Poisson Distribution¶

Overview¶

The Poisson distribution models the number of events occurring in a fixed interval of time or space, given a known average rate. It is widely used in finance (trade arrivals, default counts), insurance (claim frequency), and queueing theory.

Definition¶

A random variable \(X\) follows a Poisson distribution with rate parameter \(\lambda > 0\):

\[ X \sim \text{Poisson}(\lambda), \qquad P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}, \quad k = 0, 1, 2, \ldots \]

The parameter \(\lambda\) represents both the mean and the variance of the distribution.

Verifying the PMF Sums to 1¶

\[ \sum_{k=0}^{\infty} \frac{e^{-\lambda} \lambda^k}{k!} = e^{-\lambda} \sum_{k=0}^{\infty} \frac{\lambda^k}{k!} = e^{-\lambda} \cdot e^{\lambda} = 1 \]

using the Taylor expansion of \(e^{\lambda}\).

Properties¶

\[ \begin{aligned} E[X] &= \lambda \\ \text{Var}(X) &= \lambda \\ \text{SD}(X) &= \sqrt{\lambda} \end{aligned} \]

The equality of mean and variance is a defining characteristic of the Poisson distribution and is often used as a diagnostic check.

Derivation of Mean¶

\[ E[X] = \sum_{k=0}^{\infty} k \cdot \frac{e^{-\lambda}\lambda^k}{k!} = \lambda e^{-\lambda} \sum_{k=1}^{\infty} \frac{\lambda^{k-1}}{(k-1)!} = \lambda e^{-\lambda} \cdot e^{\lambda} = \lambda \]

Derivation of Variance¶

First compute \(E[X(X-1)]\):

\[ E[X(X-1)] = \sum_{k=2}^{\infty} k(k-1) \frac{e^{-\lambda}\lambda^k}{k!} = \lambda^2 e^{-\lambda} \sum_{k=2}^{\infty} \frac{\lambda^{k-2}}{(k-2)!} = \lambda^2 \]

Then:

\[ \text{Var}(X) = E[X^2] - (E[X])^2 = E[X(X-1)] + E[X] - (E[X])^2 = \lambda^2 + \lambda - \lambda^2 = \lambda \]

Poisson as a Limit of the Binomial¶

The Poisson distribution arises as a limit of the binomial when \(n\) is large, \(p\) is small, and \(\lambda = np\) remains constant:

\[ \lim_{n \to \infty} \binom{n}{k} p^k (1-p)^{n-k} = \frac{e^{-\lambda}\lambda^k}{k!} \qquad \text{where } p = \frac{\lambda}{n} \]

Proof Sketch¶

With \(p = \lambda/n\):

\[ \binom{n}{k}\left(\frac{\lambda}{n}\right)^k\left(1 - \frac{\lambda}{n}\right)^{n-k} = \frac{n!}{k!(n-k)!} \cdot \frac{\lambda^k}{n^k} \cdot \left(1 - \frac{\lambda}{n}\right)^n \cdot \left(1 - \frac{\lambda}{n}\right)^{-k} \]

As \(n \to \infty\): \(\frac{n!}{(n-k)! \, n^k} \to 1\), \(\left(1 - \frac{\lambda}{n}\right)^n \to e^{-\lambda}\), and \(\left(1 - \frac{\lambda}{n}\right)^{-k} \to 1\).

Rule of thumb: Use Poisson when \(n \geq 20\) and \(p \leq 0.05\) (or more conservatively, \(n \geq 100\) and \(np \leq 10\)).

Additive Property¶

If \(X_1 \sim \text{Poisson}(\lambda_1)\) and \(X_2 \sim \text{Poisson}(\lambda_2)\) are independent, then:

\[ X_1 + X_2 \sim \text{Poisson}(\lambda_1 + \lambda_2) \]

This extends to any finite sum of independent Poisson random variables.

Poisson Process Connection¶

The Poisson distribution is intimately connected to the Poisson process. If events arrive at a constant rate \(\lambda\) per unit time, and arrivals are independent, then the number of events in an interval of length \(t\) follows \(\text{Poisson}(\lambda t)\), and the time between consecutive events follows \(\text{Exponential}(\lambda)\).

Worked Example¶

Problem: A stock exchange processes an average of 3 large block trades per hour. What is the probability of observing exactly 5 block trades in a given hour? What is the probability of observing at most 2?

Solution:

\[ P(X = 5) = \frac{e^{-3} \cdot 3^5}{5!} = \frac{0.0498 \cdot 243}{120} = 0.1008 \]

\[ P(X \leq 2) = \sum_{k=0}^{2} \frac{e^{-3} \cdot 3^k}{k!} = e^{-3}(1 + 3 + 4.5) = 0.0498 \cdot 8.5 = 0.4232 \]

Python: PMF, CDF, and Sampling¶

PMF and CDF¶

import matplotlib.pyplot as plt
import numpy as np
from scipy import stats

lam = 5
x = np.arange(0, 20)

fig, ax = plt.subplots(figsize=(12, 3))
ax.bar(x - 0.15, stats.poisson(lam).pmf(x), width=0.3, label='PMF', alpha=0.7)
ax.bar(x + 0.15, stats.poisson(lam).cdf(x), width=0.3, label='CDF', alpha=0.7)
ax.set_xlabel('k')
ax.set_xticks(x)
ax.spines[['top', 'right']].set_visible(False)
ax.legend()
plt.show()

Comparing Different Rates¶

import matplotlib.pyplot as plt
import numpy as np
from scipy import stats

fig, ax = plt.subplots(figsize=(12, 3))
for lam in [1, 4, 10]:
    x = np.arange(0, 25)
    ax.plot(x, stats.poisson(lam).pmf(x), 'o-', label=f'λ={lam}', markersize=4)
ax.spines[['top', 'right']].set_visible(False)
ax.set_xlabel('k')
ax.legend()
plt.show()

Poisson as Binomial Limit¶

import matplotlib.pyplot as plt
import numpy as np
from scipy import stats

lam = 5
x = np.arange(0, 20)

fig, ax = plt.subplots(figsize=(12, 3))
ax.bar(x, stats.poisson(lam).pmf(x), alpha=0.5, label='Poisson(λ=5)')
for n in [20, 50, 200]:
    ax.plot(x, stats.binom(n, lam/n).pmf(x), 'o-', label=f'Binom(n={n}, p={lam/n:.3f})', markersize=4)
ax.spines[['top', 'right']].set_visible(False)
ax.legend()
plt.show()

Sampling and Mean-Variance Check¶

import numpy as np
from scipy import stats

np.random.seed(42)
lam = 7
samples = stats.poisson(lam).rvs(100_000)

print(f"Theoretical mean: {lam},  Sample mean: {samples.mean():.4f}")
print(f"Theoretical var:  {lam},  Sample var:  {samples.var():.4f}")
print(f"Mean ≈ Var: {np.isclose(samples.mean(), samples.var(), atol=0.1)}")

Key Takeaways¶

The Poisson distribution models rare event counts with rate parameter \(\lambda\) that equals both the mean and variance.
It arises as the limit of the binomial distribution when \(n\) is large and \(p\) is small.
The additive property makes it natural for aggregating independent event counts.
The connection to the Poisson process links discrete event counts to continuous inter-arrival times (exponential distribution).
The mean-equals-variance property is a useful diagnostic: if sample variance greatly exceeds the mean, the data may be overdispersed relative to the Poisson model.