarrow\infty$.

*$\sum^{m_n}_{m=1}\Pr[X_{n,m}\geq 2]\rightarrow0$ as
$n\rightarrow\infty$.

If $S_n=\sum^{m_n}_{m=1}X_{n,m}$, then $S_n\stackrel{n\rightarrow\infty}{\Longrightarrow} P_\lambda$.

Here, $\Longrightarrow$ stands for weak convergence of probability measures.
In the Binomial case, suppose $\{X_{n, m}:n\in\mathbb{N},m=1,\ldots,n\}$ are Bernoulli random variables and that $X_{n, m}\sim Be(\{0,1\},p_n)$, $m=1,\ldots, n$ and that $np_n\rightarrow\lambda$ (This is condition 1).  This means that $p_n=\sup_{1\leq m\leq n}P[X_{n, m}=1]\rightarrow0$ (this is condition 2). Condition 3 follows since $P[x_{n, m}\geq2]=0$ for all $n,m$.
Notice that $Y_n:=\sum^n_{m=1}X_{n, m}$ has binomial distribution with parameters $(n,p_n)$. The result says that as $Y_n$ is closed to Poisson($\lambda$).

*

*From this result, one sees that symmetry around a point is not a necessary feature of convergence to Poisson.

