Waiting times for things that happen "out of the blue", such as the next call to a help desk or the failure of a part that does not wear out, are modeled by the exponential distribution. Its mean takes an improper integral and integration by parts, building on means of densities.
Key idea
$X$ is exponential with rate $\lambda > 0$ when $f(x) = \lambda e^{-\lambda x}$ for $x \ge 0$. Then $$P(X > t) = e^{-\lambda t}, \qquad E[X] = \frac{1}{\lambda}, \qquad \operatorname{Var}(X) = \frac{1}{\lambda^{2}},$$ and $X$ is memoryless: $P(X > s + t \mid X > s) = P(X > t)$.
Why it works
Tail: $P(X > t) = \int_{t}^{\infty} \lambda e^{-\lambda x}\,dx = e^{-\lambda t}$.
Mean: by parts, $$\int_{0}^{\infty} x\lambda e^{-\lambda x}\,dx = \left[-xe^{-\lambda x}\right]_{0}^{\infty} + \int_{0}^{\infty} e^{-\lambda x}\,dx = 0 + \frac{1}{\lambda}$$ Doing it again gives $E[X^{2}] = \frac{2}{\lambda^{2}}$, so the variance is $\frac{1}{\lambda^{2}}$.
No memory: $$\frac{P(X > s + t)}{P(X > s)} = \frac{e^{-\lambda(s + t)}}{e^{-\lambda s}} = e^{-\lambda t}$$ Having waited $s$ already changes nothing about the wait ahead. The exponential is the only continuous distribution with this property.
Mean: by parts, $$\int_{0}^{\infty} x\lambda e^{-\lambda x}\,dx = \left[-xe^{-\lambda x}\right]_{0}^{\infty} + \int_{0}^{\infty} e^{-\lambda x}\,dx = 0 + \frac{1}{\lambda}$$ Doing it again gives $E[X^{2}] = \frac{2}{\lambda^{2}}$, so the variance is $\frac{1}{\lambda^{2}}$.
No memory: $$\frac{P(X > s + t)}{P(X > s)} = \frac{e^{-\lambda(s + t)}}{e^{-\lambda s}} = e^{-\lambda t}$$ Having waited $s$ already changes nothing about the wait ahead. The exponential is the only continuous distribution with this property.
Worked examples
Worked example: A light bulb
A bulb's lifetime is exponential with mean 1000 hours. Find the probability that it lasts more than 500 hours.
- The rate is $\lambda = \frac{1}{1000}$ per hour (the mean is $\frac{1}{\lambda}$, not $\lambda$).
- $$P(X > 500) = e^{-500/1000} = e^{-0.5} \approx 0.607$$
Answer: about $0.607$
Worked example: Already lasted
The same bulb has already lasted 800 hours. Find the probability that it lasts at least 500 more.
- By memorylessness, $$P(X > 1300 \mid X > 800) = P(X > 500)$$
- $= e^{-0.5} \approx 0.607$, exactly as for a new bulb.
Answer: about $0.607$
Worked example: Median and the mean
Find the median lifetime of that bulb, and the fraction of bulbs that outlast the mean.
- Median: $e^{-m/1000} = \frac{1}{2}$, so $m = 1000\ln 2 \approx 693$ hours, well below the mean of 1000.
- $P(X > 1000) = e^{-1} \approx 0.368$: a few very long lifetimes pull the mean above the typical value.
Answer: about 693 hours; about 37% outlast the mean.
Common mistake
Two slips. Using the rate as the mean: a rate of 2 per hour means a mean wait of $\frac{1}{2}$ hour. And expecting an old part to be "due" to fail: under the exponential model the past does not matter. (Real parts that wear out are not exponential; the model fits failures that come from outside shocks.) Check units: $\lambda t$ must have no units, so $\lambda$ is "per hour" when $t$ is in hours.