Poisson vs. exponential distribution
Consider events that happen independently of each other at a constant average rate $\lambda$, such as customers arriving at a shop (a Poisson process). The Poisson distribution describes the number of events in a fixed length of time, which is a whole number. The exponential distribution describes the time until the next event, which is a positive real number.
Definitions
Level:
High school
Consider events that happen independently of each other at an average rate of $\lambda$ per unit of time, such as customers arriving at a shop. The number $X$ of events in one unit of time has a Poisson distribution with mean $\lambda$:
\[P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!} \quad \text{for } k = 0, 1, 2, \dots.\]Examples
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Customers arriving at a shop independently of each other at an average rate of $3$ per hour
: The number of customers in an hour: Poisson with mean $3$
For customers arriving independently at an average rate of $3$ per hour, the number of customers in an hour has a Poisson distribution with mean $3$, because the number counts independent events at a constant rate.
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Emails arriving independently of each other at an average rate of $6$ per hour, and the event that no email arrives in the next $10$ minutes
: The number $N$ of emails in $10$ minutes is Poisson with mean $1$, so $P(N = 0) = e^{-1} \approx 0.368$
For emails arriving at $6$ per hour, the number of emails in $10$ minutes is a count in a fixed time, so the number has a Poisson distribution with mean $6 \cdot \frac{10}{60} = 1$.
Properties
- A Poisson random variable with mean $\lambda$ also has variance $\lambda$.
- The Poisson distribution is the limit of the binomial distribution with parameters $n$ and $\lambda / n$ as $n \to \infty$.
Undergraduate
Consider events that happen independently of each other at an average rate of $\lambda$ per unit of time, such as customers arriving at a shop. The waiting time $T$ until the first event has an exponential distribution with rate $\lambda$, which has the probability density
\[f(t) = \lambda e^{-\lambda t} \quad \text{for } t \geq 0.\]Examples
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Customers arriving at a shop independently of each other at an average rate of $3$ per hour
: The waiting time, in hours, until the first customer: exponential with rate $3$
For customers arriving independently at an average rate of $3$ per hour, the waiting time until the first customer has an exponential distribution with rate $3$, so the expected wait is $\frac{1}{3}$ hour, or $20$ minutes.
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Emails arriving independently of each other at an average rate of $6$ per hour, and the event that no email arrives in the next $10$ minutes
: The waiting time $T$, in minutes, until the next email is exponential with rate $\frac{1}{10}$ per minute, so $P(T > 10) = e^{-1} \approx 0.368$
For emails arriving at $6$ per hour, the waiting time until the next email is a continuous time, so the waiting time has an exponential distribution with mean $10$ minutes.
Properties
- An exponential random variable with rate $\lambda$ has mean $1 / \lambda$ and satisfies $P(T > t) = e^{-\lambda t}$.
- The exponential distribution is memoryless: $P(T > s + t \mid T > s) = P(T > t)$. The exponential distribution is the continuous counterpart of the geometric distribution.
Examples
| Example | Poisson distribution | exponential distribution | Why |
|---|---|---|---|
| Emails arriving independently of each other at an average rate of $6$ per hour, and the event that no email arrives in the next $10$ minutes | The number $N$ of emails in $10$ minutes is Poisson with mean $1$, so $P(N = 0) = e^{-1} \approx 0.368$ | The waiting time $T$, in minutes, until the next email is exponential with rate $\frac{1}{10}$ per minute, so $P(T > 10) = e^{-1} \approx 0.368$ | For emails arriving at $6$ per hour, “no email in the next $10$ minutes” is the same event as “zero emails in $10$ minutes” and “the wait is longer than $10$ minutes,” so the Poisson and exponential distributions give the same probability $e^{-1}$. |
| Customers arriving at a shop independently of each other at an average rate of $3$ per hour | The number of customers in an hour: Poisson with mean $3$ | The waiting time, in hours, until the first customer: exponential with rate $3$ | For customers arriving at $3$ per hour, the Poisson distribution gives the number of customers in an hour, with mean $3$, and the exponential distribution gives the wait until the first customer, with mean $\frac{1}{3}$ hour. |
Notes
- The two distributions are linked by $P(T > t) = P(N_t = 0) = e^{-\lambda t}$, where $N_t$ is the number of events in time $t$ and $T$ is the time of the first event.
- The Poisson distribution is discrete, and the exponential distribution is continuous.
- In a Poisson process, the mean number of events per unit of time is $\lambda$, and the mean time between events is $1 / \lambda$.