Poisson Distributions

Poisson process

A Poisson process models “purely random” counts with a known intensity or rate lambda.

lambda = average arrivals per unit of time

Poisson distribution

A Poisson distribution is determined by one parameter: lambda = expected number of counts (arrivals, events, etc.) per unit of time.

Used to make probabilistic predictions if correct probability models are known (or assumed).

ppois(x, lambda) # returns a CDF

Sample Problems

Q: What is the probability that more than 10 customers will arrive in an hour if rate lambda = 8 per hour are expected?
A:
```
1 - ppois(10, lambda = 8) 
```
0.184
Q: What is the probability of no more than 4 arrivals in the first 30 minutes if the arrival rate is lambda = 8 per hour?
A:
0.629
Q: A book has 600 pages and 250 typos. What is the probability that the first 5 pages have at most 2 typos?
A:
- Intensity (rate) = 250/600 = 0.4167 typos per page
- P(no more than 2 typos in 5 pages) =
```
ppois(2, lambda = 5 * (250/600)) 
```
  0.6541333

Plotting a Poisson PDF with plot() and dpois()

Notes:

dpois(x, lambda) returns probability of count x if counts have a Poisson distribution with mean lambda.

x = c(0:20)
plot(x, dpois(x, lambda = 5))

to list values: dpois(x, lambda=5)
to list probabilities for specific counts: dpois(count, lambda=5)

Relationship between Poisson process and distribution

The random number of arrivals in a Poisson process with rate lambda and an interval of length t has a Poisson distribution with mean lambda * t.

A poisson process with intensity or rate lambda generates mean lambda * t expected counts in an interval of length t.
The actual random number of counts has a Poisson distribution.
Times between counts have exponential distributions.
In a Poisson process with rate lambda, the expected waiting time between arrivals is 1 / lambda.

Exponential Distributions

Notes:

In a Poisson process with rate lambda, the probability distribution for the random waiting time between arrivals has an exponential distribution.
The random interval between successive counts in a Poisson process with intensity lambda has an exponential distribution with mean 1 / lambda.

pexp(m, rate) # returns an Exponential distribution function

Problem:

If arrival rate is 3 per minte (lambda = 3), then time until next count has an exponential distribution with mean 1/3 = 0.3333 minutes = 20 seconds.
Q: What is the probability of no arrivals in a minute, if 3 are expected?
A:
```
dpois(0, 3)
```
0.0497
- 1 - pexp(1, rate = 3) = 0.0497 is the probability that the next arrival takes > 1 minute.