Poisson distribution is widely used to represent occurences in an interval, mostly time but sometimes area. Examples include arrivals to queues in a day, number of breakdowns in a machine in a year, typos in a letter, oil reserve in a region.

We know from binomial distribution that \(k\) occurences in \(n\) trials with probability \(p\) has the following function.

\[P\{X = k\} = \binom{n}{k} p^k (1-p)^{n-k} = \dfrac{n!}{(n-k)!k!} p^k (1-p)^{n-k}\]

and expected value is \(E[X] = np\). Now define \(\lambda = np\).

\[\begin{align*} P\{X = k\} =& \dfrac{n!}{(n-k)!k!} \left(\dfrac{\lambda}{n}\right)^k \left(1-\left(\dfrac{\lambda}{n}\right)\right)^{n-k} \\ =& \dfrac{n (n-1) \dots (n-k+1)}{n^k} \left(\dfrac{\lambda^k}{k!}\right) \dfrac{(1-\dfrac{\lambda}{n})^n}{(1-\dfrac{\lambda}{n})^k} \end{align*}\]For very large \(n\) and very small \(p\) the resulting pdf becomes \(\dfrac{\lambda^k e^{-\lambda}}{k!}\).

PMF: \(P\{X = k\} = \dfrac{\lambda^k e^{-\lambda}}{k!}\)

CDF: \(P\{X \le k\} = \sum_{i=0}^k \dfrac{\lambda^i e^{-\lambda}}{i!}\)

\(E[X] = \lambda (due to \sum_{i=0}^\infty\dfrac{x^i}{i!} = e^x)\)

\(V(X) = \lambda\)

Rate parameter \(\lambda\) can also be defined as \(\lambda t\), \(t\) being the scale parameter. For instance, let arrivals in 30 minutes interval be \(\lambda t_{30} = 4\). If we would want to work on hourly intervals, we should simply rescale, \(\lambda t_{60} = 8\).

Suppose a machine has a probability of failure 0.001 per hour. What is the probability that the machine had failed at least three times within 2000 hours.

*Binomial solution*

```
#R version
1- sum(dbinom(0:2,2000,0.001))
```

`## [1] 0.3233236`

*Poisson solution*

```
#R version
lambda=2000*0.001
1- sum(dpois(0:2,lambda))
```

`## [1] 0.3233236`

People arrive at a bank with rate \(\lambda = 5\) every 10 minutes. What is the probability that 10 people arrive in 30 minutes?

\[\lambda t_{10} = 5\]

\[\lambda^\prime = \lambda t_{30} = 15 \]

\[P\{X = 10, t=30\} = \dfrac{e^{-15}15^{10}}{10!} = 0.049\]

`dpois(10,15)`

`## [1] 0.04861075`

A machine breaks down with a poisson rate of \(\lambda = 10\) per year. A new method is tried to reduce the failure rate to \(\lambda = 3\), but there is a 50% chance that it wonâ€™t work. If the method is tried and the machine fails only 3 times that year, what is the probability that the method worked on the machine?

\[\begin{align*} P\{Works | X = 3\} &= \dfrac{P\{Works and X = 3\}}{P\{X = 3\}} \\~\\ P\{Works\} &= 0.5\\~\\ P\{Works and X = 3\} &= 0.5 * \dfrac{e^{-3}3^3}{3!} = 0.1120209\\~\\ P\{X = 3\} &= P\{Works and X = 3\} + P\{Doesn't\ Work and X = 3\} = 0.5 * \dfrac{e^{-3}3^3}{3!} + 0.5 * \dfrac{e^{-10}10^3}{3!} \\ &= 0.96733 \end{align*}\]```
#R codes
#Probability that it works
pw = 0.5
#Probability of 3 fails if lambda is 10
ppois10 = dpois(3,10)
#Probability of 3 fails if lambda is 3
ppois3 = dpois(3,3)
#P(Works | X = 3)
(pw*ppois3)/(pw*ppois3 + (1-pw)*ppois10)
```

`## [1] 0.96733`