Modeling waiting times
The exponential distribution describes how long you wait until the next event when events occur at random at a constant average rate. It is the continuous counterpart of the Poisson distribution: if the number of events per hour is Poisson with rate λ, the time between events is exponential with the same λ.
The formulas
- Rate λ = 1 ÷ mean
- P(X ≤ x) = 1 − e^(−λx)
- P(X > x) = e^(−λx)
- P(x₁ ≤ X ≤ x₂) = e^(−λx₁) − e^(−λx₂)
- Median = ln 2 ÷ λ ≈ 0.693 × mean
- p-th percentile = −ln(1 − p) ÷ λ
Worked example
Buses arrive at random with an average gap of 10 minutes (λ = 0.1 per minute).
- Chance of waiting 5 minutes or less: 1 − e^(−0.5) ≈ 39.35%.
- Chance of waiting more than 5 minutes: 60.65%.
- Chance of waiting between 5 and 15 minutes: e^(−0.5) − e^(−1.5) ≈ 38.34%.
- Median wait: 10 × ln 2 ≈ 6.93 minutes.
- 90% of waits are under 10 × ln 10 ≈ 23.0 minutes.
The memoryless property
If you have already waited 10 minutes, the chance of waiting 5 more is exactly the same as for someone who just arrived: P(X > s + t | X > s) = P(X > t). Only the exponential distribution has this property among continuous distributions.
This makes it a good model for events with no aging: radioactive decay, random electronic failures, arrivals at a counter. It is a poor model for things that wear out, like tires or light bulbs at the end of their life, where the Weibull distribution fits better.
Common uses
- Queues: time between customer arrivals in banks, call centers or websites.
- Reliability: time to failure of components with a constant failure rate; the mean is the MTBF (mean time between failures).
- Physics: decay of radioactive atoms, where the half-life equals the median.
- Insurance and finance: time until the next claim or default in simple models.
Mean, median and skew
The distribution is skewed: most waits are short, a few are long. The mean (10 minutes) is pulled up by long waits, while the median (6.9 minutes) shows the typical case. The standard deviation equals the mean.