How can the poisson distribution be approximated by the normal distribution?

Question

I know the normal distribution will have both parameters equal to mui, the parameter of the poisson distribution. But why?

A vigorous and a less vigorous proof will both help.

Answer 1

Skewness. More specifically on the Comment of @AndreNicholas: Skewness of $Pois(\lambda)$ is $1/\sqrt{\lambda}$. (See Wikipedia on 'Poisson distribution'.) A normal distribution has $0$ skewness. A necessary condition for a 'reasonable' fit of normal to Poisson is to have $\lambda$ large enough that Poisson skewness becomes small enough to be negligible.

Limit theorem. It should be possible to find a formal proof of the appropriate limit theorem. This follows from the general Central Limit Theorem that the sum of IID random variables (for which variances exist) converge in distribution to normal. If $X_1, X_2, \dots$ are IID $Pois(\lambda)$, then $S_n = \sum_{i=1}^n X_i \sim Pois(n\lambda).$

Roughly speaking, the ides is this: As you add more and more Poisson RVs together the CLT says that $(S_n - n\lambda)/\sqrt{n\lambda}$ converges in distribution to standard normal. Simultaneously, $E(S_n) = n\lambda$ is getting larger.

Graphs. Below are graphs (made in R) showing relatively poor fit of the appropriate normal density curve to $Pois(\lambda = 2),$ possibly useful fit to $Pois(\lambda = 20),$ and more accurate fit to $Pois(\lambda=100).$

Numerical examples. Here are some numerical examples computed in R, in which all normal approximations use continuity correction, and exact probabilities and their approximations are stated to thee decimal places:

(1) For $X \sim Pois(2)$, we have $P(X = 1) = .271$, badly approximated by $.217.$

(2) For $X \sim Pois(20)$, we have $P(10 \le X \le 20) = .554$, reasonably well approximated by $.535.$

(3) For $X \sim Pois(100)$, we have $P(50 \le X \le 100) = .527$, well approximated by $.520.$

 dpois(1, 2)
 ## 0.2706706
 diff(pnorm(c(.5,1.5), 2, sqrt(2)))
 ## 0.2174146
 sum(dpois(10:20, 20))
 ## 0.5540972
 diff(pnorm(c(9.5,20.5), 20, sqrt(20)))
 ## 0.5350698
 sum(dpois(50:100, 100))
 ## 0.5265622
 diff(pnorm(c(49.5,100.5), 100, sqrt(100)))
 ## 0.5199386

Note: If printed normal tables are used for the normal approximations, results will be slightly different because of rounding errors. Because software is now available to give exact Poisson results, normal approximations are ever more rarely used in practical applications.