Poisson Probability Mass Function & Cumulative Distribution Function

Probability and Statistics

Medium

Problem

Theory

The Poisson distribution models the number of events occurring in a fixed interval of time or space, when events happen independently at a constant average rate.

It answers questions like: "How many customers will arrive in the next hour?" or "How many defects will be found in a batch?"

Named after French mathematician Siméon Denis Poisson.

When to Use Poisson

The Poisson distribution is appropriate when:

1. Events occur independently

One event does not affect the probability of another.

2. Events occur at a constant average rate

The rate \lambda is the same throughout the interval.

3. Two events cannot occur at exactly the same instant

Events are countable and discrete.

4. The probability of an event in a small interval is proportional to the interval size

Doubling the time doubles the expected count.

Real-World Examples

Many phenomena follow a Poisson distribution:

• Number of emails received per hour
• Number of cars passing a checkpoint per minute
• Number of calls to a call center per hour
• Number of typos per page
• Number of accidents at an intersection per month
• Number of mutations in a DNA strand
• Number of photons hitting a detector per second
• Number of website visitors per minute

The Parameter \lambda

The Poisson distribution has a single parameter:

\lambda = \text{average rate (expected number of events)}

Constraints:

• \lambda > 0

\lambda represents both the mean and the variance of the distribution.

The Probability Mass Function (PMF)

The probability of observing exactly k events is:

P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}

where k \in \{0, 1, 2, 3, ...\}.

Components explained:

• \lambda^k: Rate raised to power of count
• e^{-\lambda}: Normalization factor ensuring probabilities sum to 1
• k!: Factorial accounting for event ordering

Verifying the PMF Sums to 1

\sum_{k=0}^{\infty} P(X = k) = \sum_{k=0}^{\infty} \frac{\lambda^k e^{-\lambda}}{k!} = e^{-\lambda} \sum_{k=0}^{\infty} \frac{\lambda^k}{k!}

The Taylor series for e^x is \sum_{k=0}^{\infty} \frac{x^k}{k!}, so:

= e^{-\lambda} \cdot e^{\lambda} = e^0 = 1 \checkmark

Worked Example: Call Center

Setup: A call center receives an average of 5 calls per hour (\lambda = 5). What is the probability of receiving exactly 3 calls in the next hour?

Solution:

P(X = 3) = \frac{5^3 e^{-5}}{3!} = \frac{125 \cdot e^{-5}}{6}

= \frac{125 \cdot 0.00674}{6} = \frac{0.8425}{6} = 0.1404

There is about a 14% chance of exactly 3 calls.

Computing Multiple PMF Values

Setup: \lambda = 4

P(X = 0) = \frac{4^0 e^{-4}}{0!} = \frac{1 \cdot 0.0183}{1} = 0.0183

P(X = 1) = \frac{4^1 e^{-4}}{1!} = \frac{4 \cdot 0.0183}{1} = 0.0733

P(X = 2) = \frac{4^2 e^{-4}}{2!} = \frac{16 \cdot 0.0183}{2} = 0.1465

P(X = 3) = \frac{4^3 e^{-4}}{3!} = \frac{64 \cdot 0.0183}{6} = 0.1954

P(X = 4) = \frac{4^4 e^{-4}}{4!} = \frac{256 \cdot 0.0183}{24} = 0.1954

P(X = 5) = \frac{4^5 e^{-4}}{5!} = \frac{1024 \cdot 0.0183}{120} = 0.1563

The Cumulative Distribution Function (CDF)

The CDF gives the probability of observing at most k events:

F(k) = P(X \leq k) = \sum_{i=0}^{k} \frac{\lambda^i e^{-\lambda}}{i!}

Properties:

• F(k) increases as k increases
• \lim_{k \to \infty} F(k) = 1

There is no closed-form expression; it must be computed as a sum or using special functions (incomplete gamma function).

CDF Example

Setup: \lambda = 4. Find P(X \leq 3).

F(3) = P(X \leq 3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)

= 0.0183 + 0.0733 + 0.1465 + 0.1954 = 0.4335

There is about a 43% chance of 3 or fewer events.

Using CDF for Range Probabilities

P(X > k):

P(X > k) = 1 - F(k) = 1 - P(X \leq k)

P(X \geq k):

P(X \geq k) = 1 - F(k-1) = 1 - P(X \leq k-1)

P(a \leq X \leq b):

P(a \leq X \leq b) = F(b) - F(a-1)

Expected Value and Variance

For the Poisson distribution, a remarkable property is:

E[X] = \lambda

\text{Var}(X) = \lambda

The mean and variance are equal. This property helps identify Poisson-distributed data.

Standard deviation:

\sigma = \sqrt{\lambda}

Why Mean Equals Variance

Derivation of E[X]:

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

Derivation of \text{Var}(X):

Using \text{Var}(X) = E[X^2] - (E[X])^2 and computing E[X^2] similarly gives \text{Var}(X) = \lambda.

Mode of the Distribution

The mode (most likely value) is:

\text{mode} = \lfloor \lambda \rfloor

If \lambda is an integer, both \lambda and \lambda - 1 are modes.

Example: For \lambda = 4.7, mode = 4.

Shape of the Distribution

The shape depends on \lambda:

Small \lambda (e.g., 1):

• Right-skewed
• Mode at 0 or 1
• Rapid decay

Large \lambda (e.g., 10+):

• More symmetric
• Approaches Normal distribution
• Mode near \lambda

Normal Approximation

For large \lambda, the Poisson can be approximated by a Normal distribution:

X \approx N(\lambda, \lambda)

Rule of thumb: Approximation is reasonable when \lambda \geq 10.

With continuity correction:

P(X \leq k) \approx \Phi\left(\frac{k + 0.5 - \lambda}{\sqrt{\lambda}}\right)

Poisson as Limit of Binomial

The Poisson distribution arises as a limit of the Binomial:

If X \sim \text{Binomial}(n, p) with n \to \infty, p \to 0, and np = \lambda fixed:

\text{Binomial}(n, p) \to \text{Poisson}(\lambda)

This is why Poisson is good for rare events in large populations.

Example: Probability of k defects in 1000 items where each has 0.003 defect probability:

\lambda = np = 1000 \times 0.003 = 3

Use \text{Poisson}(3) instead of \text{Binomial}(1000, 0.003).

Sum of Independent Poissons

If X \sim \text{Poisson}(\lambda_1) and Y \sim \text{Poisson}(\lambda_2) are independent:

X + Y \sim \text{Poisson}(\lambda_1 + \lambda_2)

Example: If department A gets 3 calls/hour and department B gets 5 calls/hour, the total is Poisson(8).

Scaling the Interval

If events occur at rate \lambda per unit time:

• In time t, the count follows \text{Poisson}(\lambda t)

Example: 10 emails/hour means:

• Per minute: \text{Poisson}(10/60) = \text{Poisson}(1/6)
• Per 3 hours: \text{Poisson}(30)

Relationship to Exponential Distribution

The Poisson process has a dual relationship with the Exponential distribution:

• Poisson: Number of events in a fixed time
• Exponential: Time between consecutive events

If counts follow \text{Poisson}(\lambda), then inter-arrival times follow \text{Exponential}(\lambda).

Maximum Likelihood Estimation

Given observations x_1, x_2, ..., x_n from \text{Poisson}(\lambda):

\hat{\lambda} = \bar{x} = \frac{1}{n}\sum_{i=1}^{n} x_i

The MLE is simply the sample mean.

Checking if Data is Poisson

Mean-variance test:

For Poisson data, mean \approx variance.

Compute the dispersion index:

D = \frac{\text{Var}(X)}{\bar{X}}

If D \approx 1: Consistent with Poisson

If D > 1: Overdispersion (consider Negative Binomial)

If D < 1: Underdispersion

Applications in Machine Learning

Count data modeling:

• Number of clicks, purchases, page views
• Poisson regression for count outcomes

Text analysis:

• Word frequency models
• Topic modeling

Anomaly detection:

• Unusual event counts (e.g., fraud detection)

Queuing theory:

• Server load modeling
• Request arrival patterns

Examples

Hints

1. Initialize the zero-event probability with math.exp(-lam).
2. Advance from count i - 1 to i by multiplying the previous probability by lam / i.

Requirements

• Start with the probability of zero events
• Use the recurrence between consecutive Poisson probabilities
• Accumulate probabilities through k for the CDF
• Return exactly pmf and cdf in a dictionary of Python floats

Constraints

• lam is positive
• k is an integer between 0 and 50
• Use the Python standard library only

Starter Code

import math

def poisson_pmf_cdf(lam: float, k: int) -> dict:
    """
    Returns a dictionary with pmf and cdf.
    """
    # Write code here
    pass

Test Cases