AcademyProbability
Academy
Poisson Approximation
Level 1 - Math II (Physics) topic page in Probability.
Principle
A binomial count with many trials and a small success probability can often be approximated by a Poisson count. The Poisson parameter is the expected number of successes, \(\lambda=np\), not the success probability \(p\).
Notation
\(X\sim\operatorname{Bin}(n,p)\)
binomial count with n trials and success probability p
\(Y\sim\operatorname{Po}(\lambda)\)
Poisson approximation with parameter lambda
\(\lambda=np\)
expected number of binomial successes
\(k\)
fixed count being approximated
\(n(n-1)\cdots(n-k+1)\)
falling factorial for k selected successful positions
Method
Step 1: Match the mean count
Set \(\lambda=np\), so \(p=\lambda/n\). The approximation keeps the expected number of rare successes fixed while \(n\) is large.
Step 2: Split the binomial probability
For fixed \(k\), rewrite the binomial PMF into pieces that have clear limits.
Start with binomial
\[P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}\]
Substitute p=lambda/n
\[P(X=k)=\binom{n}{k}\left(\frac{\lambda}{n}\right)^k\left(1-\frac{\lambda}{n}\right)^{n-k}\]
Expand the combination
\[P(X=k)=\frac{n(n-1)\cdots(n-k+1)}{k!}\frac{\lambda^k}{n^k}\left(1-\frac{\lambda}{n}\right)^{n-k}\]
Separate the factors
\[P(X=k)=\frac{n(n-1)\cdots(n-k+1)}{n^k}\frac{\lambda^k}{k!}\left(1-\frac{\lambda}{n}\right)^n\left(1-\frac{\lambda}{n}\right)^{-k}\]
Take limits
\[\frac{n(n-1)\cdots(n-k+1)}{n^k}\to1,\quad \left(1-\frac{\lambda}{n}\right)^n\to e^{-\lambda},\quad \left(1-\frac{\lambda}{n}\right)^{-k}\to1\]
Poisson limit
\[P(X=k)\to e^{-\lambda}\frac{\lambda^k}{k!}\]
Rules
Mean matching
\[\lambda=np\]
Approximation
\[X\sim\operatorname{Bin}(n,p)\quad\approx\quad Y\sim\operatorname{Po}(np)\]
Approximate probability
\[P(X=k)\approx e^{-\lambda}\frac{\lambda^k}{k!}\]
Useful guide
\[n\ge20\quad\text{and}\quad p\le0.05\]
Excellent guide
\[n\ge100\quad\text{and}\quad np\le10\]
Examples
Question
A factory makes
\[200\]
gadgets, each defective independently with probability \[0.01\]
Approximate the probability of exactly \[3\]
defects.Answer
Use
\[\lambda=np=200(0.01)=2\]
Then \[P(X=3)\approx e^{-2}\frac{2^3}{3!}.\]
Checks
- Use \(\lambda=np\), not \(p\), as the Poisson parameter.
- The approximation is designed for large \(n\) and small \(p\).
- The count \(k\) is fixed in the limiting calculation.
- A practical guide is \(n\ge20\) and \(p\le0.05\).
- The approximation is often excellent when \(n\ge100\) and \(np\le10\).