The Exponential Distribution

The exponential distribution is the clock of random arrivals. If some kind of event, a trade printing, a customer walking in, a component failing, happens at a steady average rate but at unpredictable moments, then the gap of time until the next one is exponential. It's the continuous-time answer to "how long do I wait?", and it pairs naturally with the The Poisson Distribution, which counts how many such events land in a fixed window.

One number sets the whole distribution: the rate $\lambda$, meaning "$\lambda$ events per unit of time on average." The probability density is

$$f(x) = \lambda\, e^{-\lambda x}, \qquad x \ge 0 .$$

Here $x$ is the waiting time (never negative), $\lambda$ is the arrival rate, and the $e^{-\lambda x}$ factor makes long waits exponentially unlikely. The density is tallest at $x = 0$ and only ever falls, so short waits are the most common single outcome, exactly the shape below.

Mean, spread, and the survival curve

The headline numbers:

$$E[X] = \frac{1}{\lambda}, \qquad \operatorname{Var}(X) = \frac{1}{\lambda^2} .$$

If events come at rate $\lambda$, the average gap is $1/\lambda$, the reciprocal, which is just common sense: two trades a minute means about thirty seconds between them. The chance of waiting longer than some time $t$ has a clean closed form, the survival function:

$$P(X > t) = e^{-\lambda t} .$$

No integral to grind, one exponential and you're done.

Memorylessness — the defining trait

The exponential forgets how long you've already waited. If a bus arrives on average every ten minutes and you've already stood there five minutes, your remaining expected wait is still ten minutes, not five. Formally,

$$P(X > s + t \mid X > s) = P(X > t) .$$

A component that has survived to time $s$ is, under this model, as good as brand new. This is elegant and it's the only continuous distribution with the property, but it's also where the model can quietly mislead: real machines age and real borrowers get riskier with leverage, so a constant hazard rate can be far too optimistic.

Worked example

Trades on a small stock print at an average rate of $\lambda = 2$ per minute. Model the gap between consecutive trades as exponential.

• Average gap: $E[X] = 1/\lambda = 1/2 = 0.5$ minutes, i.e. 30 seconds.
• Chance of a gap longer than 1 minute:

$$P(X > 1) = e^{-\lambda \cdot 1} = e^{-2} \approx 0.135 .$$

So about a 13.5% chance you wait more than a full minute for the next trade.

• Chance of a gap under 15 seconds ($t = 0.25$ min):

$$P(X < 0.25) = 1 - e^{-2(0.25)} = 1 - e^{-0.5} \approx 1 - 0.607 = 0.393 .$$

Roughly 39% of gaps are shorter than a quarter-minute.

Notice how the survival formula did all the work; no calculus needed once you have $e^{-\lambda t}$ in hand.

The exponential is the continuous mirror of the memoryless The Geometric Distribution, the waiting time inside a The Poisson Process, and one of the Common Distributions worth knowing by heart, mean $1/\lambda$, memoryless, tail $e^{-\lambda t}$.