How to prove Poisson Distribution is the approximation of Binomial Distribution?

I was reading Introduction to Probability Models 11th Edition and saw this proof of why Poisson Distribution is the approximation of Binomial Distribution when n is large and p is small:

An important property of the Poisson random variable is that it may be used to approximate a binomial random variable when the binomial parameter $$n$$ is large and $$p$$ is small. To see this, suppose that $$X$$ is a binomial random variable with parameters $$(n, p),$$ and let $$\lambda=n p .$$ Then \begin{aligned} P\{X=i\} &=\frac{n !}{(n-i) ! i !} p^{i}(1-p)^{n-i} \\ &=\frac{n !}{(n-i) ! i !}\left(\frac{\lambda}{n}\right)^{i}\left(1-\frac{\lambda}{n}\right)^{n-i} \\ &=\frac{n(n-1) \cdots(n-i+1)}{n^{i}} \frac{\lambda^{i}}{i !} \frac{(1-\lambda / n)^{n}}{(1-\lambda / n)^{i}} \end{aligned} Now, for $$n$$ large and $$p$$ small $$\left(1-\frac{\lambda}{n}\right)^{n} \approx e^{-\lambda}, \quad \frac{n(n-1) \cdots(n-i+1)}{n^{i}} \approx 1, \quad\left(1-\frac{\lambda}{n}\right)^{i} \approx 1$$ Hence, for $$n$$ large and $$p$$ small, $$P\{X=i\} \approx e^{-\lambda} \frac{\lambda^{i}}{i !}$$

I can understand most part of the proof except for this equation:

$$\left(1-\frac{\lambda}{n}\right)^{n} \approx e^{-\lambda}$$

I really don't remember where it comes from, could anybody explain this to me? Thanks!.

• It's related to the definition of $e$ itself May 22 '16 at 9:22
• It's notable that Ross' A first course in probability (I checked 9th and 10th editions) misses the commas in the approximations following for $n$ large and $p$ small, which may bring others to this page, as it did myself. Sep 6 '21 at 2:49

Well, this is a basic fact of the exponential function $$e^x$$.

One definition of $$e$$ is the limit $$\lim_{n\to\infty}(1+\frac1n)^n$$. By a monotonicity argument one can prove $$\lim_{x\to\infty}(1+\frac1x)^x=e$$ where $$x$$ now ranges the real numbers.

Also note that $$1-\frac1x=\frac{x-1}x=1/\frac x{x-1}=1/(1+\frac1y)=(1+\frac1y)^{-1}$$ where $$y=x-1$$.
So, one has the following: \begin{aligned} \lim_{x\to\infty}(1-\frac1x )^x &= \lim_{y\to\infty}(1+\frac1y )^{-(y+1)} \\ &=\lim_{y\to\infty}(1+\frac1y)^{-y}\times\lim_{y\to\infty}(1+\frac1y)^{-1} \\ &=e^{-1}\times1=e^{-1}\,. \end{aligned}

From here, assuming $$\lambda>0$$, \begin{aligned} e^{-\lambda}=(e^{-1})^\lambda &= \lim_{x\to\infty}(1-\frac1x)^{\lambda x} &\to{\ z:=\lambda x} \\ &= \lim_{z\to\infty}(1-\frac\lambda z)^z\,. \end{aligned}

In consequence, we have this limit for every sequence $$z_n\to\infty$$ written in place of $$z$$ and limiting on the natural $$n\to\infty$$. In particular, this also holds for $$z_n=n$$.

Note that we had to take the turnaround for arbitrary real numbers instead of integers only because of the exponent $$\lambda$$.

• Thanks, that's a really clear explanation! May 22 '16 at 10:29