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The Poisson Distribution

SD05 — Discrete Distributions

Author

Samer El Kababji

Course: Statistics and Probability for Data Science (DS14270).
Source deck: Samer El Kababji, SD05 — Discrete Distributions.

The Poisson distribution models counts of events in a fixed interval when the events occur independently at a stable average rate.

Probability mass function

If \(X\sim\operatorname{Poisson}(\mu)\), then

\[ P(X=x)=e^{-\mu}\frac{\mu^x}{x!}, \qquad x=0,1,2,\ldots, \]

where \(\mu>0\) is the expected count per interval.

Code 
mu <- 2.4
x <- 0:10

formula_pmf <- exp(-mu) * mu^x / factorial(x)
R_pmf <- dpois(x, lambda = mu)

data.frame(x, formula_pmf, R_pmf)
    x  formula_pmf        R_pmf
1   0 0.0907179533 0.0907179533
2   1 0.2177230879 0.2177230879
3   2 0.2612677055 0.2612677055
4   3 0.2090141644 0.2090141644
5   4 0.1254084986 0.1254084986
6   5 0.0601960793 0.0601960793
7   6 0.0240784317 0.0240784317
8   7 0.0082554623 0.0082554623
9   8 0.0024766387 0.0024766387
10  9 0.0006604370 0.0006604370
11 10 0.0001585049 0.0001585049

Tile-defect example

If a tile has an average of \(\mu=2.4\) defects, then

\[ P(X=0)=e^{-2.4}\approx0.0907. \]

The probability of at least four defects is easiest through the complement:

\[ P(X\geq4)=1-P(X\leq3)\approx0.2213. \]

Code 
c(
  no_defects = dpois(0, lambda = 2.4),
  at_least_four = ppois(3, lambda = 2.4, lower.tail = FALSE),
  complement_check = 1 - ppois(3, lambda = 2.4)
)
      no_defects    at_least_four complement_check 
      0.09071795       0.22127709       0.22127709

Using lower.tail = FALSE avoids manually subtracting from 1 and can be more accurate for very small upper-tail probabilities.

Mean and variance

The Poisson parameter controls both center and spread:

\[ E(X)=\mu,\qquad \operatorname{Var}(X)=\mu,\qquad \operatorname{SD}(X)=\sqrt{\mu}. \]

Code 
mu <- 2.4
c(mean = mu, variance = mu, sd = sqrt(mu))
    mean variance       sd 
2.400000 2.400000 1.549193

Equality of the mean and variance is a distinctive Poisson property. Strong overdispersion or underdispersion in observed counts can signal that a simple Poisson model is inadequate.

Changing the interval

If the rate is 2.4 events per hour, the expected count in \(t\) hours is \(\mu_t=2.4t\).

Code 
rate_per_hour <- 2.4
hours <- c(0.5, 1, 3)
mu_by_interval <- rate_per_hour * hours

data.frame(
  hours,
  expected_count = mu_by_interval,
  P_no_events = dpois(0, lambda = mu_by_interval)
)
  hours expected_count  P_no_events
1   0.5            1.2 0.3011942119
2   1.0            2.4 0.0907179533
3   3.0            7.2 0.0007465858

Poisson approximation to the binomial

For \(X\sim\operatorname{Bin}(n,p)\), if \(n\) is large and \(p\) is small while \(np\to\mu\), then

\[ X\ \dot\sim\ \operatorname{Poisson}(\mu),\qquad \mu=np. \]

The deck’s practical rule is \(n>50\) and \(np<5\).

Code 
n <- 200
p <- 0.01
mu <- n * p
k <- 0:8

comparison <- data.frame(
  k,
  binomial = dbinom(k, size = n, prob = p),
  poisson = dpois(k, lambda = mu)
)
comparison$absolute_difference <- abs(comparison$binomial - comparison$poisson)
comparison
  k     binomial      poisson absolute_difference
1 0 0.1339796749 0.1353352832        1.355608e-03
2 1 0.2706660098 0.2706705665        4.556659e-06
3 2 0.2720330099 0.2706705665        1.362443e-03
4 3 0.1813553399 0.1804470443        9.082956e-04
5 4 0.0902197019 0.0902235222        3.820233e-06
6 5 0.0357233567 0.0360894089        3.660521e-04
7 6 0.0117273646 0.0120298030        3.024384e-04
8 7 0.0032829852 0.0034370866        1.541014e-04
9 8 0.0008000204 0.0008592716        5.925126e-05

Simulation check (fictional extension)

Code 
set.seed(14270)
simulated_counts <- rpois(100000, lambda = 2.4)

c(
  simulated_mean = mean(simulated_counts),
  simulated_variance = var(simulated_counts),
  theoretical_value = 2.4
)
    simulated_mean simulated_variance  theoretical_value 
          2.390470           2.403167           2.400000

Practice questions

  1. A help desk averages 3.5 calls per hour. Find \(P(X=2)\) in one hour.
  2. Find \(P(X\geq5)\) in one hour.
  3. Find the probability of no calls in 30 minutes.
  4. Approximate \(P(X=1)\) for \(X\sim\operatorname{Bin}(100,0.02)\) with a Poisson distribution.
  1. dpois(2, 3.5) gives approximately \(0.1850\).
  2. ppois(4, 3.5, lower.tail = FALSE) gives approximately \(0.2746\).
  3. The half-hour mean is \(1.75\), so the probability is \(e^{-1.75}\approx0.1738\).
  4. Use \(\mu=np=2\): dpois(1, 2) gives approximately \(0.2707\) (the exact binomial value is about \(0.2707\)).