Skip to content

Connection Between the Exponential Distribution and the Poisson Process

Concepts

  • Exponential distribution models the waiting time until a rare event occurs (e.g. a light bulb fails).

    \[ f(x; \lambda) = \begin{cases} \lambda e^{-\lambda t} & \text{for } t \geq 0 \\ 0 & \text{for } t < 0 \end{cases} \]

  • Poisson process models the number of rare events occurring in a given time interval.

    \[ P\{X = k\} = \frac{\lambda^k}{k!} e^{-\lambda} \]

These two are closely connected.

  • Imagine a stream of random events happening at a rate (e.g. receiving emails).
  • Even if events are frequent over time, the probability of one in a very small instant is still small.
  • Therefore, Poisson processes are valid even for frequent events like emails.

Timeline and Waiting Times

  • Events arrive at times: \(T₁, T₂, T₃, ..., Tₙ\)
  • Define waiting times between events: \(W₁ = T₂ - T₁, W₂ = T₃ - T₂\), etc.
  • Each waiting time \(W_j\):
    • Is independent
    • Follows an Exponential(λ) distribution
    • Where λ is the event rate (e.g. 5 emails per minute)
  • The waiting time (continuous) and the event count (discrete) are two sides of the same process.
  • A Poisson process unites these:
    • Exponential gaps between events
    • Poisson counts in intervals

Summary

  • Waiting time between events → Exponential(λ)
  • Number of events in time T → Poisson(λT)

Continuous vs Discrete

  • Waiting times are continuous (e.g. 1.2 minutes)
  • Number of events is discrete (e.g. 3 emails)

Example: Emails Arriving

Let emails arrive at a rate λ = 5 per minute

Question 1: Probability of waiting more than t minutes for the next email

\[ P(T > t) = e^{-5t} \]

This comes from the exponential distribution.

The longer you wait, the smaller the probability.

Question 2: Probability of getting zero emails in the next t minutes

\[ P(K = 0) = \frac{(5t)^0 e^{-5t}}{0!} = e^{-5t} \]

This is from the Poisson distribution.

Same numerical result as Question 1.

Question 3: Probability of getting exactly one email in the next t minutes

\[ P(K = 1) = \frac{(5t)^1 e^{-5t}}{1!} = 5t \cdot e^{-5t} \]

This is 5t times more likely than zero emails, for small t.