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!.

  • $\begingroup$ It's related to the definition of $e$ itself $\endgroup$
    – Dleep
    Commented May 22, 2016 at 9:22
  • $\begingroup$ 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. $\endgroup$
    – sunspots
    Commented Sep 6, 2021 at 2:49

1 Answer 1


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 $\lambda$.

  • $\begingroup$ Thanks, that's a really clear explanation! $\endgroup$
    – Demonedge
    Commented May 22, 2016 at 10:29

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