# How Binomial and Normal distributions approximate Poisson distribution respectively?

From Wikipedia:

1. In some cases, the cdf of the Poisson distribution is the limit of the cdf of the binomial distribution：

The Poisson distribution can be derived as a limiting case to the binomial distribution $\text{bin}(n,p)$ as the number $n$ of trials goes to infinity and the expected number $np$ of successes remains fixed — see law of rare events below. Therefore it can be used as an approximation of the binomial distribution if $n$ is sufficiently large and $p$ is sufficiently small.

2. In some cases, the cdf of the Poisson distribution is the limit of the cdf of the normal distribution:

For sufficiently large values of $λ$, (say $λ>1000$), the normal distribution with mean $λ$ and variance $λ$ (standard deviation $\sqrt{\lambda}$), is an excellent approximation to the Poisson distribution. If $λ$ is greater than about $10$, then the normal distribution is a good approximation if an appropriate continuity correction is performed, i.e., $P(X ≤ x)$, where (lower-case) $x$ is a non-negative integer, is replaced by $P(X ≤ x + 0.5)$. $$F_\mathrm{Poisson}(x;\lambda) \approx F_\mathrm{normal}(x;\mu=\lambda,\sigma^2=\lambda)\,$$

I wonder in what best senses these functional approximations are? Pointwise, uniformly, $L_2$, ...? Thanks and regards!

In comments above, Tim asked why it must be that if $$X\sim\mathrm{Poisson}(\lambda)$$ and $$Y\sim\mathrm{Poisson}(\mu)$$ and $$X$$ and $$Y$$ are independent, then we must have $$X+Y\sim\mathrm{Poisson}(\lambda+\mu)$$.

Here's one way to show that. \begin{align} & \Pr(X+Y= w) \\[8pt] = {} & \Pr\Big( (X=0\ \& \ Y=w)\text{ or }(X=1\ \&\ Y=w-1) \\ & {}\qquad\qquad\text{ or } (X=2\ \&\ Y=w-2)\text{ or } \ldots \text{ or }(X=0\ \&\ Y=w)\Big) \\[8pt] = {} & \sum_{u=0}^w \Pr(X=u)\Pr(Y=w-u)\qquad(\text{independence was used here}) \\[8pt] = {} & \sum_{u=0}^w \frac{\lambda^u e^{-\lambda}}{u!} \cdot \frac{\mu^{w-u} e^{-\mu}}{(w-u)!} \\[8pt] = {} & e^{-(\lambda+\mu)} \sum_{u=0}^w \frac{1}{u!(w-u)!} \mu^u\lambda^{w-u} \\[8pt] = {} & \frac{e^{-(\lambda+\mu)}}{w!} \sum_{u=0}^w \frac{w!}{u!(w-u)!} \mu^u\lambda^{w-u} \\[8pt] = {} & \frac{e^{-(\lambda+\mu)}}{w!} (\lambda+\mu)^w \end{align} and that is what was to be shown.

"In some cases, the cdf of the Poisson distribution is the limit of the cdf of the normal distribution:"

That is false. The limit of Poisson distributions is a normal distribution; the limit of normal distributions is not a Poisson distribution.

"Best senses" is a term I've never come across and I don't know what you mean by it.

Suppose $$X\sim\mathrm{Poisson}(\lambda)$$. Then the cdf of $$(X-\lambda)/\sqrt{\lambda}$$ converges pointwise to the cdf of the standard normal distribution as $$\lambda\to\infty$$.

Now suppose $$X\sim\mathrm{Bin}(n,\lambda/n)$$. Then the cdf of $$X$$ converges pointwise to the cdf of the Poisson distribution with expectation $$\lambda$$ as $$n\to\infty$$.

Generally "convergence in distribution" means a sequence of cdf's converges to a cdf pointwise except that it need not converge at points where the limiting cdf is not continuous. The reason for the exception is things like this: Concentrate probability $$1$$ at $$1/n$$. Then the value of the cdf at $$0$$ is $$0$$. But the limiting cdf has value $$1$$ at $$0$$; that's where it concentrates all the probability.

But don't write about normal distributions approaching a Poisson distribution. It's the other way around.

• Thanks! But that is what I understood from the Wikipedia article, unless it is wrong.
– Tim
Commented Oct 9, 2011 at 22:41
• @Tim, WP is right on this, you might wish to read the paragraph again.
– Did
Commented Oct 9, 2011 at 22:47
• Wikipedia says the normal distribution approximates the Poisson distribution under certain circumstances. But that doesn't mean the normal is approaching the Poisson. It's the Poisson that's approaching the normal. Commented Oct 9, 2011 at 22:55
• @Didier and Michael: Thanks! According to the Wiki article, (1) in the binomial distribution part, the convergence is when $n \rightarrow \infty$ with $np$ fixed, not $\lambda \rightarrow \infty$. Note $\lambda = np$. (2) in the normal distribution part, the convergence is $\lim_{\lambda \rightarrow \infty} \text{discrepancy}( F_\mathrm{Poisson}(x;\lambda), F_\mathrm{normal}(x;\mu=\lambda,\sigma^2=\lambda)) = 0$, i.e. the two cdfs are approaching each other under some measure of discrepancy, instead of one approaching the other.
– Tim
Commented Oct 9, 2011 at 23:17
• I would have thought that the "Now suppose ..." paragraph should end "... as $n \to \infty$" rather than "... as $\lambda \to \infty$" Commented Sep 20, 2013 at 20:18