# 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
Oct 9 '11 at 22:41
• @Tim, WP is right on this, you might wish to read the paragraph again.
– Did
Oct 9 '11 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. Oct 9 '11 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
Oct 9 '11 at 23:17
• I would have thought that the "Now suppose ..." paragraph should end "... as $n \to \infty$" rather than "... as $\lambda \to \infty$" Sep 20 '13 at 20:18